Lectures on Complex Analysis Math 502 John Roe SPRING 2015 · Lecture 3 Holomorphic functions and power series See Stein and Shakarchi 1.2.2, 1.2.3 The chief objects of study in complex

Lectures on Complex AnalysisMath 502John Roe

SPRING 2015

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AMS Open Math Notes: Works in Progress 2016-12-16 12:40:03

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• Prerequisites: Metric space language (Math 403). Lebesgue integration and Ba-nach spaces will be alluded to now and then, but not essential• Syllabus, questions, class discussion: https://piazza.com/psu/spring2015/math502/home

• Read and review lectures before class (Piazza resources page)• Post and answer questions, including questions in lecture notes• Grading: homework, midterm, final, Piazza participation


https://piazza.com/psu/spring2015/math502/home

https://piazza.com/psu/spring2015/math502/home

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This course is . . .

. . . a long but well-traveled hike through wild scenery.


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Lecture 1The complex plane

Remember that a complex number is a formal expression of the sort

z = x+ yi

where x and y are real numbers and i2 = −1. (We call x the real part of z and y theimaginary part, and we use the notation x = Re z, y = Im z.) For algebraists, the ring ofcomplex numbers is the quotient of the polynomial ring R[T ] in one indeterminate by theprincipal ideal generated by T 2 + 1. We identify each real number x with the complexnumber x+ 0i.

There is no problem in adding, subtracting or multiplying complex numbers by theusual rules. However, the following is a nontrivial fact.

Theorem 1.1. The complex numbers form a field, i.e. every non-zero complex numberhas a multiplicative inverse.

(Equivalently, for the algebraists again, the ideal generated by T 2 + 1 is maximal.)

Proof. We write an explicit formula for the inverse. If z = x + yi is a complex number,then |z| is the positive real number defined by

|z|2 = x2 + y2.

The complex conjugate of z isz = x− yi.

One computeszz = x2 + y2 = |z|2.

Thus, if z 6= 0, one has |z| > 0 and

z

|z|2=

x

x2 + y2− y

x2 + y2i

is the multiplicative inverse of z.

Exercise 1.2. Show that the modulus obeys the triangle inequality

|z ± w| 6 |z|+ |w|.

Remark 1.3. It’s often useful to represent the complex number z = x + iy in polarcoordinate form as

z = reiθ = r cos θ + ir sin θ.

Here r = |z|, and θ is a residue class modulo 2π (an element of the quotient groupR/2πZ). One calls r the modulus of z and θ the argument. In polar coordinates the lawfor multiplication of complex numbers takes the simple form

(r1eiθ1)(r2e

iθ2) = r1r2ei(θ1+θ2);

you multiply the moduli and add the arguments.

The complex plane is the set of all complex numbers x + yi, thought of as points of aEuclidean plane (a 2-dimensional real inner product space). What do complex numberoperations look like as transformations of this plane?


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(a) The operation of adding (or subtracting) a fixed complex number corresponds to atranslation of the plane, and all translations arise in this way.

(b) The operation of multiplying by a fixed complex number corresponds to a lineartransformation of the plane. However, not all linear transformations arise in thisway.

Definition 1.4. An invertible linear transformation T : R2 → R2 is conformal if it pre-serves angles: that is, for all pairs of vectors u, v, the angle between Tu and Tv is thesame as the angle between u and v.

Definition 1.5. An invertible linear transformation T : R2 → R2 is orientation-preservingif it can be represented by a matrix with positive determinant.

Proposition 1.6. The operation of multiplication by a fixed nonzero complex number is anorientation-preserving conformal linear transformation, and every orientation-preservingconformal linear transformation of the plane is given by such a multiplication.

Proof. That multiplication by a complex number z = reiθ is conformal follows from thepolar representation of complex multiplication. (See Exercise 1.7 below for another proof.)

In the other direction, suppose that the matrix

T =

(a bc d

)represents a conformal transformation. Look at the following table of vectors and theirimages under T :

Vector Image(10

) (ac

)(

11

) (a+ bc+ d

)(

01

) (bd

)Since the first and third rows must be perpendicular, ab+ cd = 0, whence there is a scalarλ with b = −λc and d = λa. Since the first and second rows make an angle of π/4, wemust have

1√2

=a(a+ b) + c(c+ d)√

(a2 + c2)((a+ b)2 + (c+ d)2)=

1√1 + λ2

.

Thus λ = ±1, and the orientation-preserving condition forces λ = 1. Now we see that Tis exactly the matrix of multiplication by the complex number a+ ci.

Exercise 1.7. Show that the dot product of two complex numbers z and w (consideredas vectors in R2) is the real part of zw. Use this to give another proof that multiplicationby a fixed complex number is a conformal linear transformation.


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Lecture 2Elementary Topology and Analysis in the Complex Plane

Because the absolute value satisfies the triangle inequality

|z + w| 6 |z|+ |w|,

the distance function d(z, w) = |z − w| makes the complex numbers into a metric space.In this lecture we shall review some of the standard notions of metric space theory withparticular reference to the complex numbers C.

Definition 2.1. A sequence zn of complex numbers converges as n→∞ to a complexnumber z if the sequence of real numbers d(zn, z) = |zn − z| tends to zero; that is, if forevery ε > 0 there is N > 0 such that |zn − z| < ε for all n > N .

Remark 2.2. It is the same thing to say that the sequence zn = xn + iyn converges toz = x+iy if the sequences xn and yn of real numbers converge to x and y respectively.

Remember that a sequence zn in a metric space is called a Cauchy sequence ifd(zn, zn′) → 0 as n, n′ → ∞. A metric space is complete if every Cauchy sequenceconverges. It is a fundamental property of the real numbers that R is complete; usingthis and the remark above, one sees that C is complete also.

We can generalize the definitions of limit of a function (rather than a sequence), con-tinuity, infinite series, uniform convergence and so on from R to C in the same way asabove.

Exercise 2.3. (A specific case of the previous sentence.) Write out (in terms of thedistance function d) what it means for a function f : C→ C to be continuous at a pointa.

Lemma 2.4 (Weierstrass M test). Let X be a set and let fn be a sequence of functionsX → C. If there exist real numbers Mn > 0 such that

(a)∑∞

n=1 Mn <∞, and(b) |fn(x)| 6Mn for all x ∈ X,

then the series∑∞

n=1 fn(x) converges uniformly on X to some function f . Moreover, ifX is a metric space and each fn is continuous, then so is f .

Proof. Fix x ∈ X, and let k, k′ ∈ N with k′ > k. Since∣∣∣∣∣k′∑

n=k+1

fn(x)

∣∣∣∣∣ 6k′∑

n=k+1

|fn(x)| 6k′∑

n=k+1

Mn → 0 as k, k′ →∞,

the partial sums of the series∑fn(x) form a Cauchy sequence, and so it converges say

to f(x). Now we have for all x∣∣∣∣∣f(x)−k∑

n=1

fn(x)

∣∣∣∣∣ 6∞∑

n=k+1

Mn → 0 as k →∞,

so the convergence is uniform. The last statement is a standard “3ε-argument”.


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Remark 2.5. A (complex) Banach space is a complex vector space E provided with anorm ‖ · ‖ : E → R+ and complete in the associated metric. The Weierstrass M -test isreally a theorem about Banach spaces. Observe that if X is any set, the collection E ofbounded functions f : X → C is a Banach space under the norm

‖f‖ = sup|f(x)| : x ∈ X,and if X is a metric space, the continuous bounded functions form a Banach subspace ofE. Now the Weierstrass M -test is a special case of the following

Exercise 2.6. Let E be any Banach space and let vn be elements of E such that∑∞n=1 ‖vn‖ < ∞. Then the series

∑∞n=1 vn converges in E. (In brief, absolute conver-

gence implies convergence in Banach spaces.)

Definition 2.7. A disc in C (with center a ∈ C and radius r > 0) is a subset of the formD(a; r) = z : |z − a| < r.

The following are standard definitions from metric space theory, specialized to the caseof C.

Definition 2.8. A subset Ω ⊆ C is open if it contains a disc around each of its points.A subset is closed if its complement is open. It is bounded if it is contained in some disc.

Warning: “open” is not the same as “not closed”, and “closed” is not the same as “notopen”.

Remark 2.9. An important theorem states that a map f : X → Y between metric spacesis continuous if and only if, for every open U ⊆ Y , the inverse image

f−1(U) := x ∈ X : f(x) ∈ Uis open in X.

Recall that a metric space X is compact if every sequence in X contains a convergentsubsequence, or, equivalently, if every open covering of X has a finite subcovering. If Xis a subset of C, it can be considered as a metric space in its own right. The Bolzano-Weierstrass theorem tells us when such a subset is compact.

Proposition 2.10. A subset X ⊆ C is compact if and only if it is closed and bounded.

The last elementary topological notion that we need is connectedness. Recall that ametric space X is connected if it cannot be expressed as a union of two disjoint nonemptyopen subsets.

Lemma 2.11. An open subset Ω ⊆ C is connected if and only if it is path connected,that is to say for all a, b ∈ Ω there is a continuous map γ : [0, 1] → Ω with γ(0) = a andγ(1) = b.

Proof. Suppose that Ω is connected and define a relation ∼ on Ω by saying that a ∼ bif there is a path γ as above with γ(0) = a and γ(1) = b. It is easy to see that ∼ is anequivalence relation. If D(a; r) ⊆ Ω then every z ∈ D(a; r) has z ∼ a, so each equivalenceclass contains a disc around each one of its points and is therefore open. By connectedness,there is only one equivalence class, which is to say that Ω is path connected.


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In the other direction, suppose that Ω is path connected. Let Ω = A ∪ B with A, Bdisjoint, nonempty and open and suppose a ∈ A, b ∈ B and γ is a path joining a tob. Then γ−1(A) and γ−1(B) are disjoint nonempty open subsets of [0, 1] whose union is[0, 1]. This contradicts the connectedness of [0, 1], which is a standard fact.


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Lecture 3Holomorphic functions and power series

See Stein and Shakarchi 1.2.2, 1.2.3The chief objects of study in complex analysis are differentiable functions (of a complex

variable). The definition is the natural one.

Definition 3.1. Let Ω be an open subset of C. A map f : Ω → C is differentiable at apoint z ∈ Ω if the difference quotient

f(z + h)− f(z)

h

tends to a limit as h→ 0. The value of this limit is then called the derivative f ′(z) of f .

Remark 3.2. The Landau notation o(h) (as h → 0) is used to denote “any function ofthe form he(h), where e(h) → 0 as h → 0”. Using this notation one can define complexdifferentiability as follows: f is differentiable at z if there is a complex number f ′(z) suchthat

f(z + h) = f(z) + f ′(z)h+ o(h)

as h → 0. This way of formulating the definition makes contact with the definition ofdifferentiation for general maps between normed vector spaces.

The hypothesis that Ω is open ensures that the difference quotient (f(z+ h)− f(z))/his defined for all sufficiently small h. It is the possibility that h approaches zero “fromany direction” that gives complex differentiability its extra force.

Definition 3.3. Let Ω ⊆ C be open and let f : Ω→ C be continuous. The function f isholomorphic on the open set Ω if it is differentiable at every point of Ω.

The usual proofs of real-variable theory work without change to show that the sum,difference, product, and quotient (where defined) of differentiable functions are againdifferentiable, with the usual expressions for their derivatives. The composite of differ-entiable functions is also differentiable, and the chain rule holds: if h(z) = g(f(z)), thenh′(z) = g′(f(z))f ′(z).

Proposition 3.4. If f is differentiable at every point of a connected open subset Ω of C,and f ′(z) = 0 for all z ∈ Ω, then f is constant on Ω.

Proof. Let γ : [0, 1]→ Ω be a differentiable path. Then f γ : [0, 1]→ C is differentiableand has derivative zero, hence it is constant by the usual form of the Mean Value Theorem(applied separately to the real and imaginary parts). Now if a ∈ Ω, then there is a discD(a; ε) ⊆ Ω (since Ω is open). Every point of this disc is connected to a by a differentiable(indeed, a straight-line) path in Ω. Thus f(z) = f(a) for all z ∈ D(a; ε). Hence, for anyconstant c, the set

f−1(c) = z ∈ Ω : f(z) = cis open. It is also closed (since f is continuous). Picking c to be a value taken by fsomewhere, connectedness of Ω now tells us that f−1(c) = Ω; that is, f is constant.


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It is important to understand that complex differentiability is a much stronger conditionthan real-variable differentiability. To say that f is differentiable as a real function wouldsay that there is a 2× 2 real matrix Df(z) such that

f(z + h) = f(z) +Df(z) · h+ o(‖h‖).

Thus the extra condition for complex differentiability is that the 2 × 2 matrix Df(z) isthe matrix of multiplication by a complex number, or in other words (thanks to Proposi-tion 1.6) it is either zero or an orientation-preserving conformal linear transformation.

According to the proof of Proposition 1.6, the matrix of such a linear transformation

must be of the form

(a −cc a

). On the other hand, if z = x + iy and f(z) = u(x, y) +

iv(x, y), then

Df =

(∂u/∂x ∂u/∂y∂v/∂x ∂v/∂y

).

Comparing these expressions we see that f = u + iv is a complex-differentiable functionof z = x + iy if and only if it is real-differentiable and, in addition, satisfies the Cauchy-Riemann equations

∂u

∂x=∂v

∂y,

∂u

∂y= −∂v

∂x.

Remark 3.5. It follows from the Cauchy-Riemann equations and the symmetry of mixedderivatives that both the real and imaginary parts of a holomorphic function satisfyLaplace’s equation1

∂2u

∂x2+∂2u

∂y2= 0.

Laplace’s equation is a fundamental one in physics and this is one reason for the closeconnection between complex analysis and many physical problems.

The most important examples of holomorphic functions are power series on C, of theform

∑∞n=0 an(z − a)n, where the an and a are complex constants.

Proposition 3.6. For any power series∑∞

n=0 an(z−a)n there is a number r (nonnegativereal number or +∞), called the radius of convergence, such that

(a)∑∞

n=0 an(z − a)n converges uniformly on compact subsets of the open disk D(a; r),(b)

∑∞n=0 an(z − a)n diverges for each z with |z − a| > r.

The behavior of the series on the circle of convergence |z − a| = r is not specified bythe proposition.

Proof. Without loss of generality we may take a = 0.Let U = ρ > 0 : the sequence (|an|ρn) is bounded and let r = supU (taken as +∞

if U is not bounded above). Then, if |z| > r, the terms of the series∑∞

n=0 anzn are

unbounded and thus the series cannot converge. This proves (b).

1At least if they are twice continuously differentiable; later we shall see that this assumption is alwayssatisfied.


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To prove (a) we shall show that the series converges uniformly (and absolutely) onevery closed disk D(0; s), with s < r. Indeed, there is ρ with s < ρ < r and the sequence(|an|ρn) bounded, say by K. Then for all z ∈ D(0; s),

|anzn| 6Mn := K

(s

ρ

)n.

Since s/ρ < 1,∑Mn is a convergent geometric series and thus

∑∞n=0 anz

n converges

uniformly on D(0; s) by the Weierstrass M-test.

Exercise 3.7. Prove Hadamard’s formula for the radius of convergence, which is

1/r = lim supn→∞

|an|1/n.

Let∑an(z−a)n be a power series with radius of convergence r, and converging within

D(a; r) to a function f(z). The series∑nan(z−a)n−1, obtained by formal differentiation,

also has radius of convergence r (one can see this from Exercise 3.7 for example). Moreover

Proposition 3.8. With notation as above, the function f is differentiable on D(a; r) andthe series

∑nan(z − a)n−1 converges on D(0; r) to the derivative

f ′(z) := limh→0

f(z + h)− f(z)

h.

Proof. Once again, we may take a = 0 without loss of generality.Start with an algebraic fact: if |w|, |z| < s then

|wn − zn − nzn−1(w − z)| 6 12n(n− 1)|w − z|2sn−2.

This follows from the identity

wn − zn − nzn−1(w − z) =

(w − z)2(wn−2 + 2wn−3z + · · ·+ (n− 1)zn−2)

which is easy to check. Let g(z) =∑nanz

n−1 and fix s < r. Then using the identityabove, provided that |z|, |z + h| < s we have∣∣∣∣f(z + h)− f(z)

h− g(z)

∣∣∣∣ 6 12|h|∑

n(n− 1)|an|sn−2

and the sum on the right converges to a finite limit since s < r. In particular, (f(z+h)−f(z))/h→ g(z) as h→ 0 and we are done.

Thus, within its radius of convergence, a power series always defines a holomorphicfunction.


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Lecture 4Some examples of holomorphic functions

See Stein and Shakarchi 1.2.3, 5.3.1, 6.1, 6.2In this lecture we are going to review some specific examples of holomorphic functions,

illustrating the many different ways in which such functions can be defined.Functions defined by power series: It follows from Proposition 3.8 that any power

series defines a holomorphic function within its disc of convergence. In particular theexponential series

ez =∞∑n=0

zn

n!

defines a function holomorphic on the whole complex plane. Such a function is called anentire function. We also use the notation exp(z) for ez. Since power series can be differ-entiated term by term, the exponential function is equal to its own derivative. Converselywe have

Lemma 4.1. If f is a holomorphic function defined on a connected open subset Ω ⊆ C,and f ′(z) = f(z) for all z ∈ Ω, then there is a constant a such that f(z) = aez for allz ∈ Ω.

Proof. Let g(z) = f(z)e−z. By the chain and product rules,

g′(z) = f ′(z)e−z − f(z)e−z = 0;

and thus, by Proposition 3.4, g is constant on Ω. Considering the special case Ω = C,f(z) = ez, gives us the fact that ez · e−z = 1 for all z, so ez is never zero and e−z = 1/ez.Now returning to the general case we get f(z)e−z = c (a constant), so f(z) = c/e−z = cez

as required.

Corollary 4.2. For z, w ∈ C, we have ez+w = ezew;.

Proof. Treat w as fixed; apply Lemma 4.1 to f(z) = ez+w.

The sine and cosine functions are defined in terms of the exponential by

sin z =eiz − e−iz

2i=∞∑n=0

(−1)nz2n+1

(2n+ 1)!,

cos z =eiz + e−iz

2=∞∑n=0

(−1)nz2n

(2n)!.

The exponential, sine, and cosine functions are real valued for real arguments, and wehave

eiz = cos z + i sin z

for all z.


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The addition law for the exponential function yields the corresponding laws for sineand cosine,

sin(z + w) = sin z cosw + cos z sinw

cos(z + w) = cos z cosw − sin z sinw.

In particular sin2 z+cos2 z = 1 — the special case w = −z of the second identity. One seesby computation that cos has a positive real zero; define π by letting π/2 be the smallestreal zero of cos. We have cos(π/2) = 0 and sin(π/2) = 1. The identities now give

sin(z + π/2) = cos(z), cos(z + π/2) = − sin(z).

Iterating these we find that cos and sin are 2π-periodic, so the exponential function is2πi-periodic. In particular we get the famous formulae

e2πi = 1, eπi = −1.

Of course we are all familiar with the properties of the exponential, sine and cosinefunctions. The point though is to show that these properties can be derived rigorouslyby purely analytic means.

Since the power series for the exponential function has real coefficients, ez = ez. Itfollows that

|ez|2 = ezez = ez+z = e2 Re z

so |ez| = eRe z, for all complex numbers z.The identity ez · e−z = 1 shows that the exponential function never takes the value 0.

However, it can take all other values. Indeed if w = r(cos θ+ i sin θ) is a nonzero complexnumber written in polar form, then z = log(r) + iθ has ez = w. There are of courseinfinitely many such z, differing by integer multiples of 2πi.

Remark 4.3. It is an extremely important fact that, although we can find a “logarithm”(indeed, infinitely many possible logarithms) for any individual nonzero complex number,there is no way to fit these together to define a holomorphic “logarithm function” onΩ = C \ 0. This fact, which is closely connected to the topology of the plane, will geta whole lecture to itself in a little while. It is, however, possible to define “holomorphiclogarithm” functions on some subsets of C \ 0:

Example 4.4. If |z| < 1 the power series

`(1 + z) =∞∑n=1

(−1)n−1zn

n

converges to a holomorphic function such that `′(1 + z) = (1 + z)−1. It follows that

d

dz

((1 + z)e−`(1+z)

)= 0

by the chain and product rules, so e`(1+z) is a constant multiple of 1+z by Proposition 3.4again; looking at z = 0 shows that the constant is 1. Thus ` gives a “holomorphic branchof the logarithm” in the disc D(1; 1).


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Functions defined by integrals: A power series is one way of taking an ‘infinitelinear combination’ of elementary functions, in this case powers of z. Another way oftaking such a ‘linear combination’ is by an integral. Here the most famous example is thegamma function, which is defined by the integral

Γ(z) =

∫ ∞0

tz−1e−tdt.

which is absolutely convergent when Re z > 0. (For a positive real number t, we definetz−1 := exp((z − 1) log t), taking the real determination of log t.) Indeed, if Re z > a > 0,then the absolute value of the integrand is dominated by the integrable function ta−1e−t,so that the integral converges. Moreover, the derivative of the integrand with respect toz, that is

(log t)tz−1e−t

is also dominated by an integrable function (for instance a multiple of t(a/2)−1e−t/2). Astandard theorem about differentiating under the integration sign now implies that thefunction Γ(z) is differentiable on the half-plane z : Re z > a for any a > 0. Our finalconclusion is that Γ is a holomorphic function in the right-hand half plane, Re z > 0.

Proposition 4.5. We haveΓ(z + 1) = zΓ(z)

whenever Re z > 0. We have Γ(1) = 1, and consequently Γ(n) = (n− 1)! for n = 1, 2, . . ..

Proof. Integrate by parts to get

Γ(z + 1) =

∫ ∞0

tze−tdt =

∫ ∞0

ztz−1e−tdt = zΓ(z)

as required.

We can rewrite the proposition as

Γ(z) =Γ(z + 1)

z=

Γ(z + 2)

z(z + 1)= . . . .

Notice that whatever z is, there is an integer n such that Γ(z + n) has positive real part.Thus we can take this equation as defining Γ for all values of z, except for nonpositiveintegers where one tries to divide by zero on the right. We say that there is an analyticcontinuation of Γ to the whole complex plane, with singularities at 0,−1,−2, . . ..

The gamma function satisfies a number of important identities, for example:

(1) Reflection formula: Γ(z)Γ(1− z) =π

sin πz.

(2) Duplication formula: Γ(2z) = π−1/222z−1Γ(z)Γ(z + 12).

We shall prove these later in the course.


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Lecture 5More examples of holomorphic functions

See Stein and Shakarchi Chapters 6, 7, 9We continue the discussion of the previous lecture of “interesting” examples of holomor-

phic functions. For this session we will assume the following fact (which will be provedlater): if a sequence fn of holomorphic functions (on an open set Ω) converges uniformlyon compact subsets to a function f , then f is holomorphic also.

Dirichlet series: A series such as

ζ(s) =∞∑n=1

1

ns

is called a Dirichlet series ; this particular example (the most important) is the Riemannzeta function. It converges for Re s > 1, and the convergence is uniform on compactsets; it follows that the limit is holomorphic. As with the gamma function, ζ(s) has ananalytic continuation to the whole complex plane. This analytic continuation satisfies thereflection formula

ζ(1− s) = 2(2π)−sΓ(s) cos(πs/2)ζ(s)

which we shall also prove later in the course.

Proposition 5.1. The zeta function can also be defined for Re s > 1 by the integral

ζ(s) =1

Γ(s)

∫ ∞0

ts−1

et − 1dt.

Proof. For t > 0 we can write

1

et − 1=

e−t

1− e−t=∞∑n=1

e−nt.

From the dominated convergence theorem, then,∫ ∞0

ts−1

et − 1dt =

∞∑n=1

∫ ∞0

ts−1e−ntdt.

But the substitution u = nt shows that∫ ∞0

ts−1e−ntdt = n−s∫ ∞

0

us−1e−udu = n−sΓ(s)

and so ∫ ∞0

ts−1

et − 1dt =

∞∑n=1

n−sΓ(s) = Γ(s)ζ(s),

as required.

Infinite products: An infinite product is a formal expression of the kind∞∏n=1

(1 + an)

and, naturally, we say that it converges if the partial products∏k

n=1(1 + an) tend to alimit as k →∞.


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Proposition 5.2. If∑|an| < ∞ then the infinite product

∏∞n=1(1 + an) converges. (In

that case we say that the infinite product is absolutely convergent.) Moreover, in this casethe infinite product is nonzero if all the (1 + an) are nonzero.

Proof. Assume without loss of generality that all |an| < 12. Use the branch of the logarithm

function ` defined in Remark 4.4. We havek∏

n=1

(1 + an) = exp

(k∑

n=1

`(1 + an)

),

so it suffices to study the convergence of the series∑`(1 + an). But if |a| < 1

2,

|`(1 + a)| =

∣∣∣∣∣∞∑n=1

(−1)n−1an

n

∣∣∣∣∣ 6 |a|∞∑n=1

2−n+1 = 2|a|.

Consequently the series∑`(1 + an) is absolutely convergent.

Remark 5.3. There is a “Weierstrass M” version of this proposition also: if you have aninfinite product

∏∞n=1(1+fn(x)), where the fn are functions on a set x, and if |fn(x)| < Mn

for all x ∈ X and∑Mn <∞, then the infinite product converges uniformly on X.

Example 5.4. Consider the expression

ϕ(z) = z∞∏n=1

(1− z2

n2

).

From the remark above, this infinite product converges uniformly on compact subsets ofC to a holomorphic function ϕ. Plainly, from its definition, ϕ(z) = 0 for every integerz ∈ Z.

Now we are familiar with another holomorphic function that vanishes for every integer,namely the function sinπz. Later in the course we will prove the fact, first intuited byEuler, that ϕ(z) = π−1 sin(πz): in other words,

sin πz = πz∞∏n=1

(1− z2

n2

).

The key connection between the Riemann zeta function and number theory is thefollowing infinite product formula:

Proposition 5.5 (Euler product formula). For Re s > 1 we have

ζ(s) =∏p

1

1− p−s

where the product is extended over all primes p.

Proof. The infinite product converges for Re s > 1 because∑p−s (indeed, even

∑n−s)

converges absolutely. This convergence can be used to justify the following manipulations:let p1, p2, . . . be the primes and write

1

1− p−sj= 1 + p−sj + p−2s

j + · · · .


17

Multiply together all these infinite series for different values of j to see that the productformula can be equated to a sum of terms of the form

((p1)m1(p2)m2(p3)m3 . . . )−s

where the mj are nonnegative integers and all but finitely many are zero. But by thefundamental theorem of arithmetic each positive integer can be represented in one andonly one way as a product of the sort appearing here. So this sum is exactly the series∑n−s defining the zeta function.

Exercise 5.6. Fill in the details of this proof.

Elliptic functions: Consider once again Euler’s infinite product defining the sinefunction (Example 5.4). By logarithmic differentiation one obtains

π cot(πz) =1

z+ 2

∞∑n=1

z

z2 − n2=∑m∈Z

′ 1

z −m.

The symbol∑′ denotes that the (non absolutely convergent) sum is to be interpreted by

first grouping +m and −m together and then summing the resulting groups (absolutelyconvergent, uniformly on compact subsets of C \ Z). This formula evidently defines aperiodic function, with period 1, that has singularities at all integer points.

Remark 5.7. By taking 1/z to the left of the above equation, expanding both sides aspower series in z, and comparing coefficients, one can obtain a closed-form expression forζ(2k) =

∑∞n=1 1/n2k in terms of powers of π and the Bernoulli numbers which are related

to the coefficients in the power series expansion of the cotangent. The familiar identityζ(2) = π2/6 is the first of these.

Suppose now that Λ is a lattice in C — that is, a free abelian subgroup of rank 2,generated by two R-linearly independent complex numbers. (The simplest, but of coursenot the only, example is the lattice of Gaussian integers m + ni for m,n ∈ Z.) Theanalogous formula

℘(z) =∑w∈Λ

′ 1

(z − w)2

is another non absolutely convergent sum, which can however be interpreted (by “group-ing” or by another summability procedure) to define a holomorphic function on C \ Λcalled the Weierstrass elliptic function based on the lattice Λ. This function is an ex-ample (in a certain sense the “simplest”) of a holomorphic function with two R-linearlyindependent periods.

These functions have deep roots in the history of analysis. We will show that ℘ satisfiesa differential equation

℘′(z)2 = 4℘(z)3 − g2℘(z)− g3,

where g2 and g3 are certain constants that can be computed in terms of the period latticeΛ. What this means is that the Weierstrass function ℘ (or more exactly its inverse) canbe used to “evaluate” an integral of the form∫

dx√p(x)

,


18

where p is a cubic — or quartic — polynomial, in the same way that inverse trigonometricfunctions can be used to evaluate such an integral when p is a quadratic. Historicallyspeaking, integrals of this kind arose in the problem of the “rectification of the ellipse”and the order of historical development — rectification of the ellipse leads to doubleperiodicity (Euler) which leads to a general theory of elliptic functions (Jacobi, Abel)which in turn leads to abstract complex analysis (Weierstrass, Riemann) — is almost theexact opposite of the order of theoretical development which is followed in most coursesthat we teach today.


19

Lecture 6Plane topology and complex analysis

See Stein and Shakarchi Appendix BEvery nonzero complex number has two ‘square roots’: if w = reiθ, then the two num-

bers z = ±r1/2eiθ/2 both satisfy z2 = w. (Notice that eiπ = −1, so the sign ambiguity isimplicit in the fact that θ is only unique modulo 2πZ.) This lecture centers around a sim-ple question: can these various individual choices for the square root of w be “assembled”into a holomorphic function,

√w, defined on the complex plane?

Definition 6.1. Let Ω be an open subset of C, and let f be a holomorphic functiondefined on Ω. Let U be an open subset of f(Ω). A branch of the inverse for f (definedon U) is a holomorphic function g : U → Ω such that f(g(w)) = w for all w ∈ U .

We’ll use the obvious terminology “branch of the square root” instead of “branch ofthe inverse of f(z) = z2”, “branch of the logarithm” instead of “branch of the inverse off(z) = exp(z)”, and so on.

Remark 6.2. Notice that if there is a branch of the logarithm defined on U , then there isa branch of the square root (or n’th root) defined there. Indeed, if g(w) is a branch ofthe logarithm then h(w) = exp( 1

ng(w)) is a branch of the n’th root.

Lemma 6.3. Every nonzero w0 ∈ C has a connected neighborhood U on which there isdefined a branch of the logarithm function, that is a holomorphic function g : U → C suchthat eg(w) = w for all w ∈ U .

Proof. Actually, U will be a disc. If w0 = 1 we take U to be the disc of center 1 andradius 1, and we define g by the power series

g(w) = `(1 + (w − 1)) =∞∑n=1

(−1)n−1(w − 1)n

n

from 4.4. As we showed, this function g is a branch of the logarithm. For any other w0,choose some z0 such that ez0 = w0 and define

g(w) = z0 + `(w/w0)

for w ∈ U = D(w0; |w0|).

It follows that every w0 ∈ C \ 0 also has a neighborhood on which there is defined abranch of the square (or n’th) root function.

We all know that there is a sign ambiguity in taking the square root of a (positive) realnumber. However, in the real case this ambiguity can be resolved by a once-and-for-allexecutive decision: to always take the positive square root. In the complex world such aonce-and-for-all decision is not possible.

Proposition 6.4. There is no branch of the square root defined on all of C, or even onC\0. That is, there is no holomorphic function g(w) defined there such that g(w)2 = w.

It follows from Remark 6.2 that there is no branch of the logarithm defined on C \ 0either.


20

Proof. Suppose that g is a branch of the square root defined on C \ 0 and let

s(t) = g(eit)g(e−it)

for t ∈ R. Notice that s(0) = g(1)2 = 1 and s(π) = g(−1)2 = −1. But

s(t)2 = g(eit)2g(e−it)2 = eit · e−it = 1,

so s(t) = ±1 for all t. This contradicts the intermediate value theorem (the connectednessof R).

Remark 6.5. We defined a “branch of the inverse” to be a holomorphic function but theabove argument makes it clear that mere continuity is the problem here — there is no“continuous branch of the square root” either.

It is natural to ask, then, which subsets of C have the property that square root (orlogarithm) functions always have continuous branches there.

We introduce some notation. U denotes the open unit disc z ∈ C : |z| < 1 and U isits closure, the closed unit disc. The boundary ∂U = S1 is the unit circle.

Definition 6.6. A connected open set Ω ⊆ C is simply-connected if every continuousmap ∂U→ Ω can be extended to a continuous map U→ Ω.

A map ∂U→ Ω is called a loop in Ω, and a loop which can be extended to a continuousmap U → Ω is called nullhomotopic or inessential. Thus, Ω is simply-connected if everyloop in it is nullhomotopic.

Example 6.7. Any convex Ω is simply-connected, by the coning construction: if h is acontinuous map S1 → Ω, pick a point z0 ∈ Ω and extend h to the map H defined by

H(ru) = rh(u) + (1− r)z0, r ∈ [0, 1], u ∈ S1.

The same construction works if Ω is merely “star shaped” about z0, rather than convex.

Let h : S1 → Ω be a loop. Then γ : t 7→ h(e2πit) is a path (a map [0, 1] → Ω) that hasγ(0) = γ(1). Conversely

Lemma 6.8. If γ : [0, 1] → Ω is a path such that γ(0) = γ(1), then there is a uniqueh : S1 → Ω with γ(t) = h(e2πit).

Because of this lemma we may also regard a “loop” simply as a “path that starts andends at the same point”.

Proof. For u ∈ S1, choose t ∈ [0, 1] such that e2πit = u and put h(u) = γ(t); the fact thatγ(0) = γ(1) shows that this is well defined. We must show that it is continuous.

For this purpose, first extend γ by periodicity (γ(t) = γ(t + 1)) to a continuous mapR→ Ω, and then note that h(u) = γ(t) for any t such that u = e2πit. Observe now thatif un → u in S1, then |un/u− 1| < 1 for almost all n. Thus we can use the holomorphicbranch of the logarithm established in 4.4 to produce a real sequence sn → 0 such thate2πisn = un/u. It follows that there exists a sequence tn = sn + t→ t such that e2πit = uand e2πitn = un. By continuity of γ, then,

h(un) = γ(tn)→ γ(t) = h(u)

as required.


21

(For those who know about the quotient topology, this is simply a rather explicit proofthat R/Z = [0, 1]/0, 1 = S1 as a quotient topological space.)

We will prove the following very non-trivial result (the Riemann mapping theorem)later in the course.

Theorem 6.9. Let Ω ⊆ C be a proper, nonempty connected open subset. The followingare equivalent.

(i) Ω is simply-connected.(ii) For every a /∈ Ω there is a holomorphic branch of z 7→

√z − a on Ω.

(iii) For every a /∈ Ω there is a holomorphic branch of z 7→ log(z − a) on Ω.(iv) There is a holomorphic bijection between U and Ω.


22

Lecture 7The winding number

In the previous discussion we used the notion nullhomotopic for loops in Ω ⊆ C. Thatis a special case of the following.

Definition 7.1. LetX be a compact metric space and Y any space. Two maps h0, h1 : X →Y are homotopic if there is a continuous map

H : [0, 1]×X → Y

with H(k, x) = hk(x) for k = 0, 1. The map H is called a homotopy between h0 and h1.

It is easy to see that homotopy is an equivalence relation.

Example 7.2. Y is path connected if any two maps from a point to Y are homotopic. Infact, with appropriate definitions of topology on mapping spaces, a homotopy is simply apath in the space Maps(X;Y ) of maps from X to Y .

Definition 7.3. Let X be some compact metric space. A (continuous) map f : X →C \ 0 lifts through the exponential function if there is a map g : X → C such thatf = exp g; in other words, if there is a commutative diagram

C

exp

X

g

<<

f // C \ 0

Remark 7.4. It follows from Example 4.4 that if |1− f(x)| < 1 for all x ∈ X, then f liftsthrough the exponential function.

Proposition 7.5. Let X be a compact metric space and f, g maps from X to C \ 0.The maps f and g are homotopic if and only if f/g lifts through the exponential function.

Proof. Suppose that f1(t)/f0(t) = exp(g(t)). Then a homotopy from f0 to f1, in C \ 0,is given by fs(t) = exp(sg(t))f0(t).

Conversely, consider a homotopy h(s, x) with h(0, x) = f(x) and h(1, x) = g(x). Be-cause [0, 1] × X is a compact space, and h never takes the value 0, there is ε > 0 suchthat |h(s, x)| > ε for all s ∈ [0, 1] and x ∈ X. As a continuous function on a compactmetric space, h is uniformly continuous. Therefore, there exists δ > 0 such that

|h(s, x)− h(s′, x)| < ε whenever s, s′ ∈ [0, 1] with |s− s′| < δ.

Thus when |s− s′| < δ the maps hs(x) = h(s, x) and hs′(x) = hs′(x) satisfy

|hs(x)− hs′(x)| < ε < |hs(x)|.Thus |1 − hs(x)/hs′(x)| < 1 for all x, and so hs/hs′ is an exponential by Remark 7.4.Choose a finite sequence sj with s0 = 0, sm = 1, and |sj − sj+1| < δ. Then

f/g =m−1∏j=0

(hsj/hsj+1

)


23

is a product of exponentials, which is an exponential.

Corollary 7.6. With the same notation, a map X → C \ 0 is itself an exponential iffit is homotopic to a constant map.

We now specialize our attention to paths and loops in the punctured plane. Recall thata path in a space Ω is just a (continuous) map γ : [0, 1] → Ω; a loop in Ω is a path withthe additional property that its initial point γ(0) is equal to its final point γ(1). We notedin Lemma 6.8 that a loop can also be considered as a map from the unit circle S1.

Proposition 7.7. Every path γ : [0, 1] → C \ 0 lifts through the exponential map to apath g : [0, 1] → C. Moreover, if g0 and g1 are two lifts of γ, then there is an integer nsuch that g0(t)− g1(t) = 2πin, for all t.

Proof. The homotopy

H(s, t) = γ((1− s)t)shows that γ is homotopic to a constant path. By Corollary 7.6, then, γ lifts through theexponential map.

Suppose now that g0, g1 are two lifts of γ. Then

exp(g0(t)) = γ(t) = exp(g1(t)),

and thus exp(g0(t) − g1(t)) = 1, which implies that (g0(t) − g1(t))/2πi is an integer. Acontinuous integer-valued function on [0, 1] must be constant, so g0(t)− g1(t) = 2πin forsome integer n.

Now suppose that γ is a loop in C \ 0, that is a path that has the same initial andfinal points. By Proposition 7.7, γ is the exponential of a path, that is to say, a mapg : [0, 1]→ C such that

(7.8) γ(t) = exp(g(t)).

But g need not be a loop! All we know about its initial and final points is that exp(g(1)) =exp(g(0)), and this tells us that g(1)− g(0) = 2πim for some integer m. Also, the secondpart of Lemma 7.7 tells us that m does not depend on the choice of lifting g for the loopγ. It is therefore an invariant of the loop γ.

Definition 7.9. If γ is a loop in C \ 0 as above, the integer m is called the windingnumber of γ about the origin, and is denoted wn(γ), or sometimes by wn(γ, 0).

Definition 7.10. If γ is a loop in C \ p, the winding number of γ about p, denotedwn(γ, p), is defined to be the winding number of the loop t 7→ γ(t)− p (in C \ 0) aboutthe origin. That is to say, wn(γ, p) = (g(1) − g(0))/2πi, where g is any path in C suchthat exp(g(t)) = γ(t)− p.

In other words, the winding number measures the extent to which a lifting of the givenloop through the exponential map fails to “close up”. In particular, if γ lifts through theexponential map to a loop, it has winding number zero. This implies

Lemma 7.11. If a loop in Ω = C \ p is nullhomotopic (as defined in the previouslecture) then its winding number around p equals zero.


24

Proof. Without loss of generality take p = 0. A loop in Ω is given by a map h : ∂U→ Ω,and “nullhomotopic” means that there is a homotopy H : [0, 1]× S1 → Ω with H(0, u) =h(u) and h(1, u) a constant (independent of u). By Corollary 7.6, the given loop h liftsthrough the exponential map as a loop: that is, there is a map S1 → C such thath(u) = exp(2πig(u)). By definition, this implies that h has winding number zero.

Lemma 7.12. Let γ1 and γ2 be loops in C\0 and let γ(t) = γ1(t)γ2(t) be their pointwiseproduct. Then

wn(γ) = wn(γ1) + wn(γ2).

Proof. If g1 and g2 are lifts of γ1 and γ2 respectively, then g1 + g2 is a lift of γ1γ2.

Theorem 7.13. A loop in C \ 0 has winding number n if and only if it is homotopicto the loop

en(t) = exp(2πint), t ∈ [0, 1].

In particular, two loops are homotopic if and only if they have the same winding number,and a loop is homotopic to a constant if and only if it has winding number 0.

Proof. It is clear from the definition that en has winding number n, since we can take alift g for en to be g(t) = 2πint.

Suppose that γ has winding number 0. Then we can write γ(t) = exp(2πig(t)) withg(0) = g(1) = 0. The homotopy

h(s, t) = exp(2πi(1− s)g(t))

is then a homotopy of loops (not just of paths) and contracts γ to a constant loop.Conversely, if γ is homotopic (through loops) to a constant loop, it has winding numberzero by Lemma 7.11. This shows that a loop is homotopic to a constant loop if and onlyif it has winding number zero. Moreover, all constant loops are homotopic to one another(since C \ 0 is path-connected).

To prove the general case we make use of Lemma 7.12 (which shows that loops γ0 andγ1 have the same winding number if and only if γ0/γ1 has winding number zero), togetherwith the observation that γ0 and γ1 are homotopic through loops if and only if γ0/γ1 ishomotopic to a constant loop (if γs is a homotopy from γ0 to γ1, then γs/γ1 is a homotopyfrom γ0/γ1 to the constant loop 1).

Lemma 7.14. Suppose that γ1 and γ2 are loops in C \ 0, both starting and ending at1. Their concatenation is the loop γ1 ∗ γ2 defined by

t 7→

γ1(2t) (t 6 1

2)

γ2(2t− 1) (t > 12)

Then wn(γ1 ∗ γ2) = wn(γ1) + wn(γ2).

Proof. Define for s, t ∈ [0, 1],

Γ1(s, t) =

γ1((1 + s)t) (1 + s)t 6 1

1 (1 + s)t > 1


25

and

Γ2(s, t) =

γ2(1− (1 + s)(1− t)) (1 + s)(1− t) 6 1

1 (1 + s)(1− t) > 1

Then H(s, t) = Γ1(s, t)Γ2(s, t) is a homotopy from the pointwise product γ1γ2 to theconcatenation γ1 ∗ γ2. So the result follows from Lemma 7.12.


26

Lecture 8Paths and integrals

See Stein and Shakarchi 2.1, 2.2A path is just a map γ from some interval [a, b] to C, and such a map can be differenti-

ated in the usual way. We’ll say that it is smooth if it is differentiable as a function from(a, b) to C = R2 and its derivative is continuous. More generally we will consider piecewisesmooth paths which are continuous and whose derivatives have only finitely many jumpdiscontinuities; this allows us to consider paths (like the boundary of a triangle) whichhave finitely many corners.

Definition 8.1. Let γ : [a, b]→ C be a piecewise smooth path. The length of γ is

Length(γ) :=

∫ b

a

|γ′(t)| dt.

(This is the standard definition from differential geometry.)

Definition 8.2. Let γ : [a, b]→ C be a piecewise smooth path and let f be a continuousfunction defined on γ∗. Then the path integral of f along γ is defined to be∫

γ

f(z) dz :=

∫ b

a

f(γ(t)) γ′(t)dt.

Formally, we obtained this expression by ‘substituting z = γ(t)’.

Notice that in both the above cases, the expression under the integral sign is indeedintegrable in the Lebesgue sense, even if γ′ has jump discontinuities. In more elementaryterms, this amounts to saying that the integral along a piecewise smooth path is definedas the sum of the integrals along the smooth sub-paths that comprise it.

Remark 8.3. Let h : [a, b]→ [a′, b′] be a smooth, increasing function with smooth inverse.If γ is a piecewise smooth path so is γ h (called a reparameterization of γ). It followseasily from the chain rule and the rule of integration by substitution that∫

γ

f(z) dz =

∫γh

f(z) dz;

thus “the path integral is independent of the choice of parameterization.”

We will need some simple estimates for path integrals. The basic one is

Proposition 8.4. Let γ be a smooth path and let f be a function defined and continuouson the image of γ. Then ∣∣∣∣∫

γ

f(z) dz

∣∣∣∣ 6M Length(γ),

where M is an upper bound for |f(z)| along γ.

Proof. Use the definition ∫γ

f(z) dz =

∫ b

a

f(γ(t))γ′(t)dt.


27

By real-variable theory∣∣∣∣∫γ

f(z) dz

∣∣∣∣ 6M

∫ b

a

|γ′(t)|dt = M Length(γ)

as required.

Example 8.5. Let γ be the circular path around the origin of radius r, and let us compute∫γzndz for various values of z. We can parameterize the path as z = reit, t ∈ [0, 2π].

Thus ∫γ

zndz =

∫ 2π

0

rneint ireitdt =

2πi (n = −1)

0 (n 6= −1)

The fundamental theorem of calculus is valid for path integrals in the following form.

Proposition 8.6. Let γ be a piecewise smooth path and let F be a holomorphic functiondefined on some open set containing γ∗. Put F ′(z) = f(z). Then∫

γ

f(z)dz = F (γ(b))− F (γ(a)).

Proof. We have

f(γ(t))γ′(t) = F ′(γ(t))γ′(t) =d

dt

(F (γ(t))

),

so the result follows from the ordinary form of the fundamental theorem of calculus.

However in complex analysis there is an additional complication: even very regularfunctions may not have antiderivatives for global (topological) reasons. Indeed, it’s obvi-ous that if a function has an antiderivative then its integral around any closed path mustvanish. Thus Example 8.5 shows that the function 1/z has no antiderivative on C \ 0,even though it is holomorphic (and smooth) there. This is related to the winding number:

Proposition 8.7. Let γ : [a, b]→ C be a closed, piecewise smooth path and let p ∈ C\γ∗.Then ∫

γ

dz

z − p= 2πiwn(γ; p)

where wn(γ; p) denotes the winding number of γ around p.

In this proposition, and elsewhere, γ∗ denotes the “trace” or image of γ, i.e. the setγ(t) : t ∈ [a, b] ⊆ C.

Proof. Fix w0 such that ew0 = γ(a)− p. Now define

w(s) = w0 +

∫ s

a

γ′(t)dt

γ(t)− p.

By the chain rule

d

ds

(e−w(s)(γ(s)− p)

)= e−w(s)(γ′(s)− γ′(s)) = 0.

So ew(s) = γ(s)−p and w(a) = w0. Thus w(s) is a lifting of γ(s)−p through the exponentialmap. By definition of the winding number (7.10), w(b)− w(a) = 2πiwn(γ; p), giving theresult.


28

Corollary 8.8. The winding number wn(γ; p) (considered as a function of p) is constanton the connected components of C \ γ∗, and is zero on the unbounded component.

Proof. The integral formula and the DCT show that wn(γ; p) is a continuous integer-valued function and tends to zero at ∞.

The following result is the key to the whole theory of holomorphic functions.

Proposition 8.9 (Cauchy’s theorem for a triangle). Let Ω be an open subset of C, andlet T be a triangle that is contained (interior and boundary) within Ω. Let ∂T denotethe closed path obtained by traversing the three sides of T counterclockwise. For anyholomorphic function f on Ω, one has∮

∂T

f(z) dz = 0.

Proof. Proof by contradiction, so suppose the result false. Define the ‘badness’ B(T ) of atriangle T to be

B(T ) =

∣∣∣∣∮∂T

f(z) dz

∣∣∣∣/ diam(T )2.

By hypothesis there is some triangle T with B(T ) > 0, say B(T ) = 5ε.Subdivide T = T0 into four sub-triangles by bisecting each side. Each sub-triangle has

half the diameter of T . On the other hand, the integrals∮f(z)dz around the boundaries of

the four sub-triangles total the integral around the boundary of T . Consequently, one suchintegral must have absolute value at least 1

4of the absolute value of the integral around

the boundary of T . We conclude that one sub-triangle, call it T1, has B(T1) > B(T ).Repeat this process obtaining a nested sequence of triangles T0 ⊃ T1 ⊃ T2 ⊃ · · · , each

half the diameter of the preceding one, and all with badness B(Tn) > B(T ) = 5ε.A compactness argument shows that

⋂Tn = z0 for some point z0.

Now since f is differentiable at z0 we have

f(z) = f(z0) + (z − z0)f ′(z0) + e(z),

with a ‘small’ error term e(z); in particular, there is δ > 0 such that if |z − z0| < δ then|e(z)| < ε|z − z0|.

The functionz 7→ f(z0) + (z − z0)f ′(z0)

is linear and thus has an antiderivative; so its integral around any closed contour is zero.Consequently ∮

∂Tn

f(z) dz =

∮∂Tn

e(z)dz

for each triangle Tn. Pick n large enough that the diameter d of Tn is less than δ. Thenby the estimate 8.4, ∣∣∣∣∮

∂Tn

e(z)dz

∣∣∣∣ 6 εd · 3d = 3d2ε,

so B(Tn) 6 3ε and this is a contradiction.


29

Lecture 9Cauchy’s Theorem

See Stein and Shakarchi 2.2, 2.4, 2.5We finished last time with the simplest version of Cauchy’s theorem (for a triangle): if

f : Ω→ C is holomorphic, and if T is a triangle whose interior and boundary are containedwithin Ω, then

∮∂Tf(z) dz = 0. In this lecture we will generalize this result in various

ways. We begin with a trivial but useful observation (a “removable singularity” theorem):

Lemma 9.1. Cauchy’s theorem for a triangle remains true if f is assumed to be contin-uous in Ω and differentiable everywhere except possibly at a single point.

Proof. First of all, consider the case where the ‘bad’ point is the vertex A of a triangleABC. Since f is continuous at A, given ε > 0 we can find points B′ (on AB) and C ′ (onAC) sufficiently close to A that ∣∣∣∣∫

∂AB′C′f(z)dz

∣∣∣∣ < ε.

Now ∫∂ABC

f(z)dz =

∫∂AB′C′

f(z)dz +

∫∂BB′C

f(z)dz +

∫∂CB′C′

f(z)dz

and the last two integrals vanish by Cauchy’s theorem (original version). Thus∣∣∣∣∫∂ABC

f(z)dz

∣∣∣∣ < ε

and letting ε→ 0 we get the result.If the bad point is on an edge or inside the triangle, subdivide into two or three smaller

triangles with the bad point at the vertex and apply the previous result.

Definition 9.2. A subset Ω of C is convex if, whenever a, a′ ∈ Ω, all the points λa+ (1−λ)a′, λ ∈ [0, 1], also lie in Ω.

Theorem 9.3 (Cauchy’s theorem for a convex set). Let Ω be a convex open subset of Cand let f be holomorphic on Ω. Then there is a holomorphic function F on Ω such thatf = F ′. Consequently,

∫γf(z)dz = 0 for every closed path in Ω.

Proof. Fix a ∈ Ω. For each z ∈ Ω let [a, z] denote the straight line path from a to z; byconvexity this lies entirely in Ω. Define

F (z) =

∫[a,z]

f(w)dw.

If z + h ∈ Ω also then by Cauchy for a triangle (8.9),

F (z + h)− F (z) =

∫[a,z+h]

f(w)dw −∫

[a,z]

f(w)dw =

∫[z,z+h]

f(w)dw.

Therefore∣∣∣∣F (z + h)− F (z)

h− f(z)

∣∣∣∣ =

∣∣∣∣1h∫

[z,z+h]

(f(w)− f(z))dw

∣∣∣∣6 sup|f(w)− f(z)| : w ∈ [z, z + h],


30

and this tends to zero as h→ 0 by continuity of f . We have shown that F is differentiableand its derivative is f . The equality

∫γf(z)dz = 0 now follows from the fundamental

theorem of calculus.

Remark 9.4. Using Lemma 9.1, we obtain the same conclusion under the weaker conditionf continuous in Ω, holomorphic except at a single point. We’ll use that in a moment.

The following amazing consequence shows that the values of a holomorphic function atall points inside a closed path can be recovered from its values on that path.

Proposition 9.5 (Cauchy integral formula). Let f be holomorphic on a convex open setΩ and let γ be a closed path in Ω. Let w ∈ Ω \ γ∗; then

wn(γ, w)f(w) =1

2πi

∫γ

f(z)dz

z − w.

Usually we are interested in the case wn(γ, w) = 1.

Proof. Define g on Ω\w by g(z) = (f(z)−f(w))/(z−w). This function is holomorphicon Ω \ w. Moreover, it extends to a continuous function on all of Ω, whose value at wis f ′(w). Thus Remark 9.4 applies, and

0 =

∫γ

g(z)dz =

∫γ

f(z)dz

z − w− 2πif(w) wn(γ, w)

as required.

Theorem 9.6. Let f be holomorphic in an open subset Ω of C. Suppose that the closeddisk D(a; r) is contained in Ω. Then there is a power series

∑∞n=0 cn(z − a)n which

converges to f(z) in the open disk D(a; r) (uniformly on compact subsets). Moreover thecoefficients are given by

cn =1

2πi

∮∂D(a;r)

f(w) dw

(w − a)n+1

The power series here is called the Taylor series for f around a.

Proof. Without loss of generality take a = 0. For fixed z ∈ D(a; r) write

f(z) =1

2πi

∮f(w)dw

w − z=

1

2πi

∮ ( ∞∑n=0

zn

wn+1

)f(w)dw,

using Cauchy’s integral formula. (The integrals are taken around ∂D(a; r).) The seriesconverges uniformly in w for z fixed (by Weierstrass’ M-test) and so we can interchangesum and integral to get

f(z) =∞∑n=0

zn(

1

2πi

∮f(w)dw

wn+1

)as asserted.

Corollary 9.7. A holomorphic function is infinitely differentiable, and we have Cauchy’sformula for derivatives

f (n)(a) =n!

2πi

∮∂D(a;r)

f(w)dw

(w − a)n+1.


31

Proof. Use the differentiability of power series.

Remark 9.8. The proof that we have given of Cauchy’s theorem is due to Goursat, andit has the great advantage that it proves the theorem without assuming any regularityproperties of the derivatives. The argument originally given by Cauchy was different,proceeding via Green’s theorem, and it required the continuity of f ′ as an assumption.It is instructive to review that argument, nevertheless. In the remainder of this lecture Iwill do that, using the modern language of differential forms.

The key observation is this.

Lemma 9.9. Let f be a holomorphic function (with continuous derivative) on an open setΩ. Then f(z)dz is a closed 1-form on Ω.

Proof. By f(z)dz we mean the complex-valued 1-form f(x + iy)(dx + idy). If we writef = u+ iv with u, v real, then

f(z)dz = (u+ iv)dx+ (iu− v)dy.

Then

d(f(z)dz) =

(− ∂

∂y(u+ iv) +

∂

∂x(iu− v)

)dx ∧ dy,

that is (−(uy + vx) + i(ux − vy)) dx∧dy. The coefficients vanish by the Cauchy-Riemannequations.

Now let M ⊆ Ω be a compact 2-dimensional submanifold with boundary ∂M . Thenby Stokes’ theorem, ∫

∂M

f(z)dz =

∫M

d(f(z)dz) = 0.

(The continuity of the derivative is needed for the validity of Stokes’ theorem.) This isa version of Cauchy’s theorem: to get the result for a triangle (8.9), you either need toextend Stokes’ theorem to submanifolds with “corners” (not too hard), or approximatethe triangle T by smooth submanifolds obtained by rounding off the corners a bit.


32

Lecture 10More about holomorphic functions

See Stein and Shakarchi 2.4, 2.5

Lemma 10.1. Let f be holomorphic in a disc D(a; r), but not identically zero there.Then there is a smaller disc D(a; ρ) within which f(z) = 0 has no solution except possiblyz = a.

Proof. Expand f in a Taylor series f(z) =∑cn(z − a)n about z = a. Since f does not

vanish identically, some Taylor coefficient cn is nonzero; let cN be the first such coefficient(so that cN 6= 0 but ck = 0 for k < N). Then

f(z) =∞∑n=N

cn(z − a)n = (z − a)N(cN + (z − a)g(z)),

where

g(z) =∞∑k=0

ck+N+1(z − a)k

is continuous at a. It follows that |(z − a)g(z)| < 12|cN | for z sufficiently close to a, and

thus cN + (z − a)g(z) does not vanish for z sufficiently close to a.

Terminology: A zero of f is a point where f(z) = 0. The exponent of the firstnonvanishing term in the Taylor series of f about a zero (i.e. the number N above) is theorder of the zero.

Let S be a subset of C. Recall that a limit point of S is the limit of a sequence ofdistinct points of S. A set is closed if and only if it contains all of its limit points. Anylimit point of limit points is itself a limit point; in particular, the set of limit points ofany set is itself closed.

Proposition 10.2 (The principle of isolated zeroes). Let Ω ⊆ C be open and connected.If f : Ω → C is holomorphic and not identically zero, then the set of zeroes of f has nolimit points in Ω.

Proof. Let A ⊆ Ω be the set of limit points of zeroes of f . Then A is closed. But bylemma 10.1, if a ∈ A then some disc D(a; ρ) is also contained in A. Thus A is open.Connectedness now gives that A is either empty or equal to Ω.

It follows that if f and g are holomorphic functions on Ω, and f(z) = g(z) for allz belonging to some set having a limit point in Ω, then f = g everywhere; this is theuniqueness of analytic continuation. (To see it, apply the previous result to f − g.) Forexample, supposing that we know that a trigonometric identity like

sin(z + w) = sin(z) cos(w) + cos(z) sin(w)

holds for all real z, w, we can deduce that it holds for complex z, w too.


33

Lecture 11Yet more about holomorphic functions

Lemma 11.1. Let f be holomorphic in a disc D(a; r) and not constant there. Then thereare points z ∈ D arbitrarily close to a at which |f(z)| > |f(a)|.

Proof. Without loss of generality, take a = 0. Using the Taylor series write

f(z) = c0 +∞∑n=N

cnzn = c0 + zN(cN + zg(z))

with g continuous, cN 6= 0. By replacing f(z) by eiϕf(eiψz) for suitable constants ϕ, ψwe may assume that c0 and cN are real and positive. But now for x > 0 real and smallenough that |xg(x)| < 1

2|cN |,

Re f(x) > c0 + 12cNx

N > c0

and so |f(x)| > c0 = |f(0)| also.

Theorem 11.2 (Maximum principle). Let f be a nonconstant holomorphic function ona connected open set Ω. Then |f | does not attain a local maximum (even non-strict)anywhere on Ω.

Proof. If it did, then according to Lemma 11.1 it would be constant on a disk around themaximum point, hence everywhere in Ω by the uniqueness of analytic continuation.

The minimum principle states that |f | attains a local minimum only at the zeroes off . To prove it, apply the maximum principle to 1/f .

Corollary 11.3 (Fundamental theorem of algebra). Every nonconstant polynomial p(z)has a root in C.

Proof. A nonconstant polynomial is holomorphic and has the property that |p(z)| → ∞as |z| → ∞. Thus |p(z)| must attain a (global!) minimum somewhere. By the minimumprinciple, the point where it does so is a root.

Proposition 11.4 (Morera’s Theorem). If a continuous function f on an open subset Ωof C has

∫∂Tf(z)dz = 0 for all triangles T in Ω, then f is holomorphic.

Proof. Without loss of generality take Ω to be a disk. Then the proof of theorem 9.3shows that f(z) = F ′(z) for some function F holomorphic in Ω. But now F is infinitelydifferentiable; so f is infinitely differentiable too — in particular it is holomorphic.

Corollary 11.5. If fn is a sequence of holomorphic functions defined on an open set Ω,and if fn → f uniformly on compact subsets of Ω, then f is holomorphic on Ω too.

Proof. The limit of a uniformly convergent sequence of continuous functions is continuous.Therefore, f is continuous. Moreover, for any triangle T ,

∫∂Tf(z)dz = lim

∫∂Tfn(z)dz =

0. The result follows from Morera’s theorem.

Corollary 11.6 (Removability of Singularities). Let f be continuous in an open set Ωand holomorphic in Ω except possibly at one point a. Then f is, in fact, holomorphic ata too.


34

Proof. Use Lemma 9.1 and Morera’s theorem.

There are several ways to use complex analysis in proving the fundamental theorem ofalgebra. Another one makes use of the Cauchy estimates for a holomorphic function andits derivatives.

Proposition 11.7 (Cauchy Estimates). Suppose that f is holomorphic in an open setcontaining the closed disk D(a; r). Then

|f (n)(a)| 6 n!r−n sup|f(z)| : |z − a| = r.

Proof. This follows immediately from the integral formula

f (n)(a) =n!

2πi

∮∂D(a;r)

f(w)dw

(w − a)n+1

of Corollary 9.7.

Corollary 11.8 (Liouville’s Theorem). Let f : C → C be an entire function (that is,holomorphic on the whole complex plane). If f is bounded, then it is constant.

Proof. Say f is bounded by M . Cauchy estimates give

|f ′(a)| 6Mr−1

for any r. Letting r →∞ we find that f ′ = 0, implying that f is constant.

Now we can prove the fundamental theorem of algebra again. If p is a nonconstantpolynomial, then 1/p is holomorphic everywhere except at the roots of p, and it tendsto zero at infinity. Thus, if p had no roots, 1/p would be a constant, so p would be aconstant too; a contradiction.

Terminology: Let Ω be an open subset of C and let a be a point of Ω. If f isa holomorphic function defined on Ω \ a, we sometimes say that f has an isolatedsingularity at a.

We are now going to give a counterpart to the Taylor series expansion valid for functionswith an isolated singularity, or more generally functions holomorphic in any annulus.

Theorem 11.9. Let f be holomorphic in the annular region Ω = D(a; r1)\D(a; r2). Thenf has a Laurent expansion convergent in Ω, uniformly on compact subsets, of the form

f(z) =∞∑

n=−∞

cn(z − a)n,

where

cn =1

2πi

∮f(w)dw

(w − a)n+1,

the integral being taken around a circular contour in Ω with center a.

Proof. Assume without loss of generality that a = 0. Use Cauchy’s integral formula towrite

f(z) =1

2πi

∫Γ

f(w)dw

w − z,


35

where the “cycle” Γ is the union of two circles, one of radius R larger than |z| and traversedcounterclockwise, one of radius r smaller than |z| and traversed clockwise2 . Now on thelarger circle expand

1

w − z=∞∑k=0

zk

wk+1

with uniform convergence in w; on the smaller circle expand

1

w − z= −

∞∑k=0

wk

zk+1

again with uniform convergence in w. Substitute these expansions into the Cauchy integralformula and interchange summation and integration (justified by uniform convergence).This gives

f(z) =∞∑k=0

(1

2πi

∮f(w)dw

wk+1

)zk +

∞∑k=0

(1

2πi

∮wkf(w)dw

)z−k−1;

here the integrals in the first sum are around the circle of radius R, the ones in thesecond sum are around the circle of radius r, but by Cauchy’s theorem we obtain thesame integral around either of these circles and indeed around any circle around a in Ω.Rearranging the displayed equation gives Laurent’s theorem as stated.

There is some standard terminology about isolated singularities. If f has an isolatedsingularity at a then there is a Laurent series expansion

f(z) =∞∑

n=−∞

cn(z − a)n

about a. If there are only finitely many negative values of n for which cn 6= 0, one saysthat f has a pole at a; in that case the order of the pole is the number N such thatc−N 6= 0 but cn = 0 for all n < −N . Otherwise, if there are infinitely many negative nfor which cn 6= 0, one says that f has an essential singularity at a.

Example 11.10. The function cot z has a pole of order 1 (a.k.a. a simple pole) at zero.The function 1/(cos z − 1) has a pole of order 2 (double pole) at zero. The functionsin(1/z) has an essential singularity at zero.

2To do this, imagine making three radial “cross-cuts” between the larger and smaller circles, applyingthe Cauchy formula to each of the three wedge-shaped closed curves so formed, and then adding theresults.


36

Lecture 12The Global Cauchy Theorem

In this lecture we shall prove the most general form of Cauchy’s theorem, removingthe restriction to convex regions Ω. First, we introduce some jargon which belongs to therealm of homology theory. Let Ω be an open subset of C.

Definition 12.1. An 1-chain in Ω is a finite formal linear combination (with integercoefficients) of piecewise smooth paths in Ω. (Formally speaking, this is an element of thefree abelian group generated by the piecewise smooth paths in Ω.)

We can extend the definition of integral to chains in the obvious way.

Definition 12.2. If Γ =∑n

j=1 kjγj is a 1-chain, we define∫Γ

f(z)dz =n∑j=1

kj

∫γj

f(z)dz

whenever f is defined and continuous on Γ∗ =⋃j γ∗j .

An equivalence relation ∼ on chains (we just call it “equivalence”) is generated by thefollowing three rules.

(a) If γ is the concatenation of γ1 and γ2, then [γ] ∼ [γ1] + [γ2].(b) If γ\ is the reverse of γ (traversed in the backwards direction), then [γ] ∼ −[γ\].(c) If γ1 is a reparameterization of γ2 (Remark 8.3) then [γ1] ∼ [γ2].

Notice that (a) and (c) imply that if γ is a constant path, then [γ] ∼ 0. Note also thatif Γ1 ∼ Γ2, then ∫

Γ1

f(z)dz =

∫Γ2

f(z)dz

for all functions f defined on Γ∗1 ∪ Γ∗2.

Definition 12.3. A chain Γ in Ω is a simple cycle if it is equivalent to [γ], where γ is aclosed, piecewise smooth loop. It is a cycle if it is a Z-linear combination of simple cycles.

Remark 12.4. An 0-chain in Ω is a formal linear combination of points of Ω. The boundarymap from 1-chains to 0-chains is the linear map which takes each path γ : [a, b]→ Ω to theformal difference of its ending and starting points, [γ(b)] − [γ(a)]. Clearly, every simplecycle has boundary zero. In fact, this condition is reversible:

Exercise 12.5. Show that a 1-chain is a cycle if and only if its boundary is zero.

This exercise links our definition of cycle with the classical one from homology theory,but we will not need to make use of it in what follows.

If Γ is a 1-cycle and p /∈ Γ∗, the number

wn(Γ, p) =1

2πi

∫Γ

dz

z − pis a Z-linear combination of the winding numbers of piecewise smooth loops (by Proposi-tion 8.7) and therefore an integer; we call it the winding number of Γ about p.


37

Definition 12.6. A chain Γ in Ω is a simple boundary if it is equivalent to [γ], where γis a closed, piecewise smooth loop for which γ∗ is contained in a convex subset of Ω. It isa boundary if it is a Z-linear combination of simple boundaries.

Remark 12.7. As in the case of cycles, we can show that this definition is the same asthe standard one from homology theory. In homology theory, one defines a (singular) 2-simplex to be a smooth map of a triangle into Ω, and its boundary to be the chain whichis the sum of its three edges, taken with appropriate signs. One then has the followingresult:

Exercise 12.8. Show that a chain Γ is a boundary (as defined in Definition 12.6) if it is alinear combination of boundaries of singular 2-simplices.

Again, we won’t actually need this for the further development, but it shows that ourdefinition matches the standard one from homology theory, when applied to the specialcase of open subsets of C.

Definition 12.9. Two 1-cycles Γ11 and Γ2 are homologous if [Γ1] − [Γ2] is a boundary.The homology group H1(Ω) is the group of homology classes of 1-cycles — that is, thequotient group

Cycles/Boundaries.

With our definitions the global Cauchy theorem is now a gimme.

Theorem 12.10. Let f : Ω → C be a holomorphic function, and let Γ be a cycle in Ω.Then

∫Γf(z)dz depends only on the homology class of Γ. In particular,∫

Γ

f(z)dz = 0

if Γ is a boundary.

Proof. We simply need to show that∫

Γf(z)dz = 0 if Γ is a simple boundary; but this

follows from Cauchy’s theorem for convex sets, Theorem 9.3.

What gives the homology version of Cauchy’s theorem legs is that there is a simplecriterion in terms of winding numbers for computing when a given cycle is a boundary. Icall this Artin’s criterion: in fact, it is a manifestation of Poincare-Lefschetz duality fromalgebraic topology, but it is OTT to invoke that here.

Proposition 12.11 (Artin’s criterion). Let Γ and Γ′ be 1-cycles in the open set Ω. ThenΓ and Γ′ are homologous if and only if wn(Γ, p) = wn(Γ′, p) for all p /∈ Ω.

We will prove this next time. Notice that “only if” follows immediately from the globalCauchy theorem and the integral formula for the winding number. The problem is toprove that if a cycle has zero winding numbers about all points outside Ω, then it is aboundary.

Exercise 12.12. Give an example of a cycle (in fact a closed path) in Ω = C\0, 1 thatis homologous to zero but is not homotopic to zero (i.e., cannot be continuously deformedinside Ω to a constant path).

Here is a typical application of Artin’s criterion, to generalize the Cauchy integralformula.


38

Theorem 12.13. Let f be holomorphic on an open set Ω and let Γ be a 1-cycle in Ω thatis homologous to zero. Then for every a ∈ Ω \ Γ∗, we have the integral formula

wn(Γ, a)f(a) =1

2πi

∫Γ

f(z)dz

z − a.

Cauchy’s formula for derivatives also holds,

wn(Γ, a)f (n)(a) =n!

2πi

∫Γ

f(z)dz

(z − a)n+1.

Proof. Let m = wn(Γ, a). Choose ε > 0 so small that B(a; 2ε) ⊆ Ω and let Γ′ be thecycle m[γ], where γ : [0, 1] → Ω is defined by γ(t) = a + εe2πit. By construction, Γ andΓ′ have the same winding number (namely m) around a, and they also have the samewinding number (namely 0) about all points outside Ω, so by Artin’s criterion they arehomologous in Ω′ = Ω \ a. Since the function z 7→ f(z)/(z − a)n+1 is holomorphic inΩ′, the general Cauchy theorem tells us that∫

Γ

f(z)dz

(z − a)n+1=

∫Γ′

f(z)dz

(z − a)n+1= m

∫γ

f(z)dz

(z − a)n+1,

and the last integral can be evaluated by the version of Cauchy’s formula for derivativesthat we have already proved.

Example 12.14. In the proof of Laurent’s theorem 11.9 we considered a cycle Γ in thepunctured disc Ω = D(a; r) \ a which was the union of two circles around a, the outerone traversed in the positive direction and the inner one in the negative direction. Thiscycle has zero winding number about a and about all points outside D(a; r), so it ishomologous to 0 in Ω. Theorem 12.13 now gives

f(z) =1

2πi

∫Γ

f(w)

w − zdw,

as required in the proof of Theorem 11.9.


39

Lecture 13The proof of Artin’s criterion

We begin with a convenient way to compute winding numbers. Let Γ be a cycle. ThenC \ Γ∗ is an open set, which falls into a number of connected components called thecomplementary cells of Γ. One and only one of these is unbounded. The winding number

wn(Γ, p) =1

2πi

∫Γ

f(z)

z − pdz

is defined on the complement of Γ∗ and is constant on each complementary cell (as it’s acontinuous Z-valued function). The winding number vanishes on the unbounded comple-mentary cell.

Definition 13.1. A piecewise linear cycle is one whose constituent paths (edges) areoriented straight line segments meeting (if at all) only at their endpoints: the weight ofan edge is the coefficient with which it appears.

The following gives an effective way to compute winding numbers for PL cycles.

Lemma 13.2. The weight of an edge e in a PL cycle Γ is equal to the difference of thewinding numbers of Γ about the complementary cells to the “left” and “right” of Γ.

Proof. Let m be the weight of e; let p and q be points near to the midpoint of e, with p tothe left of e and q to the right. Define a new cycle Γ′ to be the same as Γ except that theedge e is replaced by the sum of five edges (all with weight m), following e except for asmall diversion around three sides of a square3 that leaves q to the left of Γ′. Then Γ andΓ′ are homologous in C \ p, and in C \ q the cycle Γ′ − Γ is equivalent to m times asquare winding once around q. Moreover, p and q are in the same connected componentof C \ (Γ′)∗. Consequently

wn(Γ, p) = wn(Γ′, p) = wn(Γ′, q) =

wn(Γ′ − Γ, q) + wn(Γ, q) = m+ wn(Γ, q),

as required.

In order to present the proof of Artin’s criterion, let’s introduce some terminology. Afinite collection of horizontal and vertical lines in C will be called a grid. The grid linesof a given grid G subdivide one another into line segments (some finite and some not),and a cycle which is a finite linear combination of finite line segments of this type willbe called a grid cycle for G. The complement of the grid G in C is a disjoint union ofopen rectangles (some finite and some not), and a given grid cycle will have a well-definedwinding number around each of these complementary rectangles; let us call the collectionof these numbers the grid winding numbers of the grid cycle in question. We use thenotation wn(Γ;R) where Γ is a grid cycle and R a complementary rectangle, and wenotice that wn(Γ;R) = 0 if the rectangle R is not finite. See the figure below

3To be completely specific: Suppose wlog that e runs from −a to a, and that p = iεa and q = −iεafor ε sufficiently small. Then Γ′ is obtained from Γ by replacing the edge e with five edges, running from−a to −εa to −(ε + 2i)a to (ε− 2i)a to εa to a.


40

−1

+2

+1

0

Ω

Lemma 13.3. Every cycle in an open subset Ω ⊆ C is homologous to a grid cycle.

Proof. It is enough to consider the case of a piecewise smooth loop γ. Using compactness,there is ε > 0 such that any disc of radius ε centered on a point of γ is contained inΩ. By uniform continuity, we may subdivide the parameter interval [0, 1] into finitelymany subintervals over each of which γ varies by less than ε. Now the restriction of γ toeach such subinterval is homologous to the concatenation of a horizontal and a verticalsegment.

Lemma 13.4. A grid cycle all of whose grid winding numbers are zero is equivalent to 0.

Proof. By Lemma 13.2, the multiplicity of an edge e in a grid cycle is equal to the differenceof the grid winding numbers on the two sides of e. Thus all multiplicities are zero if allgrid winding numbers are.

Proof of Artin’s criterion 12.11. If Γ is a boundary, then Cauchy’s theorem shows thatthe winding numbers

wn(Γ, p) =1

2πi

∫Γ

1

z − pdz

must vanish for p /∈ Ω.Suppose conversely that these winding numbers all vanish. By Lemma 13.3, there is

no loss of generality in assuming that Γ is a grid cycle based on some grid G.Define a new grid cycle

(13.5) Γ′ =∑R

wn(Γ;R)[∂R],


41

where the sum is extended over all finite complementary rectangles R to G. Notice thatif R′ is a fixed finite complementary rectangle, then

wn([∂R];R′) =

1 (R = R′)

0 (R 6= R′).

Consequently wn(Γ;R) = wn(Γ′;R) for all R, and therefore, according to Lemma 13.4, Γis equivalent to Γ′.

It remains to show that Γ′ is a boundary in Ω, and this will follow if we can show thatevery R appearing with nonzero coefficient in Equation 13.5 has closure contained in Ω(since then R will have a convex neighborhood contained in Ω, namely

⋃w∈RD(w; ε) for

some sufficiently small ε > 0). But if R meets the complement of Ω, then there is astraight line path in R linking one of its interior points (say p) to a point of C \ Ω (sayq), and this path does not meet Γ∗. Thus

wn(Γ, R) = wn(Γ, p) = wn(Γ, q) = 0

by hypothesis. This completes the proof.


42

Lecture 14Poles and Zeroes

See Stein and Shakarchi 3.1–3.4Recall that a function f holomorphic on Ω\a, where a ∈ Ω, Ω open, is said to have an

isolated singularity at a. Such a function has a Laurent expansion f(z) =∑∞

n=−∞ cn(z −a)n valid in a punctured disc around a.

Definition 14.1. If f has an isolated singularity at a, as above, the residue of f at a isthe coefficient c−1 in its Laurent expansion. We write it Res(f, a).

From the formula in Theorem 11.9 for the Laurent coefficients it follows that

Res(f, a) =1

2πi

∮f(w)dw

where the integral is taken around a small circular contour enclosing a. Thus the residuemeasures the ‘local failure of Cauchy’s theorem’ at a.

Theorem 14.2 (The residue theorem). Let Ω be an open subset of C and let Γ be acycle homologous to zero in Ω. Let f be a function holomorphic on Ω except for isolatedsingularities at points a1, . . . , am ∈ Ω \ Γ∗. Then

1

2πi

∫Γ

f(z)dz =m∑j=1

wn(Γ, aj) Res(f, aj).

Proof. Let γj be a closed circular contour in Ω around aj of radius smaller than 12

min|aj−ak| : j 6= k. Then, by Artin’s criterion 12.11, the cycle Γ is homologous in Ω\a1, . . . , amto

Γ′ =m∑j=1

wn(Γ, aj)γj.

Thus by the homology form of Cauchy’s theorem

1

2πi

∫Γ

f(z)dz =1

2πi

m∑j=1

wn(Γ, aj)

∫γj

f(z)dz

as required.

There is some standard terminology about isolated singularities. If f has an isolatedsingularity at a then there is a Laurent series expansion

f(z) =∞∑

n=−∞

cn(z − a)n

about a. If there are only finitely many negative values of n for which cn 6= 0, one saysthat f has a pole at a; in that case the order of the pole is the number N such thatc−N 6= 0 but cn = 0 for all n < −N . Otherwise, if there are infinitely many negative nfor which cn 6= 0, one says that f has an essential singularity at a.

Example 14.3. The function cot z has a pole of order 1 (a.k.a. a simple pole) at zero.The function 1/(cos z − 1) has a pole of order 2 (double pole) at zero. The functionsin(1/z) has an essential singularity at zero.


43

A special case is a pole of order zero, or removable singularity. Here the Laurent seriescontains no negative powers of z, so it actually converges at z = a. In other words, f canbe extended to a holomorphic function on Ω; the singularity can be ‘removed’.

It is sometimes convenient to make the convention4 that a ‘zero of order −N ’ means apole of order N , and a ‘pole of order −N ’ means a zero of order N . With this convention,the order of the zero of f at a is the unique N (positive or negative) such that we canwrite f(z) = (z − a)Ng(z) near a, with g holomorphic near a and g(a) 6= 0. From this,it follows that if f1 and f2 have zeroes of orders N1 and N2 at a, then f1f2 has a zeroof order N1 + N2 and f1/f2 has a zero of order N1 − N2. Therefore the quotient of twoholomorphic functions has (at worst) isolated poles as singularities.

Exercise 14.4. Suppose that f1 and f2 are as above, possibly having zeroes or poles ata. What, if anything, can you say about the order of the zero or pole of f1 + f2 at a, interms of the orders of f1 and f2?

One can classify isolated singularities in terms of the limiting behavior of the functionnear the singularity.

Proposition 14.5. Let a be an isolated singularity of the holomorphic function f . Then

(i) the point a is a removable singularity of f iff |f(z)| is bounded in some neighborhoodof a;

(ii) the point a is a pole of f iff |f(z)| tends to ∞ as z tends to a;(iii) the point a is an essential singularity iff f(z) approaches every value as z approaches

a (this means that, given any b ∈ C, we can find a sequence zn → a such thatf(zn)→ b).

Proof. First look at (i). Obviously, f is bounded near a removable singularity sinceit extends continuously across the singularity. Conversely, if f is bounded by M ata singularity, look at the Laurent expansion. Estimate the coefficients by integratingaround a path of radius ε to get |cn| 6 Mε−n. Letting ε → 0 we find that cn = 0 forn = −1,−2, . . ., so the Laurent series is a Taylor series and f has a holomorphic extension.

Now for (ii), if f has a pole at a of order N then f(z) = (z − a)−Ng(z), where g isholomorphic and g(a) 6= 0. It certainly follows that |f | → ∞ at a. The converse statementwill follow from (iii) — if f had an essential singularity than f(z) would approach everyvalue, and in particular could not tend to ∞ — once we have proved that.

As for (iii), if f(z) approaches every value as z → a, then a can’t be removable ora pole (by what we have already proved) so it must be an essential singularity. Theconverse statement (sometimes called the Casorati-Weierstrass theorem) is the meat ofthe result. Suppose that f does not approach every value and let b be a value which isnot approached, so that there exist ε, δ > 0 with |f(z) − b| > δ whenever |z − a| < ε.Then g(z) = 1/(f(z) − b) is bounded near a, so it has a removable singularity. Extendg to a function holomorphic in a disk around a by removing the singularity. Writingf(z) = b + 1/g(z), we now see that the singularity of f at a is at worst a pole, of orderequal to the order of the zero of g at a.

A function that is holomorphic on Ω except for isolated poles is called a meromorphicfunction on Ω. For a pole p one conventionally writes f(p) =∞ (unsigned); the previous

4See the discussion after Lemma 10.1 to understand what we mean by the “order” of a zero.


44

proposition gives the reason for this convention. Now we will use integration to count thezeroes and poles of a holomorphic or meromorphic function.

Lemma 14.6. Let f be holomorphic or meromorphic on an open set Ω. Then the quotientf ′/f is meromorphic; its poles occur at the zeroes and the poles of f . Moreover, the polesof f ′/f are all simple, and the residue of f ′/f at a point is equal to the order of the zero(or minus the order of the pole) of f at that point.

Proof. If f is holomorphic and nonzero near a, then f ′(z)/f(z) is holomorphic near a.Thus the only possible singularities of f ′/f are the zeroes and poles of f .

Suppose that such a point (zero or pole) occurs at z = a. Then from the Taylor orLaurent series,

f(z) = (z − a)mg(z)

near z = a, where g(z) is holomorphic and non-vanishing at a and m is the order of thezero (or minus the order of the pole). Thus

f ′(z)

f(z)=m(z − a)m−1g(z) + (z − a)mg′(z)

(z − a)mg(z)=

m

z − a+g′(z)

g(z).

The second term is holomorphic near a, so the pole is simple and has residue m.

From the residue theorem we deduce

Proposition 14.7. Let f be meromorphic in Ω, and let Γ be a cycle homologous to zeroin Ω and not passing through any zeroes or poles of f . Then

1

2πi

∫Γ

f ′(z)

f(z)dz =

∑j

mj wn(Γ, aj),

where a1, a2, . . . are the zeroes and poles of f in Ω and mj is the order of the zero (orminus the order of the pole) of f at aj.

Remark 14.8. The left hand side of Proposition 14.7 is the winding number wn(f∗Γ, 0),where f∗Γ is the cycle got by composing each of the individual paths that make up Γ withthe function f . More generally, if f is holomorphic then wn(f∗Γ, b) gives the number oftimes that f takes the value b (rather than 0) inside Γ.

Proposition 14.9 (Rouches Theorem). Let f and g be holomorphic in an open set Ω,and let Γ be a cycle homologous to zero in Ω. Suppose that neither f nor g has a zero onΓ∗. If |f(z) − g(z)| < |f(z)| + |g(z)| for all z ∈ Γ∗, then f and g have the same numberof zeroes inside Γ.

The “number of zeroes inside Γ” is, by definition, the number computed in Proposi-tion 14.7.

Proof. It is an easy exercise to see that for complex numbers a, b, the strict inequality

|a− b| < |a|+ |b|implies that the line segment [a, b] does not pass through zero. Consider then the familyof holomorphic functions

ft(z) = tg(z) + (1− t)f(z)


45

for t ∈ [0, 1]. For each t and z, the point ft(z) lies on the line segment [f(z), g(z)], so bythe observation above the functions ft have no zeroes on Γ∗. Thus the integral

1

2πi

∫Γ

f ′t(z)

ft(z)dz

is a continuous integer-valued function of t, hence constant. For t = 0 we get the numberof zeroes of f , for t = 1 the number of zeroes of g.

We can use this to give yet another proof of the Fundamental Theorem of Algebra.

Corollary 14.10. A polynomial equation of degree n must have n roots.

Proof. In Rouches theorem, let g(z) = a0 + · · · + anzn be our polynomial equation and

let f(z) = anzn. Let Γ be a circle, center the origin. If the radius of Γ is large then

|f − g| < |f | on Γ. Therefore, f and g have the same number of roots inside Γ. But fobviously has n roots counted with multiplicity (namely, an n-tuple root at the origin);so g does too.

Proposition 14.11 (Open mapping principle). A non-constant holomorphic map on anopen set Ω in C is an open map; it takes open subsets (of Ω) to open subsets of C.

Proof. It is enough to show that, given any a ∈ Ω, there is a neighborhood W of a in Ωsuch that f(W ) is a neighborhood of b = f(a).

Since f is non-constant there is a disk D(a; ε) on which f(z) does not take the value bexcept at z = a. Let Γ be a circular path around a of radius ε/2. Then wn(f∗Γ; b) > 0and so (by continuity) wn(f∗Γ; b′) > 0 for b′ close enough to b, say |b′ − b| < δ. Thus ftakes the value b′ within Γ. But this is to say that if W = D(a; ε/2), then f(W ) containsD(b; δ), and so is a neighborhood of b, as required.


46

Lecture 15Applying the residue theorem

See Stein and Shakarchi 3.1, 3.2To apply the residue theorem, it is necessary to be able to evaluate residues efficiently.

Often, the simplest way to proceed is to manipulate Laurent series.

Example 15.1. Find the residue of the function

1− z2

1 + z2

at z = i.Answer. We must find a Laurent expansion in powers of (z − i). We can write

1− z2

1 + z2=

2 +O(z − i)(z − i)(2i+ (z − i))

=1

i(z − i)+O(1).

So the residue is 1/i = −i.

Example 15.2. Find the residue of the function

ez

ez − 1− zat z = 0.

Answer. The function ez − z − 1 has a Taylor series beginning with z2/2, so it has adouble zero at z = 0. The function ez is nonzero at z = 0. Thus the quotient has a doublepole. To find the residue, we need to work out the beginning of the Laurent series:

ez

ez − z − 1=

1 + z +O(z2)

(z2/2)(1 + z/3 +O(z2))=

2

z2

(1− z

3

)(1 + z

)+O(z2) =

2

z2+

4

3z+O(1).

Thus the residue is 4/3.

There are some general results that can be used to calculate residues.

Lemma 15.3. Let f be holomorphic near z = a; then the residue of f(z)/(z − a)n at ais f (n−1)(a)/(n− 1)!.

Proof. From the Taylor series for f .

Lemma 15.4. Let f and g be holomorphic near z = a, and suppose that g has a simplezero at a. Then the residue of f(z)/g(z) at a is f(a)/g′(a).

Proof. Write g(z) = (z− a)h(z), where h is holomorphic and nonvanishing at a, and notethat g′(a) = h(a) (by elementary calculus). Then f(z)/g(z) = (f(z)/h(z))/(z−a), wheref(z)/h(z) is holomorphic near a; applying the previous lemma gives the result.

One can use this lemma to shorten Example 15.1, for instance. There are analogousformulas for multiple poles, but they’re not worth trying to remember.


47

Exercise 15.5. Derive the formula for the residue at a double pole: if g has a doublezero at a, then the residue of f/g at a is

2

(f ′(a)

g′′(a)− f(a)g′′′(a)

3g′′(a)2

).

Verify that this is consistent with our solution to Example 15.2.

The residue theorem gives a computational method for evaluating certain definite in-tegrals, which is known as contour integration. The idea of the method is to considera suitable meromorphic function f(z), and evaluate its integral over the cycles in somehomology class by using the residue theorem. Then, usually, one lets the cycles approachsome limit in which it can be seen that the integral around the cycle tends to some definiteintegral that we want to evaluate. It is easier to give examples than to try to explain thisin general.

Example 15.6. Evaluate

I =

∫ ∞−∞

1

1 + x2dx.

Of course we know from elementary calculus that the answer is π, but let’s pretend for amoment that we don’t know this. Let f(z) = 1/(1+z2), which is a meromorphic functionwith simple poles at ±i. Let ΓR be the closed path which proceeds from −R to R alongthe real axis, and then returns via a semicircle in the upper half plane. From the residuetheorem we have ∫

ΓR

f(z)dz = 2πiRes(f, i) = π

so long as R > 1. Now consider what happens as R approaches ∞. The integral alongthe straight portion of ΓR approaches I. The integral around the semicircular portion isbounded by

πR · 1

R2 − 1

by the estimate 8.4, and this tends to zero as R → ∞. Thus the integral∫

ΓRf(z)dz

approaches I; but on the other hand, it is the constant π. Therefore, I = π.

Remark 15.7. We could equally well have used a semicircular “return contour” around thelower half plane. We would have obtained the same result. Generalizing this remark, onecan show that if f(z) = p(z)/q(z) is a rational function, the quotient of two polynomials,and if deg(q) > deg(p) + 2, then the sum of all the residues of f is zero.

Example 15.8. Evaluate

I =

∫ ∞−∞

cosx

1 + x2dx.

This is very similar to the previous example; we use the same semicircular contour. Put

f(z) =eiz

1 + z2.


48

(It is important that we use the function eiz rather than the tempting cos z! The reasonis that eiz is bounded in the upper half plane.) Now from the residue theorem∫

ΓR

f(z)dz = 2πiRes(f, i) =π

e.

The real part of the integral along the straight portion of Γ approaches I as R → ∞.Around the semicircular portion, the integral tends to zero as before (since |eiz| 6 1there). Thus, I = π/e.

Remark 15.9. In this example we could not have integrated using the lower half planeinstead; the function eiz is not bounded there.


49

Lecture 16Contour integration

In Examples 15.6 and 15.8 we have seen that the Residue Theorem, together withsimple estimates, can be used to evaluate the integrals of certain real-valued functions.The bag of tricks that are used for this purpose is called “evaluation of integrals by residuecalculus” or something like that. Typical steps in the evaluation of an integral I by residuecalculus will include the following:

(1) Choose a meromorphic function f(z), defined on an appropriate open set Ω, whichis usefully related to the integrand of I. In the simplest cases like Example 15.6,this may just be a matter of “writing z for x”. However, as Example 15.8 suggests,this does not always work and a more judicious choice may be required.

(2) Choose a cycle Γ in Ω which is homologous to 0. We are going to apply the ResidueTheorem to this cycle. To be a bit more precise, we often don’t just choose onecycle, but a whole family of cycles depending on some parameters (like the R inthe examples above). We’re then going to get the desired conclusion by lettingthe parameters tend to a suitable limit (typically 0 or ∞).

(3) The cycle Γ is typically made up of a number of separate arcs (like the interval ofthe real axis and the semicircle in the examples above). For each such arc γ, wewant to know that, as the parameters approach their limiting values, the integral∫γf(z)dz approaches some limiting value which is either related somehow to the

integral we want to evaluate or which we can calculate explicitly (most often byusing estimates to show that the limiting value is 0). In the two examples we havestudied the integral over the real interval and the semicircle provide examples ofthose two types.

(4) Now we apply the residue theorem to evaluate the integral∫

Γf(z)dz. Equating

this with the sum of the limiting values from the previous step gives us a calculationof I. Important Note: Often, it is enough to compare real (or imaginary) parts inthe resulting equation; this can reduce the amount of work that needs to be done.See HW3, exercise 5.

It is easiest to explain this by examples.

Example 16.1. Show that ∫ ∞−∞

eat dt

1 + et=

π

sin πa,

for 0 < a < 1.Let I denote the integral we want to evaluate. Let f(z) = eaz/(1 + ez), and notice that

(1) f(z + 2πi) = e2πiaf(z).(2) f has a simple pole at z = πi, with residue −eiπa.(3) If x = Re(z) > 1, then |f(z)| 6 eax/(ex − 1), which tends to 0 as x→∞.(4) If x = Re(z) < −1 then |f(z)| 6 2eax, which tends to 0 as x→ −∞.

Consider the integral∫

ΓRf(z)dz, where ΓR is now a rectangular contour whose vertices

are at −R, R, R + 2πi, and −R + 2πi. By the residue theorem, this integral equals−2πieiπa. On the other hand, as R → ∞ the integral along the lower long side of therectangle tends to I, and the integral along the upper long side tends to −e2πiaI (by


50

point 1 above). The integrals along the short sides tend to zero (by points 3 and 4). Weconclude

(1− e2πia)I = −2πieiπa.

Therefore

I =2πi

eiπa − e−iπa=

π

sinπa.

Notice, in the above example, that by the substitution u = et we obtain∫ ∞u=0

ua−1

1 + udu =

π

sinπa,

an important result in the theory of the Γ function.

Example 16.2. Show that

limR→∞

∫ R

0

sinx

xdx =

π

2.

Remark 16.3. It is traditional to write this as∫ ∞0

sinx

xdx =

π

2

where the “improper integral” on the left is simply shorthand for the limit described above.But this may conceal the fact that the function x 7→ (sinx)/x is not Lebesgue integrableover [0,∞). The point is that if a function is Lebesgue integrable, then so is its absolutevalue, and the integral

∫∞0| sinx|/x dx is infinite. This is the same phenomenon as appears

with the harmonic series:∑

1/n is divergent, but the limit limN→∞∑N

n=1(−1)n/n exists.This conditional convergence is responsible for the need for more delicate estimates in thediscussion below.

We are going to integrate the function f(z) = z−1eiz around a cycle Γ which is a “dentedsemicircle”. More precisely, Γ is the sum of four paths:

(1) γ1: the segment of the real axis from ε to R (these being parameters; eventuallyε will go to 0 at R to ∞)

(2) γ2: the “large” semicircle γ2(t) = Reit, t ∈ [0, π](3) γ3: the segment of the real axis from −R to −ε.(4) γ4: the “small” semicircle γ4(t) = Rei(π−t), t ∈ [0, π] (note that this one is traversed

backwards).

Notice that f is in fact holomorphic inside the contour (though it does have a pole at 0— we made the small “dent” to avoid this). We will need estimates of integrals over boththe large and small semicircles. For the large one we make use of

Lemma 16.4 (Jordan’s Lemma). The integral∫ π

0

e−R sin θdθ

tends to zero as R→∞.


51

Proof. The integrand is dominated by 1 and tends to 0 almost everywhere as R→∞, sothis follows from the dominated convergence theorem.

It is traditional to give the following elementary proof instead: on the range θ ∈ [0, π/2]one has sin θ > 2θ/π (by convexity), and thus∫ π

0

e−R sin θdθ = 2

∫ π/2

0

e−R sin θdθ 6 2

∫ π/2

0

e−2Rθ/πdθ 6π

2R

This shows that the integral is in fact of order 1/R.

On the small side we use

Lemma 16.5 (Small arc lemma). Let f(z) be meromorphic near a, with a simple pole ata. Let γε denote the arc segment

γε(t) = a+ εeit, θ0 6 t 6 θ1.

Then as ε→ 0, the integral∫γεf(z)dz tends to i(θ1 − θ0) Res(f, a).

We emphasize that this lemma works only for simple poles. Also, note the consistencywith the residue theorem when θ0 = 0 and θ1 = 2π.

Proof. Because f has a simple pole we can write

f(z) =r

z − a+ h(z)

where r is the residue and h(z) is holomorphic (and hence bounded) near z. Thus∫γε

f(z)dz =

∫γε

r

z − adz +

∫γε

h(z)dz.

The second integral on the right vanishes as ε→ 0, by our standard estimate 8.4, and thefirst can be explicitly computed and yields i(θ1 − θ0)r.

Now, to complete the proof of example 16.2, let I denote the desired integral. ByCauchy’s theorem, ∫

γ1

+

∫γ2

+

∫γ3

+

∫γ4

f(z)dz = 0.

As ε→ 0 and R→∞, the imaginary parts of the first and third terms tend to the desiredintegral I, the second term tends to 0 by Jordan’s lemma, and the fourth term tends to−iπ by the small arc lemma and the easy calculation Res(f ; 0) = 1. Thus we get

2I − π = 0,

whence I = π/2.Here are some further examples – I will do as many as we have time for.∫ ∞

0

sinmx

x(x2 + a2)dx =

π

2a2(1− e−ma), m > 0∫ 2π

0

dx

a+ b cosx=

2π√a2 − b2

, a > b > 0.∫ ∞0

(sinx

x

)3

dx =3π

8.


52

∫ ∞0

sin ax

sinh bxdx =

π

2btanh

aπ

2b, a, b > 0


53

Lecture 17Integrals involving multiple-valued functions

We can expand our repertoire of contour integration techniques by using suitably chosenbranches of “multiple-valued functions” (such as powers and logarithms) as integrands.The basic outline is the same as before: choose a suitable function meromorphic on anopen set Ω, and a suitable contour or family of contours in Ω, and apply the residuetheorem. However, it is now necessary to restrict the open set Ω — typically by making“cuts” — in order that a single-valued branch of the function that we are interestedin should be defined. For example, we have seen in Proposition 6.4 that there is nosingle-valued branch of the function

√z or log z defined on Ω = C \ 0. But there are

single-valued branches of these functions defined on Ω′ = C \ R, where R is a ray from0 to ∞ (for example, the positive real axis). If we want to apply contour integration toan integrand involving log z, we may want to use a domain such as Ω′ rather than Ω. Inturn, this will govern the choice of contour that is available to us.

Lemma 17.1. Let Ω ⊆ C be a connected open set and f : Ω → C \ 0 a non-vanishingholomorphic function. Suppose that for every piecewise smooth loop γ in Ω, the windingnumber wn(f γ; 0) is equal to 0. Then there exists a holomorphic branch of log f on Ω;in other words, a holomorphic function g such that exp(g(z)) = f(z).

Note that the winding number hypothesis is automatic for any f if, for example, Ω isstar-shaped about a point. More generally, it is automatic for any Ω which is homologicallytrivial (every cycle is a boundary), by Cauchy’s theorem 12.10. But for a particular fthe hypothesis may be satisfied even for a non-trivial Ω. Notice also that a branch oflog f automatically gives a branch of the n’th root of f too — or indeed of any (real orcomplex) power of f .

Proof. Fix z0 ∈ Ω and some w0 such that exp(w0) = f(z0). Now define

g(z) = w0 +

∫ z

z0

f ′(w)

f(w)dw,

where the integral is taken along an arbitrary path in Ω from z0 to z. The winding numberhypothesis says exactly that this integral is well-defined (does not depend on the choiceof path). Now arguing as in the proof of 9.3 we find that g is holomorphic and thatg′(z) = f ′(z)/f(z). Using the product and chain rules this implies that

d

dz

(f(z)e−g(z)

)= 0,

whence f(z) is a constant multiple of exp(g(z)) and considering z = z0 shows that theconstant is 1.

I will work these examples at the chalkboard.

Example 17.2. ∫ ∞0

xa−1

1 + xdx =

π

sin aπ, 0 < a < 1.


54

We have encountered this integral before in Example 16.1. There, we made a substitutionwhich avoided the multiple-valued function za−1. This time, we will tackle the multi-function directly.

Outline of argument: cut the plane along the positive real axis, and define a branchof f(z) = za−1/(1 + z) in the cut plane. Integrate around a “keyhole” contour (a largeand a small circle, and two straight segments just above and just below the cut). Thetwo straight segments give integrals related to the desired one, the large and small circlesvanish in the limit, and there is a simple pole at z = −1.

Note that the substitution z = ew transforms not merely the integral, but the contourand our entire argument, into Example 16.1.

Example 17.3. ∫ ∞0

log x

1 + x4dx = − π2

8√

2.

Outline of argument: cut the plane along the positive real axis and define a branch off(z) = (log z)/(1 + z4) in the cut plane. Integrate this function around a semi-circularcontour with bottom on the real axis. Over the positive part of the real axis we get theintegral I we want; over the negative part we get

I + iπ

∫ ∞0

dx

1 + x4;

the integral over the semi-circle vanishes in the limit. There are two poles inside thecontour, at eiπ/4 = (1 + i)/

√2 and e3iπ/4 = (1− i)/

√2, and we can compute their residues

and apply the residue theorem. Consider real parts to evaluate I. As a bonus, we findthe value of

∫∞0dx/(1 + x4) by considering imaginary parts.

A tricky alternative contour is to use a quarter -circle with sides on the real and imagi-nary axes. This works because z4 = (iz)4 in this case.

Example 17.4. ∫ 1

0

dx

(a− x)√x(1− x)

=π√

a(a− 1), (a > 1).

Outline of argument: cut the plane between 0 and 1 and define a single-valued branchof (z − a)

√z(1− z) on this cut plane, by an ingenious application of Lemma 17.1. We

apply the residue theorem to the cycle Γ = [γ1]− [γ2], where γ1 is a ‘dumb-bell’ contourencircling the cut and γ2 is a circle of large radius. There is one singularity enclosed by Γ,the pole at a. On the one hand,

∫γ2

tends to 0 as the radius tends to∞; on the other,∫γ1

approaches −2 times the integral we want, because the sign change across the cut meansthat the integrals along the long sides of the dumb-bell are equal.


55

Lecture 18The gamma and zeta functions

See Stein and Shakarchi 6.1,6.2We will now develop the further properties of the Γ function. Recall that we defined

this function by

Γ(z) =

∫ ∞0

tz−1e−tdt

when Re(z) > 0. The iteration formula Γ(z + 1) = zΓ(z) allows us to find an analyticcontinuation of Γ(z) to C \ 0,−1,−2, . . ., and this continuation has simple poles at thenon-positive integers.

The next result is called the reflection formula for the Γ function.

Theorem 18.1. We have Γ(z)Γ(1− z) = π/ sin(πz).

Proof. By analytic continuation it is enough to check this for z = a, 0 < a < 1. Write

Γ(1− a)Γ(a) =

∫ ∞s=0

∫ ∞t=0

s−ata−1e−(s+t)dtds.

In the t-integral substitute t = us to obtain∫ ∞s=0

∫ ∞u=0

ua−1e−s(1+u)duds.

Reverse the order of integration (Fubini’s theorem!) and perform the inner s-integral toobtain ∫ ∞

u=0

ua−1

1 + udu.

The substitution u = et gives ∫ ∞−∞

eat dt

1 + et=

π

sin πa,

as we worked out in Example 16.1.

Notice in particular the consequence Γ(12) =√π. A simple substitution gives

Γ(12) =

∫ ∞0

t−1/2e−tdt = 2

∫ ∞0

e−u2

du,

so we get the standard result ∫ ∞−∞

e−u2

du =√π.

Exercise 18.2. Prove the (Fourier integral) formula∫ ai+∞

ai−∞e−πuz

2+2πiwz dz = u−1/2e−πw2/u

for u real and positive, and w ∈ C.


56

Definition 18.3. Euler’s constant γ is the limit

γ = limn→∞

( n∑k=1

k−1 − log n).

Since 1n6 log(n) − log(n − 1) 6 1

n−1one easily checks that the expression under the

limit is decreasing in n and bounded below, so convergent. The numerical value of Euler’sconstant is approximately 0 · 577215665.

Proposition 18.4. The Γ function Γ(z) satisfies

1

Γ(z)= zeγz

∞∏n=1

[(1 +

z

n

)e−z/n

].

The infinite product converges to an entire function on C. In particular, Γ(z) nevervanishes.

Proof. Because of the uniqueness of analytic continuation it suffices to check the resultfor z real and positive. From homework we have the limit formula

1

Γ(z)= lim

n→∞

z(z + 1) · · · (z + n)

n!nz.

Rewrite the expression under the limit as

ze−z logn

n∏k=1

(1 +

z

k

)= zez(

∑nk=1

1k−logn)

n∏k=1

[(1 +

z

k

)e−z/k

].

Letting n→∞ we obtain the result.

We have already defined the Riemann zeta function

ζ(s) =∞∑n=1

1

ns.

The sum on the right hand side converges uniformly on every half-plane Re s > 1 + ε,where ε > 0, by the Weierstrass M-test. By Corollary 11.5, the limit ζ(s) is a holomorphicfunction of s on the half-plane Re s > 1.

The evaluation ζ(2) = π2/6 is due to Euler. Our results on the Γ function can be usedto give a quick evaluation of ζ at every even positive integer.

Lemma 18.5. We have the infinite product expansion

sin πz = πz∞∏n=1

(1− z2

n2

)for the sine function. Moreover, we have the series expansion

π cot πz =1

z+∞∑n=1

2z

z2 − n2= lim

n→∞

n∑k=−n

1

z − k

for the cotangent.


57

Proof. Write the reflection formula for the Γ function as

sin πz = −π/(zΓ(z)Γ(−z)).

Now use the infinite product 18.4 for 1/Γ(z) to obtain the result. The second resultfollows from the first by logarithmic differentiation.

Recall that the Bernoulli numbers are defined by z/(ez − 1) =∑Bnz

n/n!; we haveB2 = 1

6, B4 = − 1

30, B6 = 1

42and so on (the odd ones vanish after B1). Then we have

Proposition 18.6. For each even integer 2m,

ζ(2m) =∞∑n=1

1

n2m=

(−1)m+122m−1B2mπ2m

(2m)!.

Proof. Write

z cot z = izeiz + e−iz

eiz − e−iz= iz +

2iz

e2iz − 1= 1−

∞∑m=1

(−1)m+1B2m22mz2m

(2m)!.

On the other hand, using our infinite sum for the cotangent,

z cot z = 1 + 2∞∑n=1

z2

z2 − n2π2.

Expand the summand in a binomial series and rearrange as a power series in z (justifiedby absolute convergence). The coefficient of z2m is

−∞∑n=1

2

π2mn2m.

Comparing coefficients with those appearing in the previous expansion of z cot z (justifiedby uniqueness of Taylor coefficients) we get the result.

We will now give Riemann’s amazing proof that the function ζ(s) has an analyticcontinuation and that it exhibits a certain “symmetry” about the critical line Re s = 1

2.

This proof depends on the properties of another special function, the theta function defined(for real, positive u) by

ϑ(u) =∞∑

n=−∞

e−πn2u.

The relationship between the ϑ and ζ functions is the following.

Lemma 18.7. For Re s > 1 we have

π−s/2Γ(s/2)ζ(s) =

∫ ∞0

us/2−1

(ϑ(u)− 1

2

)du

where the integral exists in the Lebesgue sense.

The expression on the left of this equality occurs frequently and is denoted ξ(s). Thus,by definition, ξ(s) = π−s/2Γ(s/2)ζ(s).


58

Proof. Assume first that s is real and > 1. A simple substitution gives∫ ∞0

us/2−1 e−πn2udu = π−s/2Γ(s/2)n−s.

Summing over n, and using the monotone convergence theorem, we obtain the result ofthe lemma in this case (observe that (ϑ(u)− 1)/2 =

∑∞n=1 e

−πn2u.) The result for general(complex) s with Re s > 1 now follows, since we know that |us/2−1| = uRe(s/2)−1.

Lemma 18.8. The theta function satisfies the functional equation

ϑ(1/u) = u1/2ϑ(u).

Proof. We will employ contour integration. Consider the meromorphic function

f(z) =e−πuz

2

e2πiz − 1.

This function has simple poles at each integer n, and the residue at the n’th pole is(2πi)−1e−πun

2by Lemma 15.4. Now let ΓL be the rectangular contour with vertices at

±L± i, where L is a half-integer. From the residue theorem we deduce

ϑ(u) = limL→∞

∫ΓL

f(z)dz.

Simple estimates show that the integrals along the two “short edges” tend to zero asL→∞ through half-integers, so we are left with the long sides, or to abbreviate

ϑ(u) =

∫ −i+∞−i−∞

f(z)dz +

∫ i−∞

i+∞f(z)dz.

Consider the second of these integrals for a moment. When Im z = i, |e2πiz| < 1 and sowe may expand f in a convergent series

f(z) = −e−πuz2∞∑k=0

e2πikz.

Interchanging summation and integration,∫ i−∞

i+∞f(z)dz =

∞∑k=0

∫ i+∞

i−∞e−πuz

2+2πikz dz =∞∑k=0

u−1/2e−πk2/u

by Exercise 18.2. Similar arguments show that∫ −i+∞−i−∞

f(z)dz =−1∑

k=−∞

u−1/2e−πk2/u

and adding these two equations gives the result.

Proposition 18.9. The Riemann ξ function is given by the integral formula

ξ(s) =1

s(s− 1)+

∫ ∞1

(u−(s+1)/2 + u(s−2)/2

)(ϑ(u)− 1

2

)du.


59

Corollary 18.10. The ξ function extends meromorphically to the entire plane, with itsonly singularities being simple poles at s = 0, 1. Moreover, it satisfies the reflectionformula (a.k.a. functional equation)

ξ(s) = ξ(1− s).

Proof. Start with the formula

ξ(s) =

∫ ∞0

us/2−1

(ϑ(u)− 1

2

)du

of Lemma 18.7. Split the range of integration into [0, 1] and [1,∞), and in the first rangemake the substitution u = 1/v. This gives

ξ(s) =

∫ ∞1

us/2−1

(ϑ(u)− 1

2

)du+

∫ ∞1

v−s/2−1

(ϑ(1/v)− 1

2

)dv.

Substitute the functional equation for ϑ in the second integral and collect terms to obtain∫ ∞1

(u−(s+1)/2 + u(s−2)/2

)(ϑ(u)− 1

2

)du+

1

2

∫ ∞1

(u−s/2−1 + u−s/2−1/2

)du

which gives the result.

Exercise 18.11. Using the functional equation and the Euler product formula, show thatthe only zeroes of ζ(s) outside the critical strip 0 6 Re s 6 1 are at the negative evenintegers. (These are the so-called trivial zeroes of ζ.)


60

Lecture 19Holomorphic equivalence

See Stein and Shakarchi 8.1

Definition 19.1. Let U and V be open subsets of C and let f : U → V be a holomorphicmap. If there is a holomorphic g : V → U such that f g(v) = v and g f(u) = u for allu ∈ U , v ∈ V , then f is called a holomorphic equivalence. (Alternative terminologies arebiholomorphic equivalence, biholomorphism, and (traditionally) conformal equivalence.)The two sets U and V that are related in this way are called holomorphically equivalent(or conformally equivalent).

Our project in the next part of the course will be to understand the possible holo-morphic equivalences between various kinds of open subsets of C (and, later, betweenmore general Riemann surfaces). This is similar to the project in algebra of trying tounderstand groups “up to isomorphism”. Notice that it is possible to have a non-trivialholomorphic equivalence from an open set U to itself (just as a group can have non-trivial automorphisms). For instance, we proved in an early homework assignment thatthe Mobius transformations

z 7→ z − a1− az

, |a| < 1,

are holomorphic equivalences from the unit disc to itself.

Proposition 19.2. Any holomorphic bijection f : U → V is a holomorphic equivalence(i.e., the inverse function is automatically holomorphic); in particular it is a homeomor-phism (the inverse function is continuous). Moreover, f ′(a) 6= 0 for all a ∈ U .

Proof. We prove the second statement first. Suppose that f ′(a) = 0 for some a ∈ U andlet b = f(a). Choose ε > 0 so small that the disc D(a; ε) is included in U and does notcontain any zeroes of f ′ or of f − b except a itself. Let γ be a circular path around a ofradius ε/2. Then the integral

1

2πi

∮γ

f ′(z)

f(z)− bdz,

which counts the number of times f takes the value b inside γ, is at least 2. By continuity,for c sufficiently close to b, the integral

1

2πi

∮γ

f ′(z)

f(z)− cdz,

which counts the number of times f takes the value c inside γ, is greater than 1. No rootof f − c inside γ can be multiple (because there are no zeroes of f ′ inside γ except at a)so there must be at least two distinct solutions z1, z2 of f(z) = c. This contradicts theinjectivity of f .

Now that we know that f ′(a) 6= 0, let us prove that g is holomorphic at b. From thedefinition of differentiability it follows that, when z − a is sufficiently small,

|f(z)− b| > 12|f ′(a)||z − a|.


61

Thus (putting f(z) = w) we have that w → b if and only if g(w)→ a. Now write

f ′(a) = limz→a

f(z)− f(a)

z − a= lim

w→b

w − bg(w)− g(b)

,

which shows that the limit limw→b(g(w) − g(b))/(w − b) defining g′(b) exists and equals1/f ′(a).

Corollary 19.3. (The inverse function theorem) Let f : Ω → C be any holomorphicfunction and let a ∈ Ω be a point such that f ′(a) 6= 0. Then there exists a neighborhood Uof a in Ω such that the restriction of f to U is a holomorphic equivalence U → V = f(U).

Proof. The same argument as in the first part of the proof above shows that there is aneighborhood U of a on which f takes each of its values exactly once, i.e., f is a bijectionfrom U onto V = f(U). By the open mapping theorem (Proposition 14.11), V is an openset. Now Proposition 19.2 shows that f gives a holomorphic equivalence from U to V .

Let U be an open subset of C. A holomorphic equivalence from U to itself is called aholomorphic automorphism of U . Clearly, the holomorphic automorphisms of U form agroup (the automorphism group of U). If U and V are holomorphically equivalent, theirautomorphism groups are isomorphic (why?).

Proposition 19.4. The only holomorphic automorphisms of C are the affine maps z 7→az + b, a ∈ C \ 0, b ∈ C.

Proof. Let f : C → C be a holomorphic equivalence. Then f is an entire function, so ithas a Taylor expansion

f(z) =∞∑n=0

cnzn.

On the other hand consider the function g(w) = f(1/w). This function has a singularityat w = 0. But since f is a homeomorphism C → C, f(z) → ∞ as z → ∞; that is,g(w) → ∞ as w → 0. Thus the singularity of g is at worst a pole, by the Casorati-Weierstrass theorem (Proposition 14.5), and we have

f(1/w) = g(w) =∞∑

m=−M

dmwm, or f(z) =

N∑n=−∞

d−nzn.

The expansions must agree, since they are both valid in the open set C \ 0, so cn = dnfor all n and f is a polynomial. Since it is injective, it must be of degree 1.

Proposition 19.5. The complex plane C and the open unit disc U are not holomorphicallyequivalent.

Proof. Any holomorphic equivalence C → U would be a bounded entire function, henceconstant by Liouville’s theorem.

The Riemann mapping theorem states that any simply-connected open subset of C,except for C itself, is holomorphically equivalent to U. We will prove this deep result overthe next few lectures but first we will get to grips with some examples.


62

Lecture 20Examples of conformal equivalences

See Stein and Shakarchi 8.1In this lecture we shall discuss a cook-book of techniques for constructing explicit confor-

mal equivalences from various simple regions of the plane to the unit disk U. (Rememberthat, according to the Riemann mapping theorem, any simply connected open subset ofC, apart from C itself, is conformally equivalent to U.) Our regions will be bounded by asmall number of circlines. This ugly word refers to either a circle or a straight line in thecomplex plane.

Remark 20.1. It is convenient for this discussion to introduce an “ideal” point ∞ and tosay that a straight line is a “circline passing through∞”. For now you can just treat thisas a way of speaking, but we will give it more concrete meaning in a few lectures whenwe discuss the Riemann sphere.

Definition 20.2. A Mobius transformation (also called a fractional linear transforma-tion) is a map C ∪ ∞ → C ∪ ∞ defined by

z 7→ az + b

cz + d

where ad− bc 6= 0.

You can think of Mobius transformations as associated to matrices ( a bc d ) with nonzerodeterminant. Then it is easy to check that the composition of Mobius transformationscorresponds to multiplication of matrices.

Proposition 20.3. Mobius transformations preserve circlines: the image of a circlineunder a Mobius transformation is another circline.

Proof. There are several ways to prove this; here is one. The equation of any circline mustbe of the form

α|z|2 + βz + γz + δ = 0

for constants α, β, γ, δ with α, δ real, γ = β, and αδ − βγ 6 0. (This is an exercise inCartesian coordinate geometry.) Let us call this the circline corresponding to the matrixM = ( α β

γ δ ); the conditions on the coefficients translate to M = M∗ and det(M) 6 0,where the star denotes conjugate transpose. The equation of the circline is ZMZ∗ = 0,where Z =

(z 1

).

Suppose now that w is the image of z under a Mobius transformation, say

w =az + b

cz + d, z =

dw − b−cw + a

.

Substitute this expression for z in the equation of the circline to get (WT )M(WT )∗ = 0,where W =

(w 1

)and T = ( d −c

−b a ). This equation can be written

W (TMT ∗)W ∗ = 0

so it is also the equation of a circline, corresponding to the matrix TMT ∗ (we note thatthis matrix also is self-adjoint and has negative determinant, as expected).


63

We start our review of conformal equivalences by considering regions bounded by asingle circline.

Example 20.4. The interior of a disk. This can always be moved onto the unit disk bya suitable affine transformation z 7→ az + b. For some purposes it may be useful to use amore general Mobius transformation. Recall that points A and B are inverse relative toa circle C of center P and radius r if P,A,B are collinear and d(P,A).d(P,B) = r2. If thecomplex numbers a and b represent inverse points with respect to the circle C then theequation |z − a| = λ|z − b|, where λ is the positive constant such that |a− p| = λ|b− p|,represents C. (This is easy to prove using elementary coordinate geometry; write z = x+iyand expand the equation |z − a|2 = λ2|z − b|2.) Thus the Mobius transformation

z 7→ z − az − b

send the given circle to the circle center the origin and radius λ.

Example 20.5. A half-plane can be mapped to the unit disk by finding a pair of inversepoints as in the example above. In this case, λ = 1, and the points a and b are inversewith respect to a line L if a is the reflection of b in L.

Now we consider regions bounded by two circlines. There are three cases to consider:the circlines do not meet at all; they meet in two points; or they are tangent at a singlepoint.

Example 20.6. First consider the case of a region bounded by two nonintersecting cir-clines. Any two nonintersecting circles in the plane, one of which is entirely contained inthe other, can be mapped to two concentric circles by a Mobius transformation. One needonly find a common pair a, b of inverse points on the line joining the centers of the twocircles, which necessitates solving a quadratic equation. Then the Mobius transformationz 7→ (z − a)/(z − b) maps the region between the two circles to an annulus (the regionbetween two concentric circles). One can argue similarly for the region bounded by onecircle and a nonintersecting straight line.

Remark 20.7. There is an elegant application to Steiner’s porism, see http://www.maths.gla.

ac.uk/~wws/cabripages/inversive/steiner.html for details and a Java animation.

Now we consider the case where the circlines meet in two points.

Example 20.8. Consider the wedge-shaped region. z = reiθ : r > 0, 0 < θ < α, where0 < α 6 2π. (This is the region bounded by two circlines that meet at the points 0 and∞ on the Riemann sphere.) The mapping

z 7→ zπ/α

(which may be a multiple-valued function; if so, define it on the plane cut along thepositive real axis) will take this wedge to a half-plane, which can then be mapped to thedisk by a Mobius transformation.

Example 20.9. Any region bounded by two circlines can be mapped to a wedge-shapedregion as above by a Mobius transformation. Indeed, the two circlines will meet at pointsa and b; the transformation

z 7→ z − az − b


http://www.maths.gla.ac.uk/~wws/cabripages/inversive/steiner.html

http://www.maths.gla.ac.uk/~wws/cabripages/inversive/steiner.html

64

will send one point of intersection to 0 and the other to ∞, so the enclosed region will bemapped to the wedge between two rays from 0 to ∞.

Finally we consider the special case where the two circlines are tangent at a single point.

Example 20.10. A strip such as z : 0 < Im z < c is bounded by two circlines that aretangent at ∞. This can be mapped to a half-plane by the complex exponential function.In this case the required functions is z 7→ eπz/c. A region bounded by two circles tangentat a finite point p can be mapped to a strip by any Mobius transformation that sends thepoint p to ∞; for example, z 7→ 1/(z − p).

Example 20.11. Some half-strips can be handled by the same technique. For instance,consider z : 0 < Im z < c,Re z > 0. The exponential eπz/c maps this to the setw : Imw > 0, |w| > 1. This is a region bounded by two circlines and the previoustechniques apply.

The next stage in complexity is to consider the Schwarz-Christoffel transformationswhich map the upper half-plane conformally onto a polygon. If the polygon has exteriorangles α1, α2, . . . , αn then a Schwarz-Christoffel transformation is of the form∫ z

z0

dw

(w − x1)α1/π · · · (w − xn)αn/π

where x1, . . . , xn are points on the real axis (which will map to the vertices of the polygon).These transformations are of real practical use in solving potential theory problems —whole books have been written about them — but we shall not consider any problemsrequiring them in this course.


65

Lecture 21The geometry of the unit disc

In this lecture we will study the automorphism group of the unit disc U and the ge-ometry associated with it. The rotations z 7→ eiϕz, ϕ ∈ R, are obvious examples ofautomorphisms of U. Further examples are supplied by the next lemma (which we al-ready worked out as a homework problem)

Lemma 21.1. Let a ∈ U. Then the Mobius transformation

z 7→ z − a1− az

is an automorphism of U and maps a to the origin.

Proof. Let Ta(z) = (z − a)/(1 − az). It is easy to check that the inverse of the trans-formation Ta is T−a. Thus we need to prove that Ta maps U to itself. To see this, notethat

|1− az|2 − |z − a|2 = 1 + |a|2|z|2 − |a|2 − |z|2 = (1− |z|2)(1− |a|2) > 0.

So |1− az| > |z − a|, which gives |Ta(z)| < 1.

Corollary 21.2. The group Aut(U) is transitive.

By definition, Aut(S) is transitive if for any two points p, q ∈ S there is an automor-phism sending p to q.

Theorem 21.3 (Schwarz’ Lemma). Let f : U→ U be a holomorphic map (not necessarilybijective) and suppose that f(0) = 0. Then |f(z)| 6 |z| for all z, and |f ′(0)| 6 1.Moreover, equality occurs5 only if f(z) = eiϕ(z) for some real ϕ.

Proof. Let g(z) = f(z)/z; since f has a zero at the origin, g extends to a holomorphicfunction on U. For each r < 1, g is holomorphic on D(0; r) and

|g(z)| 6 1/r for z ∈ ∂D(0; r).

By the maximum principle (Theorem 11.2), |g(z)| 6 1/r for all z ∈ D(0; r). Let r 1to find |g(z)| 6 1 which gives the inequality. If equality occurs, then |g| attains itsmaximum value 1 at some interior point of U, so g is a constant (by the maximumprinciple again).

Proposition 21.4. Every automorphism of U is of the form

z 7→ eiϕTa(z)

where Ta is defined as above.

Proof. Let f ∈ Aut(U) and suppose that f−1(0) = a. Then g = f (Ta)−1 is an automor-

phism mapping 0 to 0. By Schwarz’ Lemma the only such automorphisms are of the formw 7→ eiϕw (proof: |g(z)| 6 |z| by Schwarz’ lemma, and |z| 6 |g(z)| by Schwarz’ lemmaapplied to g−1. Thus we have equality in Schwarz’ lemma and the result follows). Thusf (Ta)

−1(w) = eiϕw. Put z = Ta(w); we get

f(z) = eiϕTa(z)

5I mean, |f(z)| = |z| for some nonzero z, or |f ′(z)| = 1.


66

as required.

Remark 21.5. The automorphisms of U are therefore those Mobius transformations of theform

z 7→ λ+ µz

µ+ λz.

These are exactly the Mobius transformations that map U to itself. The group of suchMobius transformations is denoted PSU(1, 1).

Exercise 21.6. Show that the automorphism group of the upper half-plane is PSL(2,R),the group of Mobius transformations of the form z 7→ (az + b)/(cz + d) with a, b, c, d ∈ Rand ad− bc > 0.

It is a consequence of our discussion that if a, b ∈ U the quantity∣∣∣∣ a− b1− ab

∣∣∣∣is preserved by every automorphism of U. For it is equal to |ϕ(b)| where ϕ is any auto-morphism that maps a to 0. We can think of this quantity as a sort of ‘distance’ from ato b.

Definition 21.7. If a, b ∈ D then the hyperbolic distance between them is

d(a, b) = 2 tanh−1

∣∣∣∣ a− b1− ab

∣∣∣∣ .By construction, the hyperbolic distance is invariant, that is, d(f(a), f(b)) = d(a, b) for

any automorphism f .

Lemma 21.8. If a, 0, and b are collinear (meaning that a is a positive real multiple of−b) then

d(a, 0) + d(0, b) = d(a, b).

Proof. Use the formula tanh(x+ y) = tanh(x)+tanh(y)1+tanh(x) tanh(y)

.

We are going to show that the hyperbolic distance is a metric, i.e. satisfies the triangleinequality. To see this we need to study some hyperbolic trigonometry.

Definition 21.9. A hyperbolic line in U is a circline that meets the unit circle at rightangles.

Thus the diameters of U are hyperbolic lines; indeed, they are the only hyperbolic linesthrough the origin. Since the automorphisms of U are Mobius transformations (henceconformal) and map the unit circle to itself, they map hyperbolic lines to hyperboliclines.

Lemma 21.10. There is exactly one hyperbolic line through two given points of U.

Proof. Because of transitivity we may assume that one of the points is 0, and then theresult is obvious.


67

We define the (hyperbolic) angle between two hyperbolic lines to be the ordinary (Eu-clidean) angle between them. Note that these angles are preserved by automorphisms(since the automorphisms are conformal transformations).

Now consider a hyperbolic triangle ABC consisting of three points A,B,C ∈ D and thehyperbolic line segments between them. Let the sides have hyperbolic length AB, BC,CA and the angles at the vertices be γ, α, β.

Proposition 21.11 (Hyperbolic cosine law). In the above situation

coshAB = coshAC coshBC − sinhAC sinhBC cos γ.

Proof. Because of transitivity there is no loss of generality in assuming that C is theorigin. Let A, B be represented by the complex numbers a, b respectively.

Remember some more trigonometric identities: if t = tanhx/2 then

cosh(x) =1 + t2

1− t2, sinh(x) =

2t

1− t2.

Thus

cosh(AB) =1 +

∣∣∣ a−b1−ba

∣∣∣21−

∣∣∣ a−b1−ba

∣∣∣2=|1− ba|2 + |a− b|2

|1− ba|2 − |a− b|2

=(1 + |a|2)(1 + |b|2)− 2(ab+ ab)

(1− |a|2)(1− |b|2).

But ab + ab = 2|a||b| cos γ (see exercise 1.7). Substituting into the expression above weobtain

coshAC coshBC − sinhAC sinhBC cos γ

as required.

Corollary 21.12. If A,B,C are three points in U we have

d(A,C) 6 d(A,B) + d(B,C)

with strict inequality unless A,B,C lie on the same hyperbolic line (with B between Aand C).

Proof. This follows from the cosine law and the formula cosh(x+ y) = cosh(x) cosh(y) +sinh(x) sinh(y).

Proposition 21.13 (Hyperbolic sine law). Continuing with the same notation,

sinα

sinh a=

sin β

sinh b=

sin γ

sinh c.


68

Proof. Consider first the special case of a triangle with a right angle at C. Write d(A,B) =c, d(B,C) = a, d(C,A) = b.

We have

cosh c = cosh a cosh b

cosh b = cosh a cosh c− sinh a sinh c cos β

Rewrite the second equation to give

sin2 β = 1− cos2 β =sinh2 a sinh2 c− (cosh b− cosh a cosh c)2

sinh2 a sinh2 c.

Substitute cosh a = cosh c/ cosh b from the first equation and sinh2 a = cosh2 a− 1 (stan-dard identity). After canceling a common factor of cosh2 c− cosh2 b we get

cosh2 b− cosh2 c+ sinh2 c

sinh2 c=

sinh2 b

sinh2 c

so sin β = sinh b/ sinh c. This is the sine rule for a right triangle. The general case followsby dropping a perpendicular from a vertex of the triangle to the opposite side, and thusdividing it into two right triangles (exactly as in Euclidean geometry).

The Schwarz lemma has a beautiful invariant form relative to the hyperbolic metric.

Proposition 21.14 (Ahlfors). Any holomorphic map f : U→ U decreases all hyperbolicdistances: the hyperbolic distance from f(a) to f(b) is no greater than the hyperbolicdistance from a to b.

Proof. If a = f(a) = 0 this is the ordinary Schwarz lemma 21.3. The general case can bereduced to this one by replacing f by

g = Tf(a) f (Ta)−1

which is a holomorphic map sending 0 to 0. Thus

|g(Ta(b))| 6 |Ta(b)|.But 2 tanh−1 |Ta(b)| is the hyperbolic distance from a to b, and (by construction) 2 tanh−1 |g(Ta(b))| =2 tanh−1 |Tf(a)(f(b))| is the hyperbolic distance from f(a) to f(b).

Corollary 21.15. Let F : U→ U be any holomorphic map. Then |F ′(0)| 6 1, with strictinequality unless F is a rotation.

This is a corollary because, by the invariant form of the Schwarz lemma, the hyperbolicdistance between 0 and dz is greater than the hyperbolic distance between a = f(0) andf(0 + dz). But the hyperbolic distance between these two points is in turn at least aslarge as the euclidean distance. Rather than try to make this precise (which would leadus in the direction of Riemannian geometry) we write out an independent proof based onthe same idea.

Proof. Let F (0) = a. Then G = Ta F is a holomorphic map U → U and G(0) = 0. Bythe Schwarz lemma, |G′(0)| 6 1. But G′(0) = T ′a(a)F ′(0) by the chain rule. Calculationshows that

T ′a(a) =1

1− |a|2


69

so |F ′(0)| = (1− |a|2)G′(0) 6 1, with equality only if a = 0 and F = G is a rotation.

Remark 21.16. One can show that hyperbolic geometry obeys all the axioms of Euclideangeometry except for the parallel axiom. In hyperbolic geometry there are many linesthrough a given point that do not meet a given line.


70

Lecture 22The Arzela-Ascoli theorem

See Stein and Shakarchi 8.3.2We are about to embark on the proof of the Riemann mapping theorem, which requires

constructing a holomorphic function with certain specified properties. As is often thecase, the desired function will be constructed using a limit process. To show that thelimit exists, we need certain compactness properties of sets of (holomorphic) functions.This lecture describes a result in general functional analysis which can be used to provesuch compactness properties in function spaces.

Recall that a metric space X is compact if every sequence in X has a convergentsubsequence (or, equivalently, if every open cover of X has a finite subcover). Closed,bounded subsets of Euclidean space are compact. However, we will need to look atcompactness properties in spaces of functions. These spaces are much “larger” thanEuclidean space and, in general, closedness and boundedness do not imply compactnessin these spaces.

Definition 22.1. Let X be a compact metric space and let Y be any metric space. Thespace C(X;Y ) of continuous functions from X to Y is itself a metric space when equippedwith the supremum distance

d(f, g) = supd(f(x), g(x)) : x ∈ X.(Notice that the compactness of X implies that this supremum is defined.) If Y = C we’llomit it from the notation and write C(X) for the space of continuous, complex-valuedfunctions on X.

Remark 22.2. A metric space is complete if every Cauchy sequence in that space is con-vergent. It is known that, if Y is complete, then C(X;Y ) is complete also. In particular,the space C(X) of complex -valued functions on X is complete.

Example 22.3. Consider the metric space C[0, 2π]. The functions

fn(t) := sin(2nt), n = 1, 2, . . .

belong to the closed unit ball of C[0, 2π] and d(fn, fm) > 1 if n 6= m. Thus the sequence(fn) has no convergent subsequence, so the closed unit ball in C[0, 2π] is not compact

Notice that the fn oscillate more and more rapidly. We are going to prove a theoremwhich says that the only way compactness can fail for a closed bounded subset of C(X)is through this ‘rapid oscillation’.

Definition 22.4. Let F be a collection of functions from a metric space X to a metricspace Y . Say F is equicontinuous if for every x ∈ X and ε > 0 there is δ > 0 such that

∀f ∈ F , d(x, x′) < δ ⇒ d(f(x), f(x′)) < ε.

(The “equi” part of equicontinuity comes from the requirement that the same δ shouldwork for all the functions in the set F . )

Example 22.5. The subset of C[0, 1] consisting of all functions f such that |f ′(t)| 6 1for all t ∈ [0, 1] is an equicontinuous set.


71

Theorem 22.6. (Arzela-Ascoli) Let X be compact metric. A subset of C(X) is compactif and only if it is closed, bounded and equicontinuous.

Proof. Let F ⊆ C(X) be a bounded, equicontinuous set. It suffices to show that everysequence (fn) of members of F has a subsequence that converges in the metric of C(X)— that is, converges uniformly.

The key to the proof is the following Claim: Let gm be any bounded, equicontin-uous sequence in C(X), and let η > 0. Then gm has a subsequence gmk

such thatd(gmk

, gmk′) < η for all k, k′ (an “η-close subsequence”).

First let’s see why this claim implies the Arzela-Ascoli-theorem, and then we will provethe claim. Granting the claim for a moment, the sequence (fn) has a 1-close subsequence.

We’ll denote this subsequence by (f(0)n ). But now (f

(0)n ) is an equicontinuous sequence

too, so it has a 12-close subsequence which we denote f

(1)n . Proceeding in this way we

obtain a sequence of subsequences f(r)n of the original sequence, with the properties that

(i) (f(r)n ) is 2−r-close, and

(ii) (f(r)n ) is a subsequence of (f

(r−1)n ).

Now let gn = f(n)n . This is also a subsequence of the original sequence and it has the

property that d(gn, gm) 6 2−minm,n. Thus (gn) is a Cauchy sequence, hence convergentby the completeness of C(X).

Now let us prove the claim. Given η > 0, for each x ∈ X there exists by equicontinuitya δx > 0 such that

d(x, x′) < δx =⇒ d(gm(x), gm(x′))) < η/3 ∀m = 1, 2, . . . .

By compactness the metric space X has a finite cover by balls B(xj; δj) of radius δj = δxj ,where j = 1, . . . , J . Consider the sequence of vectors

(gm(x1), . . . , gm(xJ)) ∈ CJ , m = 1, 2, . . . .

This is a bounded sequence in a Euclidean space so it has a convergent subsequence. Inparticular, there is a sequence mk such that |gmk

(xj) − gmk′(xj)| < η/3 for all k, k′. But

by construction any point x ∈ X lies within distance δj of some xj, so for all x

|gmk(x)− gmk′

(x)| 6|gmk

(x)− gmk(xj)|+ |gmk

(xj)− gmk′(xj)|+ |gmk′

(xj)− gmk′(x)|

6 η/3 + η/3 + η/3 = η

as asserted in the claim.

Remark 22.7. We’ve only proved the “if” assertion in the Arzela-Ascoli theorem. The“only if” assertion is also true, but it is less often used (and it is also much easier), so I’llleave that as an exercise.

The usual application to complex analysis is through Montel’s lemma, which is next.Let Ω be an open subset of C and let F be a family of holomorphic functions on Ω.

Definition 22.8. We say that F is a normal family if, from any sequence fn of membersof F , one can extract a subsequence that converges uniformly on compact subsets. (Afancy way of saying this is “F is precompact in the topology of uniform convergence oncompact subsets”.)


72

Lemma 22.9 (Montel). A uniformly bounded family is normal.

Proof. Let |f | 6 M for all f ∈ F . By a diagonalization argument similar to that usedabove, it suffices to show that for some fixed compact subset K ⊆ Ω, the restriction ofF to C(K) forms a precompact subset of C(K). But, given K, there is r > 0 such thatthe closed disk D(z; r) is contained in Ω for all z ∈ K. Now if f ∈ F and z, z′ ∈ K with|z − z′| < r/2,

f(z)− f(z′) =1

2πi

∮∂D(z;r)

z − z′

(w − z)(w − z′)f(w)dw

by Cauchy’s integral formula. Remembering that |f | 6M , this gives

|f(z)− f(z′)| 6 (2M/r)|z − z′|for every f ∈ F . Thus the restrictions of the functions f to K form an equicontinuoussubset of C(K), which is then precompact by the Arzela-Ascoli theorem.


73

Lecture 23The Riemann mapping theorem

See Stein and Shakarchi 8.3There are many examples of open subsets of C that are conformally (a.k.a. biholomor-

phically) equivalent to U. For example, the Mobius transformation

z 7→ z − iz + i

is a conformal equivalence from the upper half plane z : Im(z) > 0 to U. Such aconformal equivalence Ω→ U gives a 1 : 1 correspondence between holomorphic functionson Ω and holomorphic functions on U.

Any open subset Ω of C that is biholomorphically equivalent to U must be homeomor-phic to C and therefore must have trivial homology (it is connected and every cycle is aboundary). Conversely

Theorem 23.1 (Riemann Mapping Theorem). Any proper, open, homologically trivialsubset of C is biholomorphically equivalent to U.

Remark 23.2. Artin’s criterion 12.11 provides an easy way to recognize homologicallytrivial subsets of C. It tells us that if C \ Ω is unbounded and connected, then Ω ishomologically trivial. (Exercise: prove the converse implication.)

We are going to prove this following an argument of Koebe. This proof depends onlyon the following property of homologically trivial regions.

Lemma 23.3. Let Ω be a homologically trivial open subset of C. If f : Ω → C is holo-morphic and nowhere zero, then there are holomorphic branches of log f and

√f de-

fined on Ω. In other words, there are holomorphic functions g, h : Ω → C such thatg(z)2 = eh(z) = f(z) for all z ∈ Ω .

Proof. Clearly it suffices to define h, the branch of log f , and then put g = eh/2. To defineh, choose a base point z0 ∈ Ω, choose w0 such that ew0 = f(z0), and define

h(z) = w0 +

∫γz

f ′(z)

f(z)dz

where γz denotes any path in Ω from z0 to z. The choice of path does not matter, becausethe difference of two such paths forms a cycle Γ which is homologous to zero (because ofhomological triviality every cycle is homologous to zero), and hence∫

Γ

f ′(z)

f(z)dz = 0

by Cauchy’s theorem. Now we have

h′(z) = f ′(z)/f(z), d/dz(f(z)e−h(z)) = 0,

so eh(z) is a constant multiple of f(z) and that constant is verified to be 1 by checking atz = z0.


74

Now let Ω be a simply connected open subset of C and let a /∈ Ω. Let S be the class ofinjective6 holomorphic functions Ω→ U. We are going to construct the Riemann mappingas a certain extremal element of S.

Lemma 23.4. The class S is nonempty.

Proof. Let g(z) be a branch of√z − a on Ω. Clearly g is injective Ω → C. Moreover, it

is impossible that both w and −w should belong to the image of g. Now, by the openmapping theorem, the image of g contains some disk D, so it must omit the disk −D.Let the disk −D have center b and radius r; then

z 7→ r

g(z)− bis an injective map sending Ω into U.

By Montel’s lemma, the set S is precompact in the topology of uniform convergence oncompact subsets of Ω. Now fix a point p ∈ Ω.

Lemma 23.5. The derivatives |f ′(p)| are bounded above for f ∈ S, and the least upperbound is actually attained by some f ∈ S.

Proof. Let fn ∈ S be a sequence such that |f ′n(p)| tends to its least upper bound (possibly+∞). By the previous lemma, we may assume by passing to a subsequence that fn con-verges uniformly on compact sets to a holomorphic function f . From Cauchy’s formula forderivatives, f ′n converges uniformly on compact sets also. In particular, |f ′n(p)| convergesto |f ′(p)|, which is the least upper bound. (Notice in particular that f ′(p) 6= 0, so f isnot constant.)

It remains to show that f ∈ S. Since each fn maps Ω into U, f maps Ω into U. Butf , being holomorphic and nonconstant, is an open map, so it actually maps Ω into theinterior of U, which is U.

To show that f is injective, we use Rouche’s theorem. Suppose that there are distinctpoints q, q′ ∈ Ω such that f(q) = f(q′) = c. Let the cycle Γ be the union of twosmall disjoint circles around q and q′ in Ω on which f does not take the value c. Thensup|f(z)− c| : z ∈ Γ∗ = δ > 0. There is an n such that |fn−f | < δ on Γ∗. By Rouche’stheorem, fn− c and f − c have the same number of zeroes inside Γ∗. Thus fn− c has twozeroes, contradicting its injectivity.

Completion of the proof of Theorem 23.1. Let f be the extremal element of S constructedby Lemma 23.5. We shall show that f maps Ω onto U.

Suppose not, and say c ∈ U does not belong to the image of f . Then Tc f is in S anddoes not take the value 0, so by Lemma 23.3 there is g ∈ S such that Tc f = g2. Putg(p) = d and let h = Td g ∈ S. By construction we have

f = F h, where F (w) = T−c([T−d(w)]2

).

By the chain rule we have

f ′(p) = F ′(0)h′(p)

6In this context it is traditional to say univalent, but it means the same thing.


75

since h(p) = 0. But since F is a holomorphic map U → U and is not an automor-phism, Corollary 21.15 shows that |F ′(0)| < 1. Thus |h′(p)| > |f ′(p)|, contradicting theextremality of f .


76

Lecture 24Riemann Surfaces

See Stein and Shakarchi pages 87–89

Definition 24.1. Let X be a Hausdorff topological space. A Riemann surface atlas Afor X is a collection of homeomorphisms ϕα : Uα → Vα, where

(a) The Uα are open subsets of X, and they form a covering (that is,⋃α Uα = X);

(b) The Vα are open subsets of C;(c) For each pair of indices α, β the transition function

ϕαβ = ϕα (ϕβ)−1 : Vβ ∩ ϕβ(Uα ∩ Uβ)→ Vα ∩ ϕα(Uα ∩ Uβ)

is a holomorphic equivalence.

Two such atlases are said to be equivalent if their union is also an atlas. It is easy tocheck that this “equivalence” is indeed an equivalence relation.

Remark 24.2. An atlas that includes any other atlas equivalent to it is called maximal.Any atlas is included in a unique maximal atlas, namely, the union of all the atlasesequivalent to it.

Definition 24.3. A Riemann surface is a Hausdorff space equipped with a maximal atlas.

By Remark 24.2, to equip X with the structure of a Riemann surface we need onlyspecify one atlas for X (not necessarily maximal); the corresponding maximal atlas isthen uniquely determined.

Example 24.4. The complex plane C can be considered as a Riemann surface, with atlasprovided by the single chart U = C, V = C, ϕ = identity map. Any open subset of C canbe considered as a Riemann surface in the same way.

Example 24.5. The Riemann sphere S is defined by adjoining a single point, denoted∞, to the complex plane C. We make it into a topological space by declaring that itsopen sets are

(a) The usual open subsets U ⊆ C;(b) Subsets of the form U ∪ ∞, where U is an open subset of C such that, for some

r > 0, all z with |z| > r belong to U .

This definition makes S into a Hausdorff space, which has the property that a sequencezn ∈ C tends to the point ∞ of S if and only if |zn| → ∞ in the ordinary sense ofanalysis. Topologists will recognize this as a special case of the construction of the onepoint compactification of a locally compact space. We are going to prove that S cannaturally be given the structure of a Riemann surface.

We make S into a Riemann surface by giving it an atlas of two charts.

(1) U1 is the subset C of S, and V1 = C, and ϕ1 : U1 → V1 is the identity map.(2) U2 is the subset (C \ 0) ∪ ∞ of S, and V2 = C, and ϕ2 : U2 → V2 is the map

w 7→ 1/w (extended by declaring that 1/∞ = 0). It is easy to check that this is ahomeomorphism.


77

The transition function ϕ12 : C\0 → C\0 is the map z 7→ 1/z. Since this is certainlya holomorphic equivalence, we have a Riemann surface atlas.

Definition 24.6. Let X and Y be Riemann surfaces. A continuous map f : X → Y isholomorphic if, for every chart ϕX : UX → VX for X and every chart ϕY : UY → VY forY , the composite map between open subsets of C defined by

ϕY f (ϕX)−1 : ϕX(UX ∩ f−1(UY ))→ VY

is holomorphic. A holomorphic function on a Riemann surface is simply a holomorphicmap to C.

In other words, “Holomorphicity is defined locally by charts”. Many familiar propertiesof holomorphic functions on C can be globalized to holomorphic maps between Riemannsurfaces by topological arguments involving connectedness. For instance, let’s do that forthe principle of isolated zeroes:

Proposition 24.7. Let f : X → Y be a nonconstant holomorphic map between Riemannsurfaces, with X connected. Then for each y ∈ Y , the set f−1y ⊆ X has no limit pointsin X.

Proof. Let L denote the set of limit points of f−1y. Then L is a closed set (a limitpoint of limit points is a limit point). We will show that it is also open. Let x ∈ L andlet ϕX , ϕY be charts near x and y respectively, as in the definition above. Then

g = ϕY f (ϕX)−1 : ϕX(UX ∩ f−1(UY ))→ VY

is a holomorphic function between open subsets of C, and it takes the value ϕY (y) on theset ϕX(L∩UX∩f−1(UY )), a set which has a limit point at ϕX(z). By the ordinary versionof the principle of isolated zeroes, g is constant on a small disc D containing ϕX(x). Butthen f is constant on the open set ϕ−1

X (D), so all the points of ϕ−1X (D) belong to L. It

follows that L is open, since we have shown that it contains a neighborhood of each of itspoints.

Since L is both open and closed, connectedness shows that it is either empty or thewhole of X. In the latter case f is constant, which is ruled out by hypothesis. So L = ∅and there are no limit points.

Proposition 24.8. Let f : X → C be a holomorphic function on a connected Riemannsurface X. If |f | attains a local maximum at any point, then f is constant. In particular,every holomorphic function on a compact connected Riemann surface is constant.

Proof. If |f | attains its local maximum at x ∈ X, let ϕ be a chart near x. Then f ϕ−1

is a holomorphic function from an open subset of C to C, whose absolute value attains alocal maximum. By the usual maximum principle, it follows that f is constant, say equalto c, on some open set. Thus the set f−1c has limit points and the result follows fromProposition 24.7.


78

Lecture 25The Sphere as a Riemann Surface

Recall that the Riemann sphere S is C ∪ ∞, regarded as a Riemann surface.

Proposition 25.1. Let Ω be a connected open subset of C and let f : Ω → S be a holo-morphic function that is not identically ∞. Let P ⊆ Ω be the set f−1∞. Then

(i) P is a discrete subset of Ω;(ii) The restriction of f to a function C \ P → C is holomorphic, and the singularities

at the points of P are poles; so f can be regarded as a meromorphic function on Ω;(iii) Every meromorphic function on Ω arises in this way from a unique holomorphic

function Ω→ S.

Proof. The first statement is a special case of Proposition 24.7. For the rest, let p ∈ Ω.What does it mean for a function f : Ω → S to be holomorphic near p? It means thatϕ f is holomorphic near P where ϕ is a chart for S whose domain includes f(p). Iff(p) ∈ C, we can use the chart which is the identity map from C to C and we find thatf must be holomorphic in the usual sense. If f(p) =∞ we must use the chart defined byw 7→ 1/w. So we find that 1/f(z) must be holomorphic near z = p. But this is true ifand only if f has a pole at p: the order of the pole of f is equal to the order of the zeroof the holomorphic function 1/f .

(The last step in the proof should be a familiar argument. We used exactly the sameidea to prove the Casorati-Weierstrass theorem, Proposition 14.5. In fact, you can thinkof the proposition as a globalized version of the Casorati-Weierstrass theorem.)

Remark 25.2. Because of this result we can define a meromorphic function on a generalconnected Riemann surface X to be a holomorphic map X → S that does not take theconstant value ∞.

Let f(z) = p(z)/q(z) be a rational function on C (the quotient of two polynomials).Then f is meromorphic on C. Moreover, f(1/z) is rational (and hence meromorphic onC) as well. These facts together tell us that f extends to a holomorphic map from S to S.

Theorem 25.3. Every holomorphic map from S to S is a rational function.

This is an example of the so-called “GAGA principle”: analytic objects defined oncompact Riemann surfaces (such as the sphere) turn out in fact to be algebraic objects.7

Proof. Let f be such a map. Notice that f−1(∞) is a closed, discrete subset of S, hencefinite. Thus the restriction of f to C is a meromorphic function with finitely many poles.Let the poles in C be a1, . . . , an with orders m1, . . . ,mn. Then

g(z) = (z − a1)m1 · · · (z − an)mnf(z)

has only removable singularities in C; on removing them, we obtain an entire function.Moreover, f(1/w) has at worst a pole at w = 0, which implies that there are constants Cand N such that |f(z)| 6 C|z|N for large |z|. Then

|g(z)| 6 C ′|z|N ′

7GAGA stands for the French: “geometrie analytique, geometrie algebrique”.


79

for large |z|, where N ′ = N +m1 + · · ·+mn. Let g(z) =∑anz

n be its Taylor expansion,and consider the function

h(z) =∞∑

n=N ′

anzn−N ′ = (g(z)− p(z))/zN

′,

where p(z) is a polynomial of degree < N ′. We see that h(z) is bounded and entire, henceconstant (Liouville’s theorem), so g(z) is a polynomial and we are done.

Let us ask what are the automorphisms of S (holomorphic bijections S→ S). From thediscussion above, we see that Mobius transformations

z 7→ az + b

cz + d,

where ad − bc 6= 0 are automorphisms of S. As in the examples of the disc U and theplane C, it turns out that these are all the automorphisms.

Proposition 25.4. The automorphism group of the Riemann sphere is exactly the groupof all Mobius transformations (denoted PGL(2,C)).

Proof. Any holomorphic map S → S is a rational function p(z)/q(z), where p and q arepolynomials. Thus all we have to do is to prove that a rational function is bijective asa map S → S only if deg(p), deg(q) 6 1. But this is easy: the equation p(z)/q(z) = wtranslates to the polynomial equation

p(z)− wq(z) = 0

and this equation will in general have as many roots (for z) as its degree maxdeg(p), deg(q).

Remark 25.5. The words “in general” in the argument above conceal an algebraic discus-sion which is necessary. What about the possibility that p(z)−wq(z) = 0 has a repeatedroot? Suppose that for some nonzero value of w, the equation p(z) − wq(z) = 0 has arepeated root (for z). Then we have simultaneously p′(z) − wq′(z) = 0. Eliminating w,we get

p(z)q′(z)− q(z)p′(z) = 0.

The polynomial appearing on the left of this equation cannot be identically zero, becausethis would imply that p is a constant multiple of q. Therefore there are only finitely manyvalues of z for which the displayed equation is satisfied, and each corresponds to at mostone possible value of w.

A Mobius transformation that fixes the point ∞ (that is, maps ∞ to ∞) is simply anaffine transformation of the form z 7→ az + b, a 6= 0. The affine transformations form asubgroup of the group of all Mobius transformations. As we have already seen, the auto-morphism group of the complex plane is exactly the group of all affine transformations.To put this another way, every automorphism of C extends to an automorphism of S.

Exercise 25.6. Show that if X is a compact Riemann surface and x ∈ X, then everyautomorphism of X \ x extends to an automorphism of X. Does this remain true if Xis not compact?


80

Lecture 26Elliptic Curves: A Sampler

See Stein and Shakarchi Chapter 9The sphere is the simplest example of a compact (orientable) surface and it is a fact —

though we haven’t proved it — that there is only one way to make it into a Riemann sur-face, namely as the Riemann sphere S. The next simplest example of a compact orientablesurface is the torus, and the study of the various different Riemann surface structures ona torus leads into the rich and classical theory of elliptic curves, with connections toanalysis, algebra, number theory, cryptography (and probably other things). This lecturegives a brief sketch.

Definition 26.1. A lattice Λ in C is the additive subgroup generated by two R-linearlyindependent complex numbers.

For instance, the collection m+ ni : m,n ∈ Z of Gaussian integers is a lattice in C.Any lattice has a fundamental parallelogram (like the unit square in this example) suchthat each element of C is equal modulo Λ to an element of the fundamental parallelogram.

Definition 26.2. An elliptic function for the lattice Λ is an entire meromorphic functionf that is periodic for Λ, that is, f(x+ ω) = f(x) for all ω ∈ Λ.

We proved using Liouville’s theorem that there are no nonconstant holomorphic ellipticfunctions. This is a special case of Proposition 24.8, because of the next result:

Proposition 26.3. Let X be the quotient space C/Λ. Then X is a Riemann surface, andthere is a one-to-one correspondence between elliptic functions on C and meromorphicfunctions on the Riemann surface X.

The Riemann surface X is called an elliptic curve. Notice that as well as being aRiemann surface it is also an abelian group: we’ll come back to the group structure later.

An elliptic function must therefore have some poles. We can define the residue of anelliptic function at a point of X = C/Λ in the usual way, and we have

Lemma 26.4. The sum of the residues of an elliptic function (at all the poles in X) isequal to zero.

Proof. Apply the residue theorem (in C) where we integrate around the boundary of afundamental parallelogram (shifted so as not to pass through any poles). The integralsover opposite sides of the period parallelogram cancel because of periodicity.

It follows that there is no elliptic function with just one simple pole on X. The lowest-order possibilities are therefore two simple poles (with opposite residues) or one doublepole. The Weierstrass theory of elliptic functions takes the second case as the fundamentalone.

Definition 26.5. The order of an elliptic function is its total number of poles (countedwith multiplicity), i.e. the sum of the orders of its poles.

Proposition 26.6. An elliptic function takes every value in S the same number of times(counted with multiplicity), which is equal to its order.


81

Proof. Let f be an elliptic function. By Proposition 14.7, the integral

1

2πi

∫Γ

f ′(z)

f(z)dz,

taken around the boundary of a fundamental parallelogram, gives the difference betweenthe number of zeroes and the number of poles of f on X. But this integral equals zero,by the same periodicity argument as above. Thus the number of zeroes of f is equal tothe order (the number of poles). Replacing f by f − c for arbitrary a ∈ C we get theresult as stated.

There is another important result that we can obtain from this same integration tech-nique.

Theorem 26.7. (Abel’s theorem) Let f be an elliptic function, and consider its zeroesand poles (with multiplicity) as elements of the abelian group X = C/Λ. The sum of allthe zeroes and poles is equal to the zero element of X.

Proof. Consider the integral1

2πi

∫zf ′(z)

f(z)dz

taken around the boundary of a fundamental parallelogram (that does not have zeroes orpoles on its boundary). By Lemma 14.6 and the residue theorem, this integral is the sum(in C) of the zeroes and poles counted according to multiplicity. On the other hand, thecontributions to the integral from each pair of opposite sides of the parallelogram are ofthe form

ω ·(

1

2πi

∫γ

f ′(z)

f(z)dz

)= ωwn(f γ; 0),

where ω is a period (a member of Λ) and γ is a path traversing one of the pair of oppositesides. Thus the integral, being a sum of two terms of this sort, belongs to Λ, so itrepresents the zero element in X = C/Λ.

It is not obvious that there exist any elliptic functions at all. Here is one way toconstruct them.

Proposition 26.8. Let Λ be a lattice in C. Then the series

℘(z) =1

z2+

∑ω∈Λ\0

(1

(z − ω)2− 1

ω2

)converges (uniformly on compact sets) to an elliptic function, whose derivative is

℘′(z) = −2∑ω∈Λ

1

(z − ω)3. Λ

The function ℘ is called the Weierstrass elliptic function defined by the lattice Λ. Bydirect calculation and manipulation of series, one obtains.

Proposition 26.9. The Laurent expansion of the Weierstrass ℘ function near the originis

℘(z) =1

z2+∞∑n=1

(2n+ 1)G2n+2z2n,


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where the Eisenstein series Gn(Λ) are defined by

Gn =∑

ω∈Λ\0

1

ωn, (n > 3). Λ

Let g2 = 60G4, g3 = 140G6.

Theorem 26.10. The function ℘ satisfies the nonlinear differential equation

[℘′(z)]2 = 4℘(z)3 − g2℘(z)− g3.

Moreover, the three roots of the cubic 4p3 − g2p − g3 on the right hand side are distinct,and are equal to the values of ℘ at the half periods (the center of the period parallelogramand the midpoints of its two sides).

Proof. Compare the Laurent expansions of both sides to see that the difference is holo-morphic (hence constant, since it is an elliptic function) and vanishes at z = 0. (This justinvolves computing a finite number of coefficients.)

Notice now that ℘′(z) is an odd function of z. If k(t) is any odd function with periodp, and k(p/2) is finite, then it must be zero, since

k(p/2) = −k(−p/2) = −k(p− p/2) = −k(p/2).

Since ℘′ does not have poles at the three half-periods e1, e2, e3, it must vanish there.It cannot vanish anywhere else because it has order 3. Therefore (from the differentialequation) ℘(e1), ℘(e2) and ℘(e3) are exactly the roots of the polynomial 4p3 − g2p− g3.

From the differential equation, for each m = 1, 2, 3 the elliptic function ℘(z) − ℘(ek)has a double zero at z = ek. It follows (since ℘ has order 2) that it does not take thevalue ℘(ek) anywhere other than z = ek. Therefore the three roots ℘(e1), ℘(e2), ℘(e3) ofthe cubic are distinct.

Corollary 26.11. The discriminant (g2)3 − 27(g3)2 is nonzero.

Proof. This is the standard criterion for a cubic equation to have distinct roots.

Remark 26.12. You should think of this ODE as an elliptic-function counterpart of thedifferential equation

[f ′(x)]2 = 1− f(x)2

satisfied by the trigonometric functions f(x) = sinx or f(x) = cosx. Notice that thisproperty of the trigonometric functions is what is responsible for the usefulness of inversetrigonometric functions in evaluating the integral∫

dx√1− x2

,

as we teach our students to do in Math 141. Analogously, the inverse function of ℘ isuseful in evaluating integrals like ∫

dx√4x3 − g2x− g3

,

and similar integrals with a 3rd or 4th degree polynomial under the square root. His-torically, such integrals were first investigated in connection with the problem of therectification (that is finding the length) of an arc of an ellipse, and for this reason theybecame known as elliptic integrals. The terminology elliptic curve comes from this history.


83

Using the ideas above it is not too hard to show

Theorem 26.13. Every elliptic function for the lattice Λ is a rational combination of ℘and ℘′.

I want to express this result in terms of algebra. For any Riemann surface X we candefine the function field C(X) of meromorphic functions on X (this is a field!). Ourtheorem 25.3 about the Riemann sphere from last time can then be expressed as: thefunction field of the Riemann sphere is C(z), the field of rational functions generated bya single transcendental element z. By contrast, the field C(X) for an elliptic curve isnot generated by a single transcendental, but Theorem 26.13 shows that it is obtain as acomposite

C(X) : C(℘) : Cwhere C(℘) : C is a simple transcendental extension and C(X) : C(℘) is an algebraicextension (of degree 2 in this case).

Definition 26.14. A function field (over C) is an algebraic extension of the field ofrational functions over C.

So we have seen that the field of meromorphic functions on an elliptic curve is analgebraic function field. In fact, it is a deep theorem that for any compact Riemannsurface X, C(X) is an algebraic function field, and in fact that the categories algebraicfunction field/compact Riemann surface are equivalent.

Moreover there is a third equivalent category lurking in the background. Let X be acompact Riemann surface, C(X) its function field. Then C(X) is generated over C bytwo elements, say v and w, between which there is an algebraic relation

p(v, w) = 0, some polynomial p.

Then the map x 7→ (v(x), w(x)) sends the Riemann surface X to an algebraic curve inCP2. Suitably interpreted, this shows that the category of complex projective algebraiccurves is also equivalent to the category of compact Riemann surfaces and the category offunction fields. Thus there is a three-way nexus between analysis, algebra and geometry.

In the case of an elliptic curve the corresponding algebraic curve is a cubic C of theform v2 = 4w3 + g2w + g3, with the map from X to the algebraic curve C being definedby z 7→ (℘′(z), ℘(z)). There is a beautiful, classical relation between the geometry of thiscubic curve and the group law on X. Recall that a straight line meets a cubic curve inthree points.

Proposition 26.15. Let C be the cubic corresponding to an elliptic curve X as above.Then three points on the cubic C are collinear if and only if the corresponding points ofX sum to zero under the group law.

Proof. Let z1, z2 ∈ X. Consider the function

f(z) =

∣∣∣∣∣∣1 ℘(z) ℘′(z)1 ℘(z1) ℘′(z1)1 ℘(z2) ℘′(z2)

∣∣∣∣∣∣ .This is an elliptic function of order 3, having a triple pole at 0 and vanishing at z1 andat z2. By Abel’s theorem 26.7, its third zero must be at −(z1 + z2). But the vanishing of


84

the determinant f(−(z1 + z2)) says exactly that the three points of C corresponding toz1, z2, and −(z1 + z2) are collinear.

If you expand the determinant and use that ℘ is even and ℘′ is odd, you can obtaina formula expressing ℘(z1 + z2) in terms of the values of ℘ and ℘′ at z1 and z2. Thisis the elliptic-function analog of the addition formula cos(z1 + z2) = cos(z1) cos(z2) −sin(z1) sin(z2). Historically, it was Euler’s discovery of a special case of this formula inconnection with the geometric problem of duplication of the arc of an ellipse that gaverise to the whole theory of elliptic functions, including their name. (Siegel’s “Topics incomplex function theory” reviews the history.)


85

Lecture 27Picard’s Theorem

Emile Picard published in 1879 a short paper (half a page) in which he proved

Proposition 27.1. Let f be a nonconstant entire function. Then the range of f containsall of C with at most one exception.

This is now known as the little Picard theorem. The big Picard theorem makes thesame statement in any neighborhood of an essential singularity. Despite the shortness ofthe paper, the result was an instant sensation: Littlewood later said that the half-pagepaper would have made an excellent PhD thesis.

We will give an elementary proof which is based on Alhfors’ form of Schwarz’ lemma 21.14.Remember, this says that a holomorphic map U→ U is automatically distance decreasingfor the hyperbolic metric.

As a warm up we will prove Liouville’s theorem again.

Proposition 27.2. A bounded entire function f is constant.

Proof. Without loss of generality assume f maps C into U. Let r > 0 and let Br be theball around f(0) of radius r in the hyperbolic metric. By the Schwarz lemma applied tothe restriction of f to U, f maps the hyperbolic ball around 0 of radius r (that is, the diskD(0; tanh 1

2r)) into Br. But we can apply the same argument to the rescaled functions

fn(z) = f(nz), n = 1, 2, . . .

to see that f maps D(0;n tanh 12r) into Br. Since n is arbitrary, Im(f) ⊆ Br. Now since

r is arbitrary, f is constant.

To prove Picard’s theorem in this same way, we want a version of the Schwarz lemmathat deals with maps into the twice punctured plane (say C \ 0, 1) rather than the discU. (Picard did this by using elliptic modular functions to exhibit a holomorphic coveringmap U→ C \ 0, 1.) We need some ideas from differential geometry.

Let Ω be an open subset of C (this also works on a Riemann surface with appropriateglobalization). A continuous function ρ : Ω→ R is a conformal weight if there is a discretesubset N ⊆ Ω such that ρ is smooth and strictly positive on Ω \ N and vanishes on N .(In other words, we allow “isolated zeroes” at the points of N .) Such a weight ρ definesa metric on Ω: we first define the length of a smooth curve γ : [0, 1]→ Ω by

`(γ) =

∫ 1

0

ρ(γ(t))|γ′(t)|dt,

and then define the distance between two points to be the infimum of the lengths ofsmooth curves joining them.

Example 27.3. With ρ = 1 on C we get the usual Euclidean metric on the plane. Withρ(z) = 2/(1 − |z|2) on U we obtain the hyperbolic metric on U. With ρ = 2/(1 + |z|2)we obtain the spherical metric on S = C ∪ ∞; this is the usual “round” metric on thesphere.


86

If f : Ω1 → Ω2 is nonconstant and holomorphic, and Ω2 has a conformal metric withweight ρ, then one can define a pullback metric on Ω1 by requiring that the length of acurve γ in Ω1 equal the length of the curve f γ in Ω2. The chain rule shows that thismetric has conformal factor

(f ∗ρ)(z) = ρ(f(z))|f ′(z)|.Note the isolated zeroes where f ′ = 0.

Example 27.4. Let λ(z) = 1/(1 − |z|2) be the hyperbolic metric on U. For any auto-morphism Ta : U → U one has T ∗aλ = λ (check this!). Once this is known, to check thatthe metric defined by λa agrees with our earlier definition of hyperbolic distance, we needonly verify the case of a straight path from 0 to a:∫ a

0

2dx

1− x2= 2 tanh−1 a = d(0, a).

We’ll need some facts about the Laplace operator ∆ = ∂2

∂x2 + ∂2

∂y2 .

(i) The representation of ∆ in polar coordinates (r, θ) is ∂2

∂r2 + 1r∂∂r

+ 1r2

∂2

∂θ2 .(ii) If f is holomorphic then ∆(u f)(z) = (∆u)(f(z))|f ′(z)|2 (this follows from the

chain rule and the Cauchy-Riemann equations).

Definition 27.5. The curvature of a metric with conformal factor ρ is

κρ =−1

ρ2∆(log ρ).

(The curvature is not defined on the zero set N .) The metric is ultrahyperbolic if itscurvature is 6 −1 everywhere it is defined.

(Ahlfors gives a more general definition allowing certain semicontinuous ρ, but this isall we need here.)

Example 27.6. The curvature of the hyperbolic metric is −1. The curvature of thespherical metric is +1. The curvature of the Euclidean metric is 0.


87

Lecture 28Picard’s theorem: conclusion

Now we give a generalized Schwarz lemma.

Proposition 28.1. Let U be equipped with its hyperbolic metric and suppose that Ω hasan ultrahyperbolic metric. Then any holomorphic map f : D → Ω is distance decreasing.

Proof. It suffices to prove the following two statements: (a) the pullback of an ultrahy-perbolic metric is ultrahyperbolic, and (b) if ρ is ultrahyperbolic on U then ρ 6 λ.

To prove (a) look at the curvature of f ∗ρ(z) = ρ(f(z))|f ′(z)|:

κf∗ρ(z) =−1

|f ′(z)|2ρ(f(z))2∆ (log ρ(f(z)) + log |f ′(z)|) .

The second Laplacian term vanishes and the first can be written as (∆ log ρ)(f(z))|f ′(z)|2,using fact (ii) about the Laplacian listed above. So we get κf∗ρ(z) = κρ(f(z)).

To prove b, let ρ be an ultrahyperbolic metric on U. Fix r < 1 (we will let it tendto 1 at the end of the argument) and let λ = 2/(r2 − |z|2) be the hyperbolic metric onD(0; r) ⊆ U. The function

t(z) = ρ(z)/λr(z)

is therefore continuous on the closed disk D(0; r) (it tends to zero at the boundary circle),so it must attain a maximum at some interior point a. We need to prove that t(a) 6 1.We may assume a /∈ N (else ρ(a) = 0 and the result is trivial). Then κρ(a) 6 −1 = κλ.Since log(t) has a maximum at a,

0 > (∆ log t)(a) = (∆ log ρ)(a)− (∆ log λ)(a)

= −κρ(a)ρ(a)2 + κλ(a)λ(a)2 > ρ(a)2 − λ(a)2.

Thus ρ(a) 6 λ(a), that is, t(a) 6 1. Since a gives the maximum value of t, ρ(z) 6 λ(z)for all z. Let r → 1 to obtain the desired result.

Lemma 28.2. Suppose that Ω admits an ultrahyperbolic metric. Then any entire functionC→ Ω is constant.

Proof. We can exactly follow the argument given earlier for Liouville’s theorem, Proposi-tion 27.2.

In order to complete the proof of Picard’s theorem it now suffices to show

Lemma 28.3. For any two points a, b ∈ C, the domain C \ a, b has an ultrahyperbolicmetric.

(Picard actually constructed a hyperbolic metric: complete, constant curvature −1.This is much more difficult.)

Proof. We may as well take our two omitted points to be 0 and 1. Also, it suffices toconstruct a metric whose curvature is 6 c for a constant c < 0; a rescaling will thenproduce an ultrahyperbolic metric.


88

We will look for such a metric of the form ρ(z) = ϕ(|z|)ϕ(|z − 1|), where the “ansatz”ϕ is of the from

ϕ(r) =(1 + rα)β

rγ.

Later, we will tweak the constants α, β, γ to get the desired properties.Using the formula for the Laplacian in polar coordinates, item (ii) of the properties of

the Laplacian listed last time, one gets

∆(logϕ)(r) = β∆ log(1 + rα) =βα2rα−2

(1 + rα)2.

Therefore

κρ = −βα2( |z|α−2−2γ|z − 1|2γ

(1 + |z|α)2+2β(1 + |z − 1|α)2β

+|z − 1|α−2−2γ|z|2γ

(1 + |z − 1|α)2+2β(1 + |z|α)2β

)Clearly this expression is < 0. Moreover, take α − 2 − 2γ = 0. Then this expressiontends to a finite limit −βα22−2β < 0 as z → 0, 1. Moreover, take γ = α(1 + 2β). Thenthe expression also tends to a finite limit −2βα2 as z →∞. Consequently the curvatureis bounded from above by a strictly negative constant, as required. This completes theproof of Picard’s theorem.

Remark 28.4. Specific examples that satisfy the criteria are α = 13, β = 3

4, γ = 5

6. There

is a lot of flexibility in the construction. For instance, we are okay if the curvature tendsto −∞ at the singularities, rather than a finite limit. This allows us to choose anyβ > 1

2(γα−1 − 1).

Let’s go on and prove the Big Picard Theorem: in any neighborhood of an essentialsingularity a holomorphic function assumes every value infinitely often, with at mostone exception. This requires a little discussion about normal families in the context ofmeromorphic functions.

We now know that a meromorphic function on a domain Ω can be regarded as aholomorphic map from Ω to the Riemann sphere S. Now S is a metric space, equippedwith the spherical metric we discussed in the previous lecture, and thus the notion ofequicontinuity makes sense for maps into S. Analogous to the holomorphic case we say thata family F of meromorphic functions is normal (or spherically normal) if the restriction ofF to any compact K ⊆ Ω is an equicontinuous family of maps K → S. The usual Arzela-Ascoli argument applies to show that a normal family is precompact in the topology of“spherical uniform convergence” on compact sets.

Proposition 28.5. (Montel’s theorem) Let Ω ⊆ C be any open set and let F be the familyof all holomorphic functions on Ω that omit the values 0 and 1. Then F is sphericallynormal.

Proof. We need to remark that a metric ρ on C\0, 1 as in lemma 28.3 can be constructedso that the spherical metric σ(z) = 1/(1+ |z|2) satisfies σ 6 Cρ for some positive constantC. (This requires the additional condition γ−αβ < 1, satisfied for the numbers we chose.)


89

Look at the restriction of f ∈ F to any disc D(a; r) ⊆ Ω, and equip D with its hyperbolicmetric λ. Then by Proposition 28.1, f is (λ, ρ) distance decreasing. Let z1, z2 ∈ D(a; r/2),

dσ(f(z1), f(z2)) 6 Cdρ(f(z1), f(z2)) 6 Cdλ(z1, z2) 63C

4r2|z1 − z2|.

Since this holds for all f ∈ F , the family F is equicontinuous on D(a; r/2). Any compactsubset of Ω can be covered by finitely many such discs, so F is equicontinuous on compactsubsets of Ω.

Theorem 28.6. (Big Picard) Let f be holomorphic on a punctured disc Ω = D(0; r)\0.If f omits two values (say 0 and 1) then the singularity of f at 0 is a pole or removable.

Proof. Consider the functions fn(z) = f(z/n) defined on Ω. By Montel’s theorem, fnis a normal family, so the sequence fn has a subsequence that converges spherically oncompact sets to a limit g which is either a holomorphic function or identically ∞. If it’sidentically∞, replace f by 1/f (legitimate because f never takes the value 0, and 1/f hasa removable singularity or pole if f has a removable singularity). Thus we may assumethat fn converges to a holomorphic g.

Let C1 be the circle of radius r/2 in Ω. The sequence of functions fn on the circle C1

converges uniformly to g. In particular, there is some constant M such that |fn(z)| 6Mfor all n and all z ∈ C1. But this tells us |f(z)| 6 M for all z ∈

⋃Cn, where Cn is

the circle of radius r/2n. Applying the maximum principle to each of the annuli betweenCn and Cn+1, we see that |f | is bounded by M on a neighborhood of 0. Therefore thesingularity is removable.


90

Lecture 29Harmonic functions

Harmonic functions are solutions of the Laplace equation ∆u = 0. As we saw in theprevious section, the study of the Laplace operator in dimension 2 is closely connected tocomplex analysis and conformal mapping.

Definition 29.1. Let Ω be an open subset of Rn. A smooth function h : Ω → R isharmonic if it satisfies the equation

∆h =∂2h

∂x21

+ · · ·+ ∂2h

∂x2n

= 0

called Laplace’s equation.

We shall consider the case n = 2 and we shall identify R2 with C. We have already seenthat the real part of a holomorphic function on Ω must be harmonic (this is a consequenceof the Cauchy-Riemann equations, see Remark 3.5). Conversely we have

Proposition 29.2. Any harmonic function on a simply connected Ω ⊆ C is the real partof a holomorphic function.

Proof. Suppose that h is harmonic on Ω and define g = u+iv = hx−ihy (we use subscriptsto denote partial derivatives). Then

ux − vy = hxx + hyy = 0, uy + vx = hxy − hyx = 0

so g satisfies the Cauchy-Riemann equations. Since g is smooth, it is holomorphic.Because Ω is simply connected, the holomorphic function g has an antiderivative on Ω,

a holomorphic f such that f ′ = g. Then

(Re f)x = Re g = hx, (Re f)y = −(Im f)x = − Im g = hy

and it follows that h − Re f is a constant c. Thus h is the real part of the holomorphicfunction f + c.

Remark 29.3. By definition, an antiholomorphic function of z is a holomorphic function ofz. Since the Laplace equation is invariant under complex conjugation (x 7→ x, y 7→ −y),the real part of an antiholomorphic function is also harmonic. We’ll use this fact in amoment.

Exercise 29.4. Show that the assumption of simple connectedness is necessary in theabove theorem. (Consider the function g(x+ iy) = log(x2 + y2) on C \ 0. )

Corollary 29.5 (Maximum principle for harmonic functions). Let h be a non-constantharmonic function on the connected open set Ω. Then h cannot attain a local maximum(or minimum) at any point of Ω.

Proof. If h has a local maximum then eh does so too. Fix a disk in Ω containing thesupposed local maximum. On this disk we can write h = Re f with f holomorphic. Butnow eh = eRe f = |ef |, so the result follows from the maximum principle for holomorphicfunctions (Theorem 11.2) applied to the function ef .


91

It follows that “a harmonic function is determined by its boundary values”. Moreprecisely, suppose that Ω is an open set with compact closure Ω, and let ∂Ω = Ω \ Ω.Then for each continuous function g on ∂Ω there is at most one harmonic function hon Ω which extends g continuously. Given there were two such functions, we take theirdifference to get a harmonic function h that is zero on ∂Ω. By the maximum principle,h must attain its maximum and minimum values on the boundary ∂Ω; so it must beidentically zero.

This is a uniqueness statement. The corresponding existence statement is called theDirichlet Problem: given a continuous function g on ∂Ω, find a harmonic function hhaving g as its boundary values. We will begin our investigation of the Dirichlet problemby looking at the simplest case — the unit disk.

Proposition 29.6. The Dirichlet problem for the unit disk U has a (unique) solution forany continuous boundary data g. In fact, the solution is given explicitly by the Poissonintegral formula

h(reiθ) =1

2π

∫ 2π

0

1− r2

1− 2r cos(θ − ϕ) + r2g(eiϕ)dϕ.

Proof. First we want to show that the expression

Pr(θ − ϕ) =1− r2

1− 2r cos(θ − ϕ) + r2,

called the Poisson kernel, is a harmonic function of w = reiθ. Put z = eiϕ, w = reiθ, andu = w/z = rei(θ−ϕ). Consider

Re1 + u

1− u=

1

2

(1 + u

1− u+

1 + u

1− u

)=

1− |u|2

|1− u|2= Pr(θ − ϕ)

This shows that the Poisson kernel is the real part of a holomorphic function of w ∈ U,so it is harmonic. By differentiation under the integral sign, it follows that the functionh defined by the Poisson integral formula is harmonic.

It remains to show that h has g as its boundary values, which is to say that h(reiθ)approaches g(eiθ), uniformly in θ, as r → 1. Notice the following properties of the Poissonkernel

(a) Pr(θ) > 0 for all r, θ.

(b) (2π)−1∫ 2π

0Pr(θ)dθ = 1.

(c) For all δ > 0,∫ 2π−δδ

Pr(θ)dθ → 0 as r → 1.

Indeed, (a) is apparent from the definition; (b) follows from the Cauchy integral formula:∫ 2π

0

Pr(θ)dθ = Re

(∫γr

z + 1

z − 1

dz

iz

)= 2π,

where γr is a circular path of radius r. To see (c), write

Pr(θ) =1− r2

(1− r)2 + 2r(1− cos θ);

thus for θ ∈ (δ, 2π − δ), we have |Pr(θ)| 6 (1 − r2)/2r(1 − cos δ) and thus tends to 0uniformly in θ as r → 1.


92

Now let ε > 0 be given. Let g : S1 → R be the given boundary data. Since g isuniformly continuous, there is δ > 0 such that if |ψ| < δ, |g(z) − g(zeiψ)| < ε. Let

M = sup |g| and using property (c) choose R such that, for r > R,∫ 2π−δδ

Pr(θ)dθ < ε/M .Now write

h(reiθ)− g(eiθ) =1

2π

∫ 2π

0

Pr(θ − ϕ)(g(eiϕ)− g(eiθ)

)dϕ,

using (a). We estimate the integral on the right over two ranges separately: |θ − ϕ| < δ(mod 2π, range R1) and the rest (range R2), getting the estimate

|h(reiθ)− g(eiθ)| 6(

1

2π· 2π · ε

)︸︷︷︸

R1

+

(1

2π· 2M · 2πε

M

)︸︷︷︸

R2

6 3ε,

which proves uniform convergence.

Remark 29.7. It is clear (by power series expansion) that for each fixed r, Pr(θ) is auniform limit of trigonometric polynomials (linear combinations of sin kθ and cos kθ, orequivalently of eikθ for k ∈ Z). Thus a corollary of the convergence proof above is theWeierstrass approximation theorem: every continuous function on the circle is a uniformlimit of trigonometric polynomials.


93

Lecture 30The mean value property

In the previous lecture we have investigated the Poisson integral formula, which pro-duces a harmonic function h on the unit disk having given continuous boundary values.A simple rescaling produces a version of the formula valid for any disk. On a disk centera radius R, this takes the form

h(a+ reiθ) =1

2π

∫ 2π

0

R2 − r2

R2 − 2Rr cos(θ − ϕ) + r2g(a+Reiϕ)dϕ

with r < R.If h is harmonic on an open set Ω and D(a;R) ⊆ Ω, then the Poisson integral formula

can be applied to h to produce a harmonic function on D(a;R) having the same boundaryvalues as h. By the maximum principle, this harmonic function actually is h. Thus weget the formula

h(a+ reiθ) =1

2π

∫ 2π

0

R2 − r2

R2 − 2Rr cos(θ − ϕ) + r2h(a+Reiϕ)dϕ

which is a harmonic counterpart of the Cauchy integral formula.In particular a harmonic function has the mean value property

h(a) =1

2π

∫ 2π

0

h(a+Reiϕ)dϕ.

Definition 30.1. A continuous function f on Ω is subharmonic if it has the sub meanvalue property : for every a ∈ Ω we have

f(a) 61

2π

∫ 2π

0

f(a+Reiϕ)dϕ

for all sufficiently small R > 0.

Thus, harmonic functions are subharmonic. Subharmonic functions also satisfy themaximum principle:

Proposition 30.2. Let Ω ⊆ C be open and connected and have compact closure. Let fbe nonconstant, continuous on Ω and subharmonic on Ω. Then the maximum value of fis attained at a point of ∂Ω = Ω \ Ω.

Proof. Suppose that f attains its maximum at a ∈ Ω. Then for all sufficiently small R,

f(a) 61

2π

∫ 2π

0

f(a+Reiϕ)dϕ 61

2π

∫ 2π

0

f(a)dϕ = f(a)

and equality must hold everywhere, so f(a+Reiϕ) = f(a) for all ϕ and f is constant ona disk around a.

It follows that the set of points a ∈ Ω where f attains its maximum value is open.Clearly it is also closed, so it is either Ω or ∅ because Ω is connected. In the first case, fwould be constant.

Proposition 30.3. A continuous function on an open subset of C that has the meanvalue property is harmonic.


94

Proof. Let f be continuous on Ω and have the MVP. Let D be any disk whose closure iscontained in Ω. Let h be the harmonic function on D (provided by the Poisson integralformula) that agrees with f on ∂D. Then f − h has the MVP and is zero on ∂D. By themaximum principle, f − h is zero on D, so f is harmonic on D. Since D is arbitrary, fis harmonic.


95

Lecture 31The Perron construction

Lemma 31.1. Lat h be a nonnegative harmonic function defined on a domain containingthe closed disk D(a;R). Then for r < R,

R− rR + r

h(a) 6 h(a+ reiθ) 6R + r

R− rh(a).

Proof. The Poisson integral formula is

h(a+ reiθ) =1

2π

∫ 2π

0

R2 − r2

R2 − 2Rr cos(θ − ϕ) + r2h(a+Reiϕ)dϕ.

Observe that (R− r)2 6 R2 − 2Rr cos(θ − ϕ) + r2 6 (R + r)2 and thus

R− rR + r

6R2 − r2

R2 − 2Rr cos(θ − ϕ) + r26R + r

R− r.

If C denotes the L1 norm of the function ϕ 7→ (2π)−1h(a+Reiϕ), simple estimates give

R− rR + r

C 6 h(a+ reiθ) 6R + r

R− rC.

But since h is nonnegative, the mean value property tells us that C is exactly h(a).

Corollary 31.2. A harmonic function on the plane that is bounded below (or, boundedabove) is constant.

Proof. We may assume the function is nonnegative. Apply the lemma and let R→∞.

This corollary is a version of Liouville’s theorem for harmonic functions.

Proposition 31.3 (Harnack’s Principle). Let hn be a monotone increasing sequence ofharmonic functions on a connected open set Ω ⊆ C. Then either hn converges uniformlyon compact subsets of Ω to a harmonic function, or hn →∞ everywhere.

Proof. Replacing hn by hn − h1 we may assume that all the hn are nonnegative. Leta ∈ Ω and suppose that D(a;R) ⊆ Ω.

If hn(a) converges to a finite limit, then by the right-hand inequality in lemma 31.1,

(hm − hn)(a+ reiθ) 6 3(hm − hn)(a)

for any m > n and any r 6 R/2. Thus hm converges uniformly on the disk of radius R/2.If hn(a) increases to infinity, a similar argument using the left-hand inequality in

lemma 31.1 shows that hm increases to infinity on the disk of radius R/2.These observations show that the set of points a where hn(a) converges to a finite limit

and the set of points a where hn(a) increases to infinity are open (and disjoint). Byconnectedness, one of them is empty.

Finally, in the case that the hn converge to a finite limit h, we have shown that eacha ∈ Ω has a neighborhood on which the convergence is uniform. Therefore the convergenceis uniform on compact sets. The limit function h is therefore continuous and satisfies themean value property. Thus, it is harmonic.


96

Let Ω be a connected open subset of C with compact closure. Recall that the Dirichletproblem is this: given a continuous function g on ∂Ω, find a function h that is continuouson Ω, harmonic on Ω, and extends g.

Example 31.4. Let Ω be the punctured disk D(0; 1)\0. If we take g to be 0 on the unitcircle and 1 at the origin, then g is continuous on ∂Ω. But there is no harmonic function hon Ω that continuously extends g, as this would contradict the maximum principle. Thisshows that some condition on ∂Ω is necessary for the solvability of the Dirichlet problem.

In this lecture we shall solve the Dirichlet problem subject to a ‘regularity’ conditionon ∂Ω. The method is due to Perron.

Definition 31.5. Let Ω be as above and let g be a continuous function on ∂Ω. ThePerron family associated to g is the collection Fg of all continuous functions f : Ω → Rthat are subharmonic on Ω and have boundary values 6 g.

Every function f ∈ Fg is bounded above by sup g, by the maximum principle. Moreover,Fg is nonempty since it contains each constant function 6 inf g. Thus the supremum inthe next definition exists:

Definition 31.6. The Perron function Pg associated to g is the function (defined on Ω)

Pg(z) = supf(z) : f ∈ Fg.Proposition 31.7. The Perron function Pg is harmonic on Ω.

Proof. Let us begin by making three observations.

(i) If f1, f2 belong to the Perron family then so does maxf1, f2. (Indeed, we only needto check that the maximum of two subharmonic functions is subharmonic.)

(ii) Let D be a closed disk in Ω and let f belong to the Perron family. Define a newfunction fD to be equal to f on Ω \ D, and to be equal on D itself to the uniqueharmonic function (provided by the Poisson integral formula) that agrees with f on∂D. We must check that this construction produces a subharmonic function. Noticefirst that fD > f (since, on D, f −fD is a subharmonic function with zero boundaryvalues, hence it is 6 0 by the maximum principle). Now to check that fD has thesub mean value property, the only points a where this is not obvious are those on∂D. But for such points fD(a) = f(a) and

f(a) 61

2π

∫ 2π

0

f(a+Reiϕ)dϕ 61

2π

∫ 2π

0

fD(a+Reiϕ)dϕ

which gives the result.(iii) We note explicitly that since the Poisson integral preserves positivity (consequence

of the maximum principle), if f ′ > f then f ′D > fD.

Now let a ∈ Ω and choose R > 0 so that D(a;R) ⊆ Ω. There is a sequence of functionsfn ∈ F with fn(a) → Pg(a), and because of (i)–(iii) above we can choose each fn tobe harmonic on D(a;R) and fn+1 > fn. By Harnack’s principle (Proposition 31.3), thesequence fn converges locally uniformly on D(a;R) to a harmonic function h, where byconstruction h(a) = Pg(a).

Let a′ ∈ D(a;R) be another point and choose a sequence of functions f ′n ∈ F suchthat f ′n(a′) → Pg(a′). Using (i)–(iii) in the same way as before, we can assume that


97

f ′n > fn, that the sequence f ′n is increasing and that it consists of functions harmonicon D(a;R). The limit function h′ is therefore harmonic on D(a;R) and has h′ > h andh′(a′) = Pg(a′). By the definition of Pg(a) as a supremum we have f ′n(a) 6 h(a) = Pg(a)for all n, so h′(a) 6 h(a) and therefore h′(a) = h(a). The harmonic function h′ − h thenhas a minimum at a, so it is identically 0 on D(a;R). We have therefore proved that

Pg(a′) = h′(a′) = h(a′).

Thus the functions Pg and h agree on D(a;R), so Pg is harmonic there. Since a wasarbitrary, Pg is harmonic everywhere.

We should like to say that the Perron function Pg solves the Dirichlet problem. Thereremain two issues in doing this: show that the Perron function is continuous on Ω, andshow that its boundary values are exactly g (obviously they are 6 g).

Definition 31.8. A boundary point a ∈ ∂Ω is regular if there exists a continuous functionϕa on Ω, subharmonic on Ω, with ϕa(a) = 0 and ϕa(z) < 0 for all z ∈ ∂Ω \ a. (Thefunction ϕa is called a barrier at a.)

Example 31.9. If there is a straight line segment [a, b] in C that meets Ω only at theboundary point a, then a is regular. For, without loss of generality, take a = 0 andb ∈ R+. Then define w to be the branch of

√z/(z − b) (defined on C \ [a, b]) which has

positive real part everywhere. Then ϕa(z) = −Rew(z) is a barrier at a.

Proposition 31.10. The Dirichlet problem is solvable for Ω if and only if every a ∈ ∂Ωis regular.

Proof. If the Dirichlet problem is solvable, its solution with boundary data g(z) = −|z−a|is a barrier.

Conversely, suppose that a is a regular boundary point. Let ϕa be a barrier at a, letg ∈ C(∂Ω) be given and let h = Pg be the associated Perron function. Let ε > 0, andchoose a neighborhood U of a such that |g(z)− g(a)| < ε for all z ∈ U ∩ ∂Ω.

Consider the functionu(z) = g(a)− ε+ Cϕa(z)

where C is a positive constant. Then u is subharmonic and u 6 g on U ∩∂Ω. By taking Csufficiently large we can arrange that u 6 g on ∂Ω \U also (because ϕa is bounded aboveon this set by a strictly negative constant). If we choose C in this way then u belongs tothe Perron family, so we have h > u. Since ϕa is continuous, we have

lim infz→a

h(z) > lim infz→a

u(z) = g(a)− ε.

Letting ε→ 0 we find thatlim infz→a

h(z) > g(a).

Now consider the function

v(z) = g(a) + ε− Cϕa(z)

which is superharmonic (i.e. −u is subharmonic); we choose C so that v > g on ∂Ω. Iff belongs to the Perron family, then f − v is subharmonic and 6 0 on ∂Ω, hence 6 0everywhere on Ω by the maximum principle. It follows that h 6 v on Ω, so

lim supz→a

h(z) 6 lim supz→a

v(z) = g(a) + ε.


98

Letting ε→ 0 we getlim supz→a

h(z) 6 g(a).

Together with the earlier inequality this implies that h is continuous at a with h(a) =g(a).


99

Lecture 32Runge’s Theorem

The Weierstrass approximation theorem tells us that any continuous function definedon a compact subset of R is a uniform limit of polynomials. However, this fact does notextend to compact subsets of C. For example, the function z 7→ z defined on the closedunit disc is not a uniform limit of polynomials there, since any such uniform limit wouldhave to be holomorphic on the interior (the open unit disc).

Runge’s theorem states that a continuous function f on a compact set K ⊆ C can beuniformly approximated by polynomials provided a holomorphicity condition (on f) anda topological condition (on K) are satisfied. As with Cauchy’s theorem, there are variousversions. The simplest is the following.

Theorem 32.1. Let K be a compact subset of C whose complement is connected. Letf be a function holomorphic on a neighborhood of K. Then there exists a sequence ofpolynomials converging uniformly to f on K.

Remark 32.2. The example of the function 1/z on the circle S1 shows that the topologicalhypothesis (connectedness of the complement) cannot be omitted.

The proof will use a series of lemmas. Let E be the Banach space C(K) of continuouscomplex-valued functions on K (with the supremum norm) and let P denote the closedsubspace spanned by the polynomials in z, so that what we are trying to prove is thatf ∈ P .

Lemma 32.3. With assumptions as in Theorem 32.1, the functions

fa(z) =1

z − abelong to P for all a ∈ C \K.

Proof. We proved this as a homework assignment. Here is a short version of the argument.First, suppose that R is such that K ⊆ B(0;R), and suppose that |a| > 2R. Then thepower series

fa(z) =1

z − a= −

∞∑n=0

zna−(n+1)

converges uniformly on K (by the M -test) and shows that fa ∈ P . Observe also that themap a 7→ fa, from C \ K to C(K), is holomorphic with derivative ga : z 7→ 1/(z − a)2.Let Λ ∈ E∗ be any continuous linear functional annihilating P . Then a 7→ Λ(fa) isholomorphic and vanishes for |a| > 2R, hence by the principle of isolated zeroes it vanishesfor all a in the connected set C \K. Thus for any such a, fa is annihilated by all Λ thatannihilate P . It follows from the Hahn-Banach theorem that fa ∈ P .

The next lemma is just “real analysis”.

Lemma 32.4. Suppose that X and Y are compact metric spaces and that k is a continuousfunction on X×Y . Let P be a closed subspace of C(Y ), and suppose that for every x ∈ X


100

the function y 7→ k(x, y) belongs to P . Then, for any finite Borel measure µ on X, thefunction

y 7→∫X

k(x, y)dµ(x)

belongs to P also.

Proof. The function k is uniformly continuous, by compactness. Hence, given any ε > 0,there exists δ > 0 such that if |x− x′| < δ then |k(x, y)− k(x′, y)| < ε for all y ∈ Y .

Partition X into finitely many disjoint Borel subsets X1, . . . , Xn each of diameter < δ,and let xj ∈ Xj for j = 1, . . . , n. Then we have

∣∣∣∫ k(x, y)dµ(x)−

∈P︷︸︸︷n∑j=1

k(xj, y)µ(Xj)∣∣∣ 6

n∑j=1

∫Xj

|k(x, y)− k(xj, y)| dµ(x) 6 εn∑j=1

µ(Xj) = εµ(X)

Since ε is arbitrary and P is closed, this shows y 7→∫Xk(x, y)dµ(x) belongs to P .

Finally a topological lemma.

Lemma 32.5. Let Ω ⊆ C be open and let K be a compact subset of Ω. Then there is acycle Γ in Ω \K such that wn(Γ; z) = 1 for all z ∈ K, and Γ is nullhomologous in Ω.

We could say that such a cycle encircles K.

Proof. Given K and Ω there is ε > 0 such that the distance from any point of K to anypoint not in Ω ia at least 2ε. Now cover C by a grid of squares of side ε. Finitely manyof these squares will contain a point of K in their interior or boundary (we say that sucha square “meets K”, for short). Each such square has a bounding cycle made up of itsfour sides. Let Γ1, . . . ,ΓN be the bounding cycles of the squares Σj meeting K. Clearly,Γ is nullhomologous in Ω.

I claim that the sum Γ =∑N

j=1 Γj has the properties claimed. By construction, Γ is acycle in Ω. Suppose that γ is an edge in the grid that intersects K. Then γ appears asa component in two of the cycles Γj, and it appears with opposite orientations. Thus itcancels from the sum defining Γ. We conclude that only edges that do not intersect Kcan appear in Γ with a non-zero coefficient. Thus, Γ is a cycle in Ω \K.

By construction, Γj has winding number 1 about any point in the interior of the squareΣj and winding number 0 about any point outside the closure of Σj. Any point a ∈ Kis either in the interior of exactly one Σj, or is a limit point of a sequence ak of suchinterior points. In the first case we get wn(Γ, a) = wn(Γj, a) = 1; in the second we havewn(Γ, ak) = 1 by the same argument, and it follows that wn(Γ, a) = 1 since ak and a arein the same connected component of Ω \ Γ∗.

Proof of Runge’s Theorem. We use Cauchy’s integral formula. Suppose that f is holo-morphic on an open set Ω containing K, and let Γ be a cycle in Ω \K encircling K, as in


101

Lemma 32.5. Then for w ∈ K,

f(w) =1

2πi

∫Γ

1

z − wf(z) dz.

Let P denote the closure of the polynomials in C(K). By Lemma 32.3, the integrandf(z)/(z − w) (considered as a function of w) belongs to P for each fixed z. By Lemma32.4, the integral also belongs to P . This completes the proof of Runge’s theorem.

Remark 32.6. A more general form of Runge’s theorem can be proved in the same way. Itstates that if K is a compact subset of C, and if we prescribe a point pj in each componentof the complement S\K, then any function f holomorphic on a neighborhood of K can beapproximated uniformly on K by rational functions having poles only at the prescribedpoints pj. (The case that we proved corresponds to the single prescribed point ∞.)

In particular, if K ⊆ Ω, where Ω is open, and if no component of the complementof K is a subset of Ω, then every function holomorphic in a neighborhood of K can beapproximated, uniformly on K, by functions holomorphic on Ω. We will use this in amoment.


102

Lecture 33Applications of Runge’s theorem

The chief application of Runge’s theorem is to construct holomorphic functions withprescribed properties. We are going to describe several examples of this.

Example 33.1. We have seen that if a sequence of holomorphic functions convergesuniformly on compact sets, then its limit function must be holomorphic. But what abouta sequence of holomorphic functions converging pointwise: must the limit be holomorphicin this case? We’ll use Runge’s theorem to construct an example where the limit is noteven continuous.

To do this, consider the compact subsets Kn = An ∪Bn of C, where

An = z : |z| 6 1/n, Bn = z = reiθ : 2/n 6 r 6 n, 1/n 6 θ 6 2π.Notice that for every nonzero z ∈ C there exists N such that z ∈ Bn for n > N .

Runge’s theorem applies to Kn and can be used to produce a polynomial pn such that|pn − 1| < 1/n on An and |pn| < 1/n on Bn. The sequence of polynomials pn(z) thenconverges pointwise to the function which is 1 at the origin and 0 everywhere else.

A more generally important application is to the ∂-equation.

Definition 33.2. Let f be a smooth (but not necessarily holomorphic) function on anopen subset Ω of C. The differential operators ∂

∂zand ∂

∂zare defined by

∂f

∂z=

1

2

(∂f

∂x− i∂f

∂y

),

∂f

∂z=

1

2

(∂f

∂x+ i

∂f

∂y

).

Notice that if f is holomorphic, then ∂f∂z

= 0 (this is a restatement of the Cauchy-

Riemann equations) and ∂f∂z

= f ′(z) as we defined it before. Moreover, for any f wehave

df =∂f

∂zdz +

∂f

∂zdz,

where we make the natural definitions

dz = dx+ idy, dz = dx− idy.For future reference, note that dz ∧ dz = −2idx ∧ dy.

Lemma 33.3. Let u ∈ C∞c (C) be a smooth function on the plane with compact support.Then there exists a smooth function ϕ on the plane satisfying

∂ϕ

∂z(z) = u(z).

Proof. The solution is given by an explicit integral formula. Notice that the functionz 7→ 1/z is integrable with respect to Lebesgue measure on the plane. Therefore, thefollowing formula makes sense

(33.4) ϕ(z) =−1

π

∫∫C

u(w)

w − zdλ(w).

To check that this indeed gives the desired solution, make use of the following observationwhich follows from Stokes’ (or Green’s) theorem: if Ω is a smoothly bounded open set


103

with compact closure (it will be an annulus in our application) and if v is smooth in aneighborhood of Ω, then∫∫

Ω

∂f(z)

∂zdλ(z) =

1

2i

∫∫Ω

df ∧ dz =1

2i

∫∂Ω

fdz.

In Equation 33.4, take Ω to be an annulus centered at 0, with inner radius ε and outerradius R large enough that D(0;R) ⊇ Supportu. Substitute w + z for w and thendifferentiate under the integral sign. Remembering that the function 1/w is holomorphicon Ω and thus satisfies the Cauchy-Riemann equations, this gives

∂ϕ(z)

∂z=−1

2πi

∫∂Ω

u(w + z)

wdw.

The integral over the outer boundary component of Ω is zero. The inner boundarycomponent is a small circle around 0 traversed once in the negative direction, so aroundthat component

∮dw/w = −2πi. Remembering that u is smooth, we may let ε → 0 to

obtain the limit u(z), as required.

Theorem 33.5. Let Ω be any open subset of C, and let u be a smooth function on Ω,not necessarily with compact support. Then there exists a smooth function ϕ on Ω with∂ϕ∂z

= u.

This is the same result as we used before, but now without the compact support con-dition on the data. We need an approximation result like Runge’s to prove this.

Proof. For simplicity consider the case when Ω = C (we’ll discuss the general case ina moment). Let K1, K2, . . . be a sequence of compact sets with connected complementexhausting Ω (for example, Kn could be the closed disc of radius n) and let αn be a “bumpfunction” equal to 1 on a neighborhood of Kn and supported in the interior of Kn+1. Wewill construct by induction a sequence of smooth functions ϕn such that

∂ϕn(z)

∂z= u(z) ∀z ∈ Kn, |ϕn(z)− ϕn−1(z)| < 2−n ∀z ∈ Kn−1.

The induction starts by using Lemma 33.3 to construct ϕ1 with ∂ϕ1

∂z= α1u. Now supposing

that ϕ1, . . . , ϕn−1 have been constructed, use Lemma 33.3 to construct ψ with ∂ϕ∂z

= αnu;then ψ − ϕn−1 satisfies the Cauchy-Riemann equations (and so is holomorphic) on aneighborhood of Kn−1, so by Runge’s theorem it can be uniformly approximated on Kn−1

by polynomials. In particular there is a polynomial p such that |ψ − p− ϕn−1| < 2−n onKn−1. Take ϕn = ψ − p.

Having constructed the sequence ϕn, we see that it converges uniformly on compact setsto a continuous function ϕ. Moreover, given any compact set K, the functions ϕn−ϕ areholomorphic on a neighborhood of K for sufficiently large n. It follows that ϕ is actuallysmooth, and satisfies the ∂-equation as required.

(For a general Ω, use a “gridding” construction to exhaust it by compact sets Kn havingno component which is a subset of Ω, and then use the more general version of Runge’stheorem given in Remark 32.6.)


104

Lecture 34The Weierstrass and Mittag-Leffler problems

Let f be a meromorphic function with a pole at a. A polynomial in (z − a)−1, withzero constant term, will be called a principal part at a, and the principal part of f at apoint a is the sum of the terms in its Laurent series involving negative powers of (z − a).Equivalently, the principal part is the equivalence class of the Laurent series of f , wheretwo Laurent series are equivalent if their difference is a Taylor series, that is, has aremovable singularity at a.

Let Ω be an open subset of C. A set of Mittag-Leffler data on Ω consists of a discretesubset A of Ω, and a principal part qa at each a ∈ A. To each meromorphic function onΩ there is associated a set of Mittag-Leffler data, consisting of its principal parts at all itspoles. The Mittag-Leffler problem asks whether the map so defined, from meromorphicfunctions to Mittag-Leffler data, is surjective. In other words, given an arbitrary set ofMittag-Leffler data, can we find a meromorphic function that realizes it?

Theorem 34.1 (Mittag-Leffler). Let Ω ⊆ C be an open set and let A be a discrete subsetof Ω. For each a ∈ A let qa be a principal part at a. Then there exists a meromorphicfunction f on Ω which has poles exactly at each a ∈ A, and principal parts qa.

Proof. For each a ∈ A choose a smooth compactly supported bump function αa which isequal to 1 near a, in such a way that the supports of the functions αa are all disjoint.(This can be done because of the discreteness of A.) Form the function

ψ(z) =∑a∈A

αa(z)qa(z)

on Ω \A. Note that ∂ψ∂z

is a smooth function vanishing near the points of A, so it can beextended to a smooth function u on Ω. By Theorem 33.5 there is a smooth ϕ on Ω with∂ϕ∂z

= u. Then f = ψ − ϕ is holomorphic on Ω \A (since it satisfies the Cauchy-Riemannequations) and f − qa is holomorphic near a for each a, so f is the desired meromorphicfunction.

Exercise 34.2. Give an alternative proof which uses Runge’s theorem directly (to as-semble Mittag-Leffler solutions on relatively compact subsets of Ω), rather than via thesolution of the ∂-equation, as here.

There is also a multiplicative version of the Mittag-Leffler problem, called the Weier-strass problem. As we observed above, a principal part near a is just an element of aquotient group Ma/Oa, where Ma denotes the additive group of functions meromorphicnear a, and Oa denotes the additive subgroup of holomorphic functions. (To be precise,Ma is the group of germs of meromorphic functions near a, and Oa the subgroup ofgerms of holomorphic functions: the germ of a function is its equivalence class modulothe relation “equality in a neighborhood of a”.) Analogously, we could consider M ∗

a /O∗a :

here M ∗a is the multiplicative group of germs of meromorphic functions near a and O∗a the

subgroup of germs of nonzero holomorphic functions near a. Any meromorphic functionon Ω will define “Weierstrass data”, a list of elements of the quotient group M ∗

a /O∗a at

its poles and zeroes A; and by analogy with the Mittag-Leffler problem we could pose theWeierstrass problem of finding a meromorphic function with prescribed Weierstrass data.


105

Weierstrass data, in fact, are rather simple things.

Lemma 34.3. The quotient group M ∗a /O

∗a defined above is naturally isomorphic to Z.

The isomorphism sends a meromorphic function f to the unique integer n such that f(z) =(z − a)ng(z) near a, where g(z) is holomorphic and nonzero at a.

In other words, n is the order of the zero (or minus the order of the pole) of f at a.

Proof. The proof is immediate once one knows that such a function g exists, which followsfrom Laurent’s theorem and the definition of a pole.

The Weierstrass problem is therefore equivalent to finding a meromorphic function on Ωwith prescribed zeroes and poles (“prescribed” here is taken to include prescribing theirorder). The list of integers attached to discrete points A which give the orders of thezeroes and poles — what we have called the “Weierstrass data” — is known as a divisor.

Remark 34.4. The version of this problem for entire functions — find an entire functionwith prescribed zeroes aj — was already solved by Weierstrass. His approach was tomodify the naive infinite product ∏

j

(1− z

aj

)(which does not converge in general) by convergence-producing exponential “weight fac-tors”. We saw an explicit example of this in the infinite product for the gamma function,Lemma 18.4.

The positive solution to the Weierstrass problem can also be derived from the solvabilityof the ∂-equation. Rather than give a proof directly, though, we’ll take a detour to explainhow all this discussion is related to sheaf theory.

Definition 34.5. Let X be a topological space. A sheaf of abelian groups over X is givenby a topological space S and a continuous surjection π : S → X such that

(i) π is a local homeomorphism.(ii) Each stalk Sx := π−1x has the structure of a (discrete) abelian group.

(iii) The group operations are continuous.

A section s of S over an open subset U of X is a continuous map s : U → S such thatπ s = 1U .

Example 34.6. Let p be a point of X and let G be any abelian group. A sheaf S canbe defined as follows: as a set, S = (G× p)∪ (0 × (X \ p) (with the obvious mapto X and group structures), and the open sets of S are those of the forms

(a) 0× U , where U ⊆ X is open and does not contain p.(b) (0× V \ p) ∪ (g, p), where g ∈ G and V ⊆ X is open and contains p.

Note that this topology is not Hausdorff. A sheaf of this kind is called a skyscraper.

Example 34.7. Let X be an open subset of C (or any Riemann surface if you prefer).The sheaf O is defined by setting Ox equal to the (additive) abelian group of germs ofholomorphic functions at x. To define the topology on O, proceed as follows. For an opensubset U ⊆ X and a holomorphic function f on U , there is a natural section sf : U → O


106

which sends each a ∈ U to the germ of f at a. The image of this map is a subset ofO which we will call the basic set determined by f and U . It is a tautology that if twobasic sets intersect, their intersection is a basic set. Therefore, we may define a topologyon O by saying that a subset of O is open if it contains a basic neighborhood of each ofits points. It is not hard to check that this makes O into a sheaf, the sheaf of germs ofholomorphic functions on X. Similarly we may define sheaves M , O∗, M ∗ as used above.

Remark 34.8. In contrast to the previous example, the sheaf O is a Hausdorff space.This is because of the uniqueness of analytic continuation. Suppose that we have distinctpoints f and g of Oa, that is, distinct germs of holomorphic functions at the same point a.We want to show that they have disjoint neighborhoods in O. Suppose they are germs ofholomorphic functions f and g respectively, and let U be a connected neighborhood of a onwhich f and g are defined. I claim that sf (U) and sg(U) are disjoint basic neighborhoodsof f and g respectively. Indeed, if they intersect (say over a point b ∈ U), that means thatf and g have the same germ near b. That is, f and g agree near b, so they agree on U bythe uniqueness of analytic continuation, and then f = g contrary to hypothesis.

We can carry out the usual sort of algebraic operations “stalk-wise” on sheaves. Forexample, we can form the short exact sequence of sheaves

0→ O →M →M /O → 0.

Even though O and M are Hausdorff, the quotient is not — it is of “skyscraper” type.A section of M /O over U is a collection of Mittag-Leffler data, so that the Mittag-Leffler problem is equivalent to the question whether the surjection of sheaves M →M /O induces a corresponding surjection on spaces of sections. Similarly, the Weierstrassproblem is equivalent to the corresponding question relative to the sheaf exact sequence

0→ O∗ →M ∗ →M ∗/O∗ → 0.

Questions of this sort belong to cohomology theory.

Theorem 34.9. Let X denote a paracompact Hausdorff space. For any sheaf S (ofabelian groups) over X there are contravariant functors X 7→ Hq(X; S ) (to abeliangroups) with the following properties:

(a) H0(X; S ) = Γ(X; S ) is the group of sections of S (over X);(b) For any short exact sequence of sheaves

0→ A → B → C → 0

there is a long exact sequence of cohomology groups

. . .→ Hq(X; A )→ Hq(X; B)→ Hq(X; C )→ Hq+1(X; A )→ . . . .

Thus the solutions to the Mittag-Leffler and Weierstrass problems depend on studyingthe groups H1(X; O) and H1(X; O∗) respectively. Next time, we’ll see how the ∂-equationhelps us do that.


107

Lecture 35Sheaves and complex analysis

For any endomorphism (of a sheaf of abelian groups) there is a natural notion of support,namely, the closure of the set of points on which the endomorphism is non-zero.

Definition 35.1. Let S be a sheaf of abelian groups over X (a paracompact Hausdorffspace). We say S is fine if for every locally finite open cover Uα of X, there existendomorphisms uα of S with Support(uα) ⊆ Uα and∑

α

uα(x) = identity at x ∀x ∈ X.

(Notice that the sum is finite for each x.)

For example, the sheaf C∞ of germs of smooth functions (on a manifold X) is fine —use a partition of unity. Similarly for other “smooth” objects, e.g. the sheaf of germs ofk-forms. The sheaf O is not fine.

Proposition 35.2. If S is a fine sheaf, then Hq(X; S ) = 0 for all q > 1.

Now let S be any sheaf. A fine resolution of S is an exact sequence

0→ S → S0 → S1 → S2 → . . . ,

where all the sheaves Sk are fine.

Proposition 35.3. If S has a fine resolution as above, then the cohomology groupsHq(X; S ) are isomorphic to the cohomology groups of the complex

Γ(X; S0)→ Γ(X; S1)→ Γ(X; S2)→ . . .

of spaces of sections of the resolving sheaves.

This is an easy consequence of Theorem 34.9 for a resolution of length two, which isactually all we’ll need for our Riemann surface application. The general is most simplyexpressed using a little extra homological trickery (spectral sequences). Notice that thePoincare lemma tells us that taking Sk to be the sheaf of germs of k-forms on a manifoldM , with the operator being the exterior derivative, gives a fine resolution of the constantsheaf. The de Rham theorem — that the de Rham complex computes cohomology withreal coefficients — therefore follows from the proposition above.

Now let X be an open subset of C and let C∞ denote the sheaf of germs of smooth(complex-valued) functions. This is a fine sheaf. The sheaf sequence

0 // O // C∞∂∂z // C∞ // 0

is exact by Lemma 33.3, so it is a fine resolution of O. Thus

H1(X; O) = Coker(∂ : C∞(X)→ C∞(X)).

By Theorem 33.5 this cokernel is 0.A similar analysis can be applied to the Weierstrass problem of constructing a mero-

morphic function with zeroes and poles of prescribed orders. By the same argument as


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above, to see that a meromorphic function on X can always be constructed with pre-scribed Weierstrass data, it suffices to see that H1(X; O∗) = 0. Now there is an exactsheaf sequence coming from the exponential map

0 // Z // Oexp // O∗ // 0

and the associated long exact cohomology sequence sandwichesH1(X; O∗) betweenH1(X; O)(which vanishes, as we have already observed, because of Theorem 33.5) and H2(X;Z)(which vanishes for any non-compact connected 2-manifold).

Remark 35.4. Notice that things are quite different here if the open subset of C is re-placed by a compact Riemann surface X. Then, in particular, there is an obstructionin H2(X;Z) = Z to constructing a meromorphic function with prescribed Weierstrassdata. This obstruction is just the difference between the total order of poles and totalorder of zeroes in the prescribed data. (For example, Proposition 26.6 established thatthis obstruction must vanish for elliptic functions—that is, meromorphic functions on thetorus.)

The sheaves O and M can also be applied to clarify the classical idea of analyticcontinuation.

Definition 35.5. Let f be holomorphic on D = D(a; r). Then a boundary point b ∈ ∂Dis regular for f if there are a disc D(b; ρ) and a holomorphic function fb on D(a; r)∪D(b; ρ)such that f = fb on D. Otherwise, b is singular for f .

Note that this is a more general use of the term “singular” than in our discussion ofisolated singularities. Clearly, the regular points form an open subset of ∂D.

Proposition 35.6. If (with the above notation) the Taylor series for f has radius ofconvergence exactly r, then there is at least one singular point on ∂D.

Proof. Suppose not. Then for each b ∈ ∂D there is a disc Db = D(b; ρb) such that f hasan analytic continuation fb to D ∪Db. Since ∂D is compact, finitely many of the Db, sayDb1 , . . . , Dbm , cover ∂D. Since Dbj∩Dbk is connected for all j, k, the analytic continuationsfbj and fbk agree wherever they are both defined; so there is an analytic continuation gof f to D ∪Db1 ∪ · · · ∪Dbk . But this is an open set that contains the closure of D, so bycompactness it contains D(a; r+ ε) for some r. It follows from Taylor’s theorem 9.6 thatthe radius of convergence of the power series is at least r+ε/2, contrary to hypothesis.

Is it possible that there be no regular points? The answer is yes.

Example 35.7. Consider the function

f(z) =∞∑n=0

z2n = z + z2 + z4 + z8 + . . . .

The radius of convergence of the power series is 1. However, this function satisfies theequation

f(z) = f(z2) + z.

Considering positive real values of z, we have

f(2−1/2) = f(2−1) + 2−1/2 > 2−1/2 > 1/2.


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But similarly f(2−2−k) > f(2−2−k+1

) + 1/2 so inductively f(2−2−k) > k/2. This shows

that f is unbounded on the ray [0, 1], so 1 is a singular point. But now it follows fromthe functional equation that f is also unbounded on the ray leading to any dyadic angleexp(2πim2−n), m,n ∈ N, so these are also singular points. Since the set of singularpoints is closed and the dyadic angles are dense on the unit circle, this implies that everyboundary point is singular. In this circumstance we say that ∂D is a natural boundaryfor f .

Because of multi-valuedness there need not be a largest domain to which a given func-tion can be analytically continued. Classically, then, one considers analytic continuationsof a function f defined on a disc D = D(a0; r0) by building a chain of discs in the follow-ing way: choose (if possible) a regular a1 ∈ ∂D(a0; r0) and continue f to D(a1; r1); thenchoose (if possible!) a regular a2 ∈ ∂D(a1; r1) and continue f to D(a2; r2); go on in thisway. One then refers to this as the analytic continuation of f along the chain of discs oralong the polygonal path whose vertices are a0, a1, a2, etc.

The discussion becomes much cleaner if we use the sheaves O and M . We’ll expresswhat follows in terms of O (analytic continuation) but one can use M (meromorphiccontinuation) instead.

Lemma 35.8. O is a Riemann surface.

Proof. We already know it is a Hausdorff space, we have to provide it with some charts.Indeed, let U ⊆ X be open and let sf (U) ⊆ O be a basic open set described by aholomorphic function f on U . Then the projection π : sf (U) → U will serve as a chartfor U ; it is trivial that the transition functions for these charts are holomorphic, becausethey are in fact the identity function on X!

Definition 35.9. Let f be the germ of a holomorphic function at a ∈ C, and let γ : [0, 1]→C be a path starting at γ(0) = a. An analytic continuation of f along γ is a lifting of γto O, beginning at f: that is, a path γ : [0, 1]→ O such that the diagram

0

f // O

π

[0, 1]

γ

>>

γ // C

commutes.

Let U ⊆ X be open and let f be a germ at a point of U . The Riemann surface of f overU is the collection of all germs that can be obtained by analytically continuing f along allpossible paths in U . That is to say

Definition 35.10. The Riemann surface Σf(U) of f over U is the (path) connected com-ponent of π−1(U) that contains f.

The map π : Σf(U) → U is a local homeomorphism, but local homeomorphisms don’thave to be covering maps (indeed, π does not even have to be surjective, as Example 35.7


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shows). If π is a covering map, we say that U is regular for f. Standard topology nowgives:

Theorem 35.11. U is regular for f if and only if f can be analytically continued alongevery path in U . When this is the case, the result of that continuation depends only onthe (endpoint-fixed) homotopy class of the path along which continuation takes place. Inparticular, continuation depends only on the endpoint if U is simply-connected.

This is the monodromy theorem. Results like Lemma 23.3, for example, are specialcases.


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