COMPLEX ANALYSIS

JONATHAN CAMPBELL

Contents

1. Complex Numbers and Their Geometry 21.1. Elementary Definitions and Facts 21.2. Useful Inequalities 41.3. Functions of a Complex Variable 41.4. Plane Geometry with Complex Numbers 42. Functions of a Complex Variable 62.1. Exponential Function 72.2. Logarithms and Branches 72.3. Complex Exponents 72.4. Sine and Cosine 82.5. Derivatives 83. Elementary Functions 103.1. Exponential Function 103.2. Logarithms and Branches 103.3. Complex Exponents 113.4. Sine and Cosine 114. Conformal Geometry 124.1. Stereographic Projection 124.2. Curves and Conformality 134.3. Linear Maps 144.4. Inversion 144.5. Fractional Linear Transformations 144.6. Other Maps 175. Integration 185.1. Integrating Along Curves 185.2. Cauchy-Goursat 215.3. Interlude: Integration and Contours 235.4. Cauchy Integral Formula and consequences 246. Practical Matters 277. Meromorphic Functions / Residue Calculus 277.1. Zeroes and Poles 277.2. Residue Formula 297.3. Fun Integrals 307.4. Integrals invovling za and log z 327.5. Summary of Integrals 327.6. Argument Principle and a Grab-bag of theoretical results 338. Sums and Products 358.1. Sums 358.2. Evaluation of Sums 379. Special Functions, Solving Equations 399.1. Gamma Function 3910. Laplace Transforms 40

1

10.1. More Complicated Laplace Transform 4410.2. Some Equations from Physics 4510.3. Confluent Hypergeometric Functions 4711. Fourier Series 48

These notes are a not particularly original explication of complex analysis. A lot of it has been drawnfrom Stein and Shakarchi. Other books consulted have been Gamelin and Ahlfors.

1. Complex Numbers and Their Geometry

First, I should say: why complex numbers? To what question are these things the answer? Well, aneasy question is: what is the square root of −1? This is kind a facile answer. The more realistic one isprobably “why don’t all polynomials have roots? The answer, of course, is that we were only allowing realroots. If we expand our notion of what a number can be (and why shouldn’t we?), then we can prove thatall polynomials of degree n have n roots, that is, they split into n linear factors. This is the FundamentalTheorem of Algebra, which we’ll prove in this course.Another way of stating this is that the complexnumbers are an algebraically closed field (don’t worry if you don’t know what that means).

1.1. Elementary Definitions and Facts. We will take it as known that you can manipulate complex num-bers with some facility. Nevertheless, we’ll review some basic concepts to fix notation and terminology.

We of course have

Definition. i will denote√−1.

A complex number will be denoted as z = x + iy where we have

Definition. In z = x + iy, we call x the real part of z and often write Re(z) to denote that. Similarly, wecall y the complex part of z and write Im(z).

Notation. The set of all complex numbers will be denoted C. Often people use C.

Remark. Much intuition for complex numbers comes from geometry and many of the definitions incomplex analysis are explicitly geometric. Of course, the reason for this is that C can be identified withR2, i.e. the plane. So, everything in complex analysis is thought of as living in the plane somehow.

Complex addition and multiplication proceed exactly as polynomial addition and multiplication do,with the one rule that i2 = −1. Everything that you think would be true, is true: both operations arecommutative and associative and multiplication and addition distribute.

Remark. One should think about what addition looks like geometrically. Multiplication is bit trickier tovisualize. However, it’s easier when |z| = 1 (see below for the definition of |z|)

Definition. The absolute value or modulus of a complex number z = x + iy is

|z| =√(x2 + y2)

Definition. The complex conjugate of z = x + iy is defined to be

z = x− iy

Remark. Note that |z|2 = zz.

We also all know that in R2 an effective thing to do is represent numbers in their polar form. Underthe identification C ∼= R2 we can do that for complex numbers as well — and in fact the structure ofmultiplication in this setting becomes much simpler.

Definition. Let z = x + iy then z can be written in the form z = reiθ . It is of course clear that

r =√

x2 + y2 = |z| θ = tan−1( y

x

).

2

Now, when we have z1 = r1eiθ1 and z2 = r2eiθ2 then

z1z2 = r1eiθ1 r2eiθ2 = (r1r2)ei(θ1+θ2)

Remark. What does multiplying by a complex number z1 with |z1| = 1 look like?

Definition. The θ above is called the argument of z, often denoted arg(z).

Note that

Definition. The principal value of the argument is the one where −π < θ < π.

Remark. Note that by definition, the following identities hold:

cos(arg z) =Re(z)|z| sin(arg z) =

Im(z)|z|

Example. One can use this useful multiplication to (re)derive and remember trigonometric identities.Consider, for example, the double angle formulae. These express cos 2θ and sin 2θ in terms of cos θ andsin θ. To rederive them note that

eiθ = cos θ + i sin θ

so that

(cos θ + i sin θ)2 = ei2θ = cos 2θ + i sin 2θ.

Computing the left hand side and separating out the real and complex parts you get

cos2 θ − sin2 θ = cos 2θ 2 sin θ cos θ = sin 2θ

In fact, note that it is easy to prove

Proposition. For any n ∈ N,

cos nθ + i sin nθ = (cos+i sin θ)n

Now, one thing that works a little differently in complex numbers is taking roots. This shouldn’t be sosurprising to you, but it requires some discussion anyway.

The thing to note is that numbers expressed in polar form have some degree of ambiguity to them.Note that

eiθ = ei(θ+2π) = ei(θ+4π) = · · ·

This means that if we have z = r0eiθ then z1/n can be any/all of

r1/n0 ei θ

n , r1/n0 ei( θ

n +2πn ), r1/n

0 ei( θn +

4πn ), . . .

To say this more formally:

z1/n = r1/n0 exp

[i(

θ

n+

2πkn

)]k = 0,±1,±2,±3, . . .

Remark. Note that the above will only be distinct for k = 0, . . . , n− 1.

This means that there is no ONE nth root for a complex number. Instead, there is a set of them.However, as with arguments, we single out one of the roots

Definition. The root with argument θn is referred to as the principal root.

Exercise. What are the nth roots of 1? What are the nth roots of z in terms of the nth roots of 1?3

1.2. Useful Inequalities. There are a number of useful inequalities that complex numbers satisfy. Ofcourse, most of them derive from geometric considerations. In particular, the following inequality isobvious if one realizes that C is a metric space with metric d(z1, z2) = |z1 − z2|. Well, it’s obvious beforethen, but that’s a nice general way of seeing it, since all metric spaces satisfy the triangle inequality.

Proposition (Triangle inequality). For complex numberes z1, z2 we have

|z1 + z2| ≤ |z1|+ |z2|and of course this holds for n complex numbers as well:∣∣∣∣∣ n

∑i=1

zi

∣∣∣∣∣ ≤ n

∑i=1|zi|

This can easily be used to derive the following.

Corollary. For z1, z2 ∈ C we have|z1 + z2| ≥ ||z1| − |z2||

Finally, the following inequality does not manifestly complex analytic, but it will be useful.

Proposition (Cauchy-Schwarz). For vector a, b ∈ Rn,

a · b ≤ |a||b|where the dot products and lengths are taken in Rn. Another way to state this is

∑i

aibi ≤√

∑ a2i

√∑ b2

i

and yet another way is (∑ aibi

)2 ≤∑ a2i ∑ b2

i

Proof. Figure out the proof or look it up.

1.3. Functions of a Complex Variable. A complex function is a function that eats a complex number andoutputs a complex number.

Definition. A complex function is a function f : C → C. That is, it is a rule associating to one complexnumber, another complex number. Since complex numbers can be written as two real numbers, one oftenwrites f (z) = u(z) + iv(z) where u, v : C→ R.

Example. f (z) = 1z or f (z) = z or f (z) = zz = |z|2.

Example. Any polynomialf (z) = anzn + an−1zn−1 + · · ·+ a1z + a0

Example. Rational functions. These are any functions of the form

f (z) =p(z)q(z)

where p, q are polynomials. As with rational functions in real variables, there are points where they arenot defined (in particular, the zeros of q(z)).

1.4. Plane Geometry with Complex Numbers. [[[section under construction]]]

Remark. The material in this subsection is not usually included in a complex analysis course, since itreally belows to the arithmetic of complex numbers. Nevertheless, it’s not usually taught in courses, andit’s kind of fun, so it’s well worth including.

It should come as no surprise that complex numbers can “algraicize” plane geometry. Let’s go throughthe exercise of encoding familiar notions of plane geometry in terms of complex numbers.Well, the notionsmay not be so familiar, so I’ll introduce some of them first.

4

Remark. In what follows, uppercase letters will be used to denote points in the Cartesian plane. Corre-sponding lowercase letters will be the corresponding complex number.

Definition. The centroid of a triangle is the point where the three medians of the triangle meet. You canthink of it as the center of gravity.

Definition. The orthocenter of a triangle is the point where the three altitudes meet.

Definition. The incenter of a triangle is the center point of an inscribed circled. It is located at theintersection of the three angle bisectors.

Definition. The circumcenter of a triangle is the center of the circumscribed circle. It is located at theintersection of the perpendicular bisectors of the triangle’s sides.

But we won’t use these quite yet, that was just a reminder. We’ll work up to these things.First, a little exercise

Example. Suppose the points of a triangle are 0, 1, and z. What is the area of the triangle? Well, the area is12 (base)(height). The base is obviously 1. The height is the imaginary part of z. But that’s just z−z

2 . Thus,the area is

14i(z− z)

Example. Given complex numbers, a, b, c let’s find the area of the triangle that these determine, and writein only in terms of a, b, c. First, things are easier if we translate so that one point is the origin. We do thisby subtracting off, say, a. Our points are now 0, b− a, c− a. It’s even easier if one of the points is 1. Toget this, we multiply the remaining points by 1

b−a to get 0, 1, c−ab−a . This of course scales the area of our

original triangle by 1|b−a|2 , so we’ll need to multiply that back in. However, we’re back in the situation of

the previous example, so the area of the triangle with vertices 0, 1, c−ab−a is

14i

(c− ab− a

− c− ab− a

)=

14i

(c− a)(b− a)− (b− a)(c− a)|b− a|2 .

But remember, this is off by a factor of |b− a|2! So, we get that the area is14i[(c− a)(b− a)− (b− a)(c− a)]

We’ll now look at some more elaborate examples. Many of these will be left as (good) exercises.We’d like to be able to verify the basic geometric properties using only complex numbers.You should

verify the below.(1) parallel lines AB is parallel to CD if

a− bc− d

∈ R

(2) perpendicular lines AB is perpendicular to CD ifa− bc− d

∈ iR

(3) cyclic points: the points ABCD lie on a circle if

b− ac− a

/b− dc− d

∈ R

(4) triangle similarity: a triangle ABC is similar to PQR ifa− cb− c

=p− rq− r

We could have also used the following little criterion to check that the triangle was equilateral

Proposition. Let ABC be a triangle. Then it is equilateral if

a + bω + cω2 = 0 ω = e2πi/3

5

Exercise. Prove the above proposition.

I’m going to elaborate a little on the conditions of colinearity and perpendicularity.. Suppose a, b lie onthe unit circle, and consider a point z on the line AB. Our condition that z lie on this line is that

z− ab− a

=z− ab− a

.

Since a and b are on the unit circle, aa = 1 = bb. Multiply the right hand side by ab in both the numeratorand denominator:

z− ab− a

· abab

=zab− |a|2b

a|b|2 − |a|2b=

zab− ba− b

Thus, we have a− z = zab− b orz + zab = a + b

This is a nice closed form condition for z lying on the line AB.Similar considerations give that if a, b are on the unit circle and c is another point, in order for z to be

on the line perpendicular to AB passing through the point C, we must have

z− abz = c− abc

Proposition. Suppose a, b lie on the unit circle. We have

a− ba− b

= −ab

Proof.

Proposition. Let AB and CD be intersecting chords. Their intersection is

ab(c + d)− cd(a + b)ab− cd

Proposition. Let ABC be a triangle on the unit circle. The orthocenter is given by

o = a + b + c

Proof.

We now present two examples of how to prove actual geometric theorems.

Example. Let S be tangent tot he circumcricle of an actue triangle ABC at B Let K be the projection of theorthocenter onto the line S. Let L be the midpoint of side AC. Then BKL is isoceles.

2. Functions of a Complex Variable

A complex function is a function that eats a complex number and outputs a complex number.

Definition. A complex function is a function f : C → C. That is, it is a rule associating to one complexnumber, another complex number. Since complex numbers can be written as two real numbers, one oftenwrites f (z) = u(z) + iv(z) where u, v : C→ R.

Example. f (z) = 1z or f (z) = z or f (z) = zz = |z|2.

Example. Any polynomialf (z) = anzn + an−1zn−1 + · · ·+ a1z + a0

Example. Rational functions. These are any functions of the form

f (z) =p(z)q(z)

where p, q are polynomials. As with rational functions in real variables, there are points where they arenot defined (in particular, the zeros of q(z)).

We now move on to more complicated functions, and slowly build up our store of them.6

2.1. Exponential Function. One of our favorite functions is the exponential functions. Since we knowhow to define it for real numbers, we can use that to define it for complex numbers.

Definition. Let z = x + iy, thenez = exeiy eiy = cos y + i sin y

The more usual definition of ez is in terms of power series. You’ve seen power series in earlier mathcourses, but we haven’t introduced them here yet. As a consequence, the following definition will bede-emphasized at the moment, but it still very important.

Definition. Let z ∈ C. Then

ez =∞

∑k=0

zk

k!= 1 + z +

z2

2!+

z3

3!+

z4

4!+ · · ·

The exponential function possesses all the properties you would hope and expect. For example, it iseasy to prove that ez1+z2 = ez1 ez2 .

2.2. Logarithms and Branches. Of course once we’ve define the exponential function, we should probablydefine the logarithm.

Definition. Let z ∈ C then definelog z = log |z|+ i arg z

Remark. Write this in terms of polar coordinates.

Note that since arg is not a unique thing, the log is not a unique thing. That is, it is a multi-valuedfunction. Or, to put it another way, we can think of it as a set.

Example. Let z = 3 + 4i. Then

log(z) = log 5 + i arg(3 + 4i) arg(3 + 4i) = tan−1(

34

)+ 2nπ

Remark. The usual formulalog(z1z2) = log z1 + log z2

holds, but one has to make sure that one is using the correct sets of logarithms.

At this point it has become use to introduce the notion of “branches.” We’ve been using them implicitlyup til this point, but we should make their use explicit.

Definition. Let f be a multi-valued function defined on Ω. Then a branch of f is a holomorphic functionf defined on some Ω ⊂ Ω and such that f (z) ∈ f (z) (recall f (z) is a set).

Remark. It is very typical that in order to define a branch, one simple removes a line or a curve from Ω.That is, if γ is a line or curve in Ω, Ω = Ω− γ.

Definition. The γ above is called a branch cut.

2.3. Complex Exponents. Once we’ve defined exponentials and logs, we can easily define arbitrary expo-nentiation. Let z ∈ C and c ∈ C.

Definition. We definezc = exp(c log z)

Example. What is ii? Well,

log i = log |i|+ i arg i = 0 + iπ

2and now

ei·i π2 = e−

π2 ≈ .2

and is a real number!7

2.4. Sine and Cosine. Consider the following equations

eiθ = cos θ + i sin θ

e−iθ = cos θ − i sin θ

so we can solve to get

cos θ =eiθ + e−iθ

2sin θ =

eiθ − eiθ

2These equations were all for real numbers. But, we take them as our guide and make the followingdefinitions.

Definition. Let z ∈ C then

cos z =eiz + e−iz

2sin z =

eiz − eiz

2

Remark. Verify that every trig identity you can think of still holds.

2.5. Derivatives. The way we define a derivative in calculus is a completely reasonable definition, and itwill be the definition of derivative we use in complex analysis, too. However, the class of functions thatare differential are much, much better behaved than in single variable calculus.

I should first point out that there is a notion of “continuous”, which is exactly what you think it shouldbe. It is not used much.

Definition. Let f : Ω→ C be a complex function. Then we define the derivative to be the limit (if it exists)

f ′(z0) = limh→0

f (z0 + h)− f (z0)

h

Definition. If the limit above exists we say that f is holomorphic at z0

Definition. If the limit exists for every z0 ∈ Ω then we say that f is holomorphic on Ω

Remark. From a complex function f : C → C one can extract a function f : R2 → R2. Can you see howthe above definition is different from demanding all directional derivatives of such a function exist? Thedefinition of holomorphic function is an expressly complex notion.

Example. In contrast to the real case, it is easy to write down a function that is nowhere holomorphic. Forexample, f (z) = z.

First, we note that because the proofs of the usual derivative rules are formal (i.e. basically just manip-ulations of symbols), we are entitled to the following theorem

Theorem. Let f , g : Ω→ C be holomorphic functions. Then

(1) f + g is holomorphic on Ω and ( f + g)′ = f ′ + g′

(2) f g is holomorphic on Ω and ( f g)′ = f ′g + f g′

(3) If g(z0) 6= 0 then f /g is holomorphic at z0 and(fg

)′=

f ′g− f g′

g2 .

This gives us a host of examples to work with. In fact, it is easy to see that if f (z) = z, then f ′(z) = 1and then from that and the above we get the following examples.

Example. If f (z) = anzn + an−1zn−1 + · · ·+ a1z + a0 then

f ′(z) = nanzn−1 + (n− 1)an−1zn−1 + · · ·+ a1

Example. If f (z) = 1z then f ′(z) = − 1

z2 .8

Let’s examine at least a necessary condition for a limit of the type above to exist. It will be useful to writef (z) = u(z) + iv(z) for this development. As with all two dimensional limits, if the limit exists, it shouldbe the same whatever direction we come from. In this case, we could come from a totally real direction,i.e.

f ′(z0) = limh→0

u(x0 + h, y0) + iv(x0 + h, y0)− u(x0, y0)− iv(x0, y0)

h

=∂u∂x

+ i∂v∂x

Or, we could come in from a purely complex direction

f ′(z0) = limh→0

u(x0, y0 + h) + iv(x0, y0 + h)− u(x0, y0 + h)− iv(x0, y0 + h)ih

=1i

(∂u∂y

+ i∂v∂y

)=

1i

∂u∂y

+∂v∂y

Since the above have to be equal, we can set the complex parts of each equal to each other. This gives theequations

∂u∂x

=∂v∂y

∂u∂y

= − ∂v∂x

Definition. The above equations are called the Cauchy-Riemann equations.

Remark. For notational ease, the following

∂

∂z:=

12

(∂x +

1i

∂

∂y

)∂

∂z=

12

(∂

∂x− 1

i∂

∂y

)We then have

Proposition. If f is holomorphic at z0 then

• We have∂ f∂z

(z0) = 0

• We have

f ′(z0) =∂ f∂z

(z0)

• If we write f as a function F : R2 → R2 then

det JF(x0, y0) = | f ′(z0)|2

Proof. These will be left as exercises.

All right. We have some necessary conditions for a function to be holomorphic, but we don’t yet havesufficient conditions. However, turns out that they are one and the same (or close to it).

Theorem. Let f : Ω→ C and write f = u + iv. Suppose that

• u, v are C1 functions• u, v satisfy the Cauchy Riemann equations.

Then, f is holomorphic on Ω.9

Proof. We have to show that a certain limit exists. To show that it exists, we treat the real and complexpart of it separately. Using our knowledge of calculus, and that u, v are each C1 functions, we are entitledto write

u(x + h1, y + h2)− u(x, y) =∂u∂x

h1 +∂u∂y

h2 + |h|ε1(h)

v(x + h1, y + h2)− v(x, y) =∂v∂x

h1 +∂v∂y

h2 + |h|ε2(h)

where ε1, ε2 → 0 as h→ 0.Now we know that ∂u

∂x = ∂v∂y and ∂u

∂y = − ∂v∂x by the Cauchy-Riemann equations, so we can write

f (z + h)− f (z) =(

∂u∂x− i

∂u∂y

)(h1 + ih2) + |h|(ε1 + ε2)

Dividing by h = (h1 + ih2), we note that the limit exists. Furthermore,

f ′(z) =∂u∂x− i

∂u∂y

=∂u∂x

+1i

∂u∂y

= 2∂u∂z

=∂ f∂z

3. Elementary Functions

I’ve defined what a complex function is, and I’ve even told you how to take derivatives, but we don’treally know any complex functions besides polynomials. We are going to quickly rectify that. However,I want to point out that you still don’t really know what complex function “look like.” That is, when wemeet real functions we often meet them in terms of graphs, and we have no idea what that looks like inthis case. You’ll have to deal with that lack of knowledge for the moment, but it will be fixed soon.

3.1. Exponential Function. One of our favorite functions is the exponential functions. Since we knowhow to define it for real numbers, we can use that to define it for complex numbers.

Definition. Let z = x + iy, thenez = exeiy eiy = cos y + i sin y

The more usual definition of ez is in terms of power series. You’ve seen power series in earlier mathcourses, but we haven’t introduced them here yet. As a consequence, the following definition will bede-emphasized at the moment, but it still very important.

Definition. Let z ∈ C. Then

ez =∞

∑k=0

zk

k!= 1 + z +

z2

2!+

z3

3!+

z4

4!+ · · ·

The exponential function possesses all the properties you would hope and expect

Lemma (Properties of the exponential function). We have(1) ez1z2 = ez1 ez2

(2) ez is holomorphic (proved later)(3) (ez)′ = ez.

3.2. Logarithms and Branches. Of course once we’ve define the exponential function, we should probablydefine the logarithm.

Definition. Let z ∈ C then definelog z = log |z|+ i arg z

Remark. Write this in terms of polar coordinates.

Note that since arg is not a unique thing, the log is not a unique thing. That is, it is a multi-valuedfunction. Or, to put it another way, we can think of it as a set.

10

Example. Let z = 3 + 4i. Then

log(z) = log 5 + i arg(3 + 4i) arg(3 + 4i) = tan−1(

34

)+ 2nπ

Remark. The usual formula

log(z1z2) = log z1 + log z2

holds, but one has to make sure that one is using the correct sets of logarithms.

At this point it has become use to introduce the notion of “branches.” We’ve been using them implicitlyup til this point, but we should make their use explicit.

Definition. Let f be a multi-valued function defined on Ω. Then a branch of f is a holomorphic functionf defined on some Ω ⊂ Ω and such that f (z) ∈ f (z) (recall f (z) is a set).

Remark. It is very typical that in order to define a branch, one simple removes a line or a curve from Ω.That is, if γ is a line or curve in Ω, Ω = Ω− γ.

Definition. The γ above is called a branch cut.

3.3. Complex Exponents. Once we’ve defined exponentials and logs, we can easily define arbitrary expo-nentiation. Let z ∈ C and c ∈ C.

Definition. We define

zc = exp(c log z)

Example. What is ii? Well,

log i = log |i|+ i arg i = 0 + iπ

2and now

ei·i π2 = e−

π2 ≈ .2

and is a real number!

3.4. Sine and Cosine. Consider the following equations

eiθ = cos θ + i sin θ

e−iθ = cos θ − i sin θ

so we can solve to get

cos θ =eiθ + e−iθ

2sin θ =

eiθ − eiθ

2These equations were all for real numbers. But, we take them as our guide and make the followingdefinitions.

Definition. Let z ∈ C then

cos z =eiz + e−iz

2sin z =

eiz − e−iz

2

Remark. Verify that every trig identity you can think of still holds.

Proposition. sin z, cos z are holomorphic.

Proof. Exercise.

11

4. Conformal Geometry

When we do calculus we do many things graphically: given the graph of a particular function you aretaught to roughly identify maxima and minima, points of inflection, points of discontinuity, etc. That is,given geometric information about the function, one can reasonably infer analytic information. You knowin your gut what continuous and discontinuous functions “look like” and you know what differentiableand not differentiable functions “look like.” You do not have this kind of facility with complex numbers,nor will you ever, probably (graphs of complex functions are necessarily 4 dimensional). There is howeverplenty of geometric intuition to be gleaned. The geometry of conformal maps turns out to be extremeleyrich.

In all of the cases below there will be two tacks we take to understand a mapping T.(1) Consider some shape, S (usually a line or cirlce) in the complex plane and see what happens to it

when we apply T, i.e. look at the image of S.(2) Consider some region, Ω (usually a circle, half-circle, half plane, or strip) and look at the image of

Ω.

4.1. Stereographic Projection. One thing that will be useful for you in visualizing maps of the complexplane is to think of the complex plane as a ball. This sounds weird, but the complex plane can becompletely identified with the unit sphere.

Consider the sphere x21 + x2

2 + x23 = 1. Let N denote the “north pole”, the point (0, 0, 1). We consider

the x1 − x2 plane as the complex plane. Consider a point x = (x, y) in the plane. We want to figure outwhere this projects to on the sphere. Let s(x) be the corresponding point on the sphere. We note that thevectors ~mb f xN and ~s(x)N are congruent, so we have

(x, y,−1) = c(x1, x2, x3)

This gives us that c = 11−x3

and

zN =

(x1

1− x3,

x2

1− x3,

11− x3

)which easily gives us that

(x1, x2, x3) 7→(

x1

1− x3,

x2

1− x3, 0)

.

Now suppose we want to go the other direction. Given z = x + iy in the “complex” plane, what does itcorrespond to on the sphere? Well, we know that

z = x + iy =x1 + ix2

1− x3.

Note also that

|z|2 =x2

1 + x22

(1− x3)2 =1− x2

3(1− x3)2 =

1 + x3

1− x3

It’s then not hard to see that

z 7→(

z + z1 + |z|2 ,

z− zi(1 + |z|2) ,

1− |z|21− |z2|

)How do various favorite constructions behave under stereographic projection?

Proposition. Stereographic projection takes lines to circles

Proof. A plane is determined by a point and a line. Draw the plane through N and the line in the plane.This intersects the sphere. The intersection of a sphere and a plane is a circle.

Proposition. Stereographic projection takes circles to circles.

Proof. Omitted for now.

An important remark for us is that the North Pole, N corresponds to the “point” ∞ in the complexplane.

12

Definition. The sphere above is often called the Riemann sphere. It corresponds to the complex planewith an extra point added in. That point is ∞.

4.2. Curves and Conformality. We’re first going to develop some easy properties of holomorphic func-tions that are quickly derived from properties we already know.

There is a chain rule for differentiating complex functions. We need a slightly different one.

Definition. A curve is a map z : [0, 1]→ C, denoted z(t).

Remark. As always, this can be viewed as a map f : [0, 1]→ R2

Proposition (The Chain Rule). Let γ(t) be a curve in C and f (z) a holomorphic function. Then

[ f (γ(t))]′ = f ′(γ(t))γ′(t)

Proof. As usual, write f (z) = u(x, y) + iv(x, y) and γ(t) = x(t) + iy(t). We then have

f (γ(t)) = u(x(t), y(t)) + iv(x(t), y(t)).

Take derivatives using the usual chain rule:

[ f (γ(t))]′ =(

∂u∂x

∂x∂t

+∂u∂y

∂y∂t

)+ i(

∂v∂x

∂x∂t

+∂v∂y

∂y∂t

)=

(∂u∂x

x′(t) +∂u∂y

y′(t))+ i(−∂u

∂yx′(t) +

∂u∂x

y′(t))

=∂ f∂z

γ′(t)

Let’s consider what happens when we apply a mapping f to z(t) and look at the derivative:

[ f (z(t0))]′ = f ′(z(t0))z′(t0)

which implies that

arg[ f (z(t0))]′ = arg( f ′(z(t0))) + arg(z′(t0))

If you notice, this means the difference in angle between f (z(t0))′ and z′(t0) is independent of the curve.

More concretely, suppose we are given another curve w(t), and further suppose that it intersects z(t) att0. The angle between z(t) and w(t) at t0 is the angle between z′(t0) and w′(t0), that is

∠1 = arg z′(t0)− arg w′(t0)

Similarly, the angle between f (z(t)) and f (w(t)) at t0 is

∠2 = arg([ f (z(t0))]′)− arg([ f (w(t0))]

′)

= arg f ′(z(t0)) + arg z′(t0)− arg f ′(w(t0))− arg w′(t0)

= arg z′(t0)− arg w′(t0)

Thus, ∠1 = ∠2. That is, when we apply a holomorphic map to two curves, it preserves the angles betweenthe two curves.

Remark. This is in marked contrast to the case of just continuous functions.

Definition. A conformal mapping is a map Ω→ C that preserves angles between curves.

Definition. A conformal equivalence is a conformal map with a conformal inverse.13

4.3. Linear Maps. One brand of holomorphic/conformal map is reasonably easy to visualize.

Definition. A linear map C→ C is given by

z 7→ Az + B

Let’s look at this bit by bit. First, we know what adding a complex number to another complex numberlooks like. That’s just vector addition. So that “+B” part is easy to visual. Now, what does multiplying zby A do? Well, A can be written as r0eiθ0 for some r0, θ0. Thus, multiplying by A scales z by a factor of r0and then rotates by an angle θ0.

Example. T(z) = (1 + i)z + 2i.

Example. Think about what this looks like on the Riemann Sphere.

4.4. Inversion. Another important transformation:

Definition. Inversion is just the transformation

z 7→ 1z

What does this look like in the complex plane? Well, let’s think about what 1z is. We can rewrite this as

1z=

z|z|2 =

(z|z|2

).

Thus, we can think of this as the composition of dividing z by it’s length squared, which has the effect ofmaking large complex numbers small and vice versa, and then applying complex conjugation. Note that

z|z|2 preserves the circle.

This isn’t such a great description, really. The better desciption is that it takes the Riemann sphere andinverts it! (See the proposition below)

Proposition. Inversion takes circles to circles and lines to lines.

Proof. Verify this yourself. Write out the most general formula for the equation of a circle and then seewhat happens when you apply z 7→ 1

z .

Proposition. On the Riemann sphere z 7→ 1z corresponds to rotating the North pole to the south pole along the

“imaginary” axis.

4.5. Fractional Linear Transformations. Fractional linear transformations are a type of map that is rea-sonably complicated, but we can still get our hands on. For this reason, they are a nice playground fordeveloping some intuition about how maps in the complex plane work geometrically. Of course, they areimportant for other reason (for the cognoscenti they model SL2(C))

Definition. A fractional linear transformation is a map of the form

z 7→ az + bcz + d

ad− bc 6= 0

What makes these things special? First, they form a group.

Proposition. Let T be a fractional linear transformation, and S another one. Then T S is also a fractional lineartransformation. Furthermore, each T has a T−1 such that T T−1 = T−1 T = Id.

Proof. The inverse is given as

T−1(z) =−dz + bcz− a

.

The rest can be verified easily.

Second, they can map three arbitrary to points to any other three arbitrary points, and they are uniquelydetermined by that property. This makes them a slightly souped up version of linear things. Recall fromlinear algebra that a linear map T : R2 → R2 is uniquely determined by where it takes two points.

14

Theorem. Given points z1, z2, z3 ∈ C∪ ∞, and points w1, w2, w3 ∈ C∪∞, there is exactly one fractional lineartransformation such that

T(z1) = w1 T(z2) = w2 T(z3) = w3

Proof. It is enough to show that we can map z1, z2, z3 to 0, 1, ∞ (why?). To this end, we can just write downthe transformation!

T(z) =z− z1

z− z3

z1 − z2

z1 − z0.

It is an easy check that this satisfies the property we want.

Example. Let’s figure out the FLT that sends (0, ∞, i) 7→ (0, 1, ∞). We could use the formula, but wemight as well just do it directly. If it sends 0 to 0 there must just be a z in the numerator. Sending i to ∞means there should be a z− i in the denominator. And ∞ to 1 means that the leading coefficients in thenumerator and denominator should be the same. This gives

T(z) =z

z− i.

Finally, fractional linear transformations are compositions of easier transformations, and in fact ourfavorite ones: dilations (scaling + rotation), translation and inversion

Theorem. All fractional linear transformations are compositions of(1) translation(2) dilations(3) inversion

Proof. Not so hard. Assume our fractional linear transformation is of the usual form. Then we just note

az + bcz + d

=ac+

bc− adc

1cz + d

and that right there is a composition of such things.

How do our various favorite shapes behave with respect to these transformations?

Theorem. Fractional linear transformations take circles to circles and lines to lines.

Proof. We know that inversion does this. Translation obviously does. Dilation also obviously does. Thus,any composition of them will. Done!

Example. Let’s examine how some things transform under our earlier examples. Let’s consider

T(z) =z

z− i• What happens to the imaginary axis?: Any fully imaginary number can be written as z = ri

where r ∈ R. Then

T(z) =ri

ri− i=

rr− 1

∈ R

Thus, any purely imaginary number is taken to a real number. Furthermore, every real numbercan be obtain this way. So the imaginary axis gets mapped to the real axis.• What happens to the real axis? We take r ∈ R. Then

T(r) =r

r− i=

r2 + rir2 + 1

.

We now appear to be an an impasse. This is not obviously anything. But! We know it has to be aline or circle. It sure as hell isn’t a line, so it has to be a circle. A little playing around (and a fairamount of algebra) will convince you that it is in fact the circle(

x− 12

)2+ y2 =

14

that is, a circle of radius 1/2 centered on (1/2, 0).15

• What happens to the circle |z| = 1? We might as well write z = eiθ here. Then

zz− i

=eiθ

eiθ − i=

ei2θ + ieiθ

ei2θ + 1=

eiθ + ieiθ + e−iθ

=cos θ + i(sin+1)θ

2 cos θ=

12+ i(

sin θ + 12 cos θ

)So, as θ varies, the complex coordinate is going between −∞ and ∞ and the real coordinate isstaying the same. Thus, the image of |z| = 1 is the line Re(z) = 1

2.

Remark. Sadly (or maybe happily), all examples have this kind of flavor — there is not a formulaic wayof dealing with them. It’s a little bit of an art form.

Sometimes, huge amounts of computation are not needed:

Example. Let’s consider the FLT that takes

(−1, 1, i) 7→ (−i, 2i, 0)

Where does this take the unit circle?

There aren’t really general ways of dealing with these fractional linear transformations. But there aremildly general classes that we can deal with, and relatively fun to visualize.

Example (Circles of Appolonius). Consider transformations of the form

T(z) = kz− az− b

.

What do these guys do? Well then take

a 7→ 0

b 7→ ∞

What this translates to is that they take circles through a, b to lines through the origin. So, the image ofcircles through a, b is lines through the origin.

Lines through the origin are, of course, a natural thing to look for when considering images. Howabout circles centered around the origin? That is, what kind of shape gets taken to circles around theorigin ? If T(z) is a circle about the origin, then we have |T(z)| = r for some r, that is∣∣∣∣ z− a

z− b

∣∣∣∣ = r|k|

What shape do these equations determine? Well, they’re not lines, so they’re definitely circles. And whatthe equations are saying is for z, the ratio of the distance from z to a to the distance from z to b is constant.One could easily write down formulas for that or make them via geometric construction. These are calledcircles of Appolonius with limit points a, b. I don’t know why.

Following Ahlfors, let’s denote family of circles that go through a, b as C1 and the family of circles ofAppolonius as C2. There are a few easy facts to see. First, C1 and C2 are automatically perpendicular. Thisis because their images are and holomorphic functions preserve angles.

Example. The above was the case when we had a mapping to 0 and b mapping to ∞. But 0 and infinityaren’t special. We could have chosen any other points a′ and b′ instead. In this case, we can write thetransformation as

w− a′

w− b′= k

z− az− b

.

In this case, the circles that go through a, b turn into circles that go through a′, b′ and the circles ofAppolonius with limit points a, b turn ito circles of Appollonius with limit points a′, b′.

16

4.6. Other Maps. What kind of behavior do our other favorite holomorphic functions have? As always,it’s good to think about what happens to circles and lines. Additionally, we’ll think about what happensto certain regions. The regions we usually consider are

(1) The disc and various half discs(2) Half planes(3) Various strips

Since we’re going to work with discs and half planes so much, let’s investigate transformations that gobetween those two

Proposition. Define functions

F(z) =i− zi + z

G(z) = i1− z1 + z

then both are conformal, F : H→ D and G : D→ H and F and G are inverse.

Proof. This will be an exercise.

Example ( f (z) = za, a ∈ R). Let’s look at circles and lines. First, za takes the unit circle to itself. It takesany other circle centered at the origin to another circle centered at the origin, but with radius scale.

Consider the ray through the origin determined by the line θ. Under za this gets turned into the linedetermined by the ray rθ.

We can also consider the region spanned by the positive real axis and the ray determined by θ. It lookslike a slice of pie [[[picture]]]. This gets mapped to the region spanned by the positive real axis and thethe ray determined by the angle αθ.

More specifically, if we consider the region θ : θ < πα then this region gets mapped to the upper half

plane.

Example. eiz. What does this bad boy do? Well, we know what it does to the real line. It takes the realline to the unit circle. So that’s one data point. What about if we add a complex bit to it? We haveei(x+iy) = eixe−y. Thus, adding in a positive complex component shrinks the number. So, if we considerthe region

z : −π ≤ Re(z) < π, Im(z) > 0then the mapping z 7→ eiz takes this to the unit disc D. One can also see that

z : −π

2< Re(z) <

π

2, Im(z) > 0

gets taken to the upper half circle.

Example. log. Log takes the upper half plane H to the strip

z : 0 < Im(z) < πthat is, the strip between the real line and line Im(z) = iπ.

Let’s also think about what log does to the unit circle. Waht does it do to the boundary? log z =log |z|+ i arg z so if z = eiθ this is just iθ. So, the unit circle gets mapped to the part of the complex axisbetween −iπ and π. Now, if a point is inside the circle, then log |z| < 0 since |z| < 1, so points insidethe circle get mapped to negative numbers. Similarly, if a point is outside the circle it gets mapped to thepositive numbers. So, one can see we can chop up this mapping and understand where it takes certaincomponents.

• We haveH 7→ z : 0 < Im(z) < iπ

• The whole plane (minus a branch) gets mapped toa strip as well:

C \ [0,−∞] 7→ z : −iπ < Im(z) < iπ• The upper half disc gets maped to a half-strip:

H ∩D = z : Re(z) < 0, 0 < Im(z) < iπ• The lower half disc gets mapped to the “positive” half strip

17

Example. We consider the mapping

f (z) = −12

(z +

1z

).

This is a conformal map from the upper half disc to the upper half plane

Example. sin z . This will be an execise.

We can start plugging these things together to kind of visualize how one maps certain regions to certainother regions.

Example. Suppose we want to map the regionθ :

7π

8< θ <

9π

8

.

to, for example, an upper-half disc, D ∩H.This is a wedge of angle π/4. We want to rotate it so that it sits on the x-axis, so first, we’d apply a

transformation T1(z) = e−7π8 to rotate it. Then, we’d expand the wedge by applying T2(z) = z4, which

maps the wedge to the upper half plane. Then we know that T3(z) = i−zi+z maps the upper half plane to

the upper half disc. Thus,T := T3 T2 T1

maps our original wedge to an upper half disc and does it in a conformal manner.

5. Integration

We’ve introduced functions, done some kind of calculus with them, learned how to “graph” them. It’stime for integration. If you haven’t thought about it, think for a second of how “integrating” a complexfunction should work. If you think about what you might want, something like a two-dimensional Rie-mann integral measuring “complex area” you might come to the conclusion that this is sort of nonsenseand may not give anything interesting. You’d be right. There is, however, a notion of integration that doesmake sense. It is basically the complex version of line integrals, which hopefully you saw in multivariablecalculus. Given a curve γ, we’ll be able to define an integral along the curve, and denote it∫

γf (z) dz.

I’ll give away the game now and tell you that the thing which will buy almost all results in complexanalysis is the result that for f holomorphic and γ a CLOSED curve, we have∫

γf (z) dz = 0.

The importance of this result to complex analysis cannot be overstated.

Remark. You might want to contemplate other instances where you know that integrals around closedcurves are zero (e.g. conservative forces in physics). There is a relationship there which we’ll get into later.

5.1. Integrating Along Curves. Before we introduce the notion of integration, we need some terminology,though. We’ve seen what a “curve” is. We need to expand our idea of this a little bit and be more precisewith our terminology.

Definition. A curve is a map z : [a, b]→ C where a, b ∈ R.

Although the following definition was implicitly used above, we should actually define it.

Definition. For a curve z(t), the derivative is defined to be

z′(c) = limh→0

z(c + h)− z(c)h

Note here that h is a real number.

Definition. A curve z(t) is smooth if z′(t) exists and is continuous on [a, b] and z′(t) 6= 0 for t ∈ [a, b]18

Remark. This conflicts with the usual definition of “smooth” as “infinitely differentiable” or C∞. Never-theless, this seems to be how the word is used in complex analysis-land.

Definition. A curve z(t) is piecewise-smooth if there are points

a = a0 < a1 < · · · < an = b

where z(t) is smooth on intervals [ai, ai+1].

Definition. A piecewise-smooth curve is closed if z(a) = z(b).

Definition. A piecewise-smooth curve is simple if it does not intersect itself

Remark. We defined all the vocabulary above, but typically I’ll just say “curve” and leave it to context todetermine the adjectives.

By now we have the means of defining integrals.

Definition. Let f be a complex function (not neccessarily holomorphic) defined on Ω and let γ be a curvein Ω (i.e. a map [a, b]→ Ω). Then we define∫

γf dz =

∫ b

af (γ(t))γ′(t) dt

We can now compute these guys like we would compute interals of single variable functions. Forexample.

Example. Let γ be the line from 0 to 1 + i. We parameterize this by

γ(t) = (1 + i)t, t ∈ [0, 1].

Then ∫γ

z3 dz =∫ 1

0((1 + i)t)3 dt =

∫ 1

0(1 + i)3t3 dt

= (1 + i)3 t4

4

∣∣∣∣10=

14(1 + i)3

It’s that easy.

Example. Let’s do an example of a closed curve to convince ourselves of the statement in the premableto this section. We’ll use the easiest holomorphic functions we know: zn and the easiest closed curves weknow γ being the unit circle parameterized by cos t + i sin t, t ∈ [0, 2π].∫

γzn dz =

∫ 2π

0(cos t + i sin t)n dt =

∫ 2π

0(cos tn + i sin tn) dt

=∫ 2π

0cos tn dt + i

∫ 2π

0sin tn dt = 0

One more example of a closed curve.

Example. Let γ be the unit circle. Let us compute∫

γ1z dz. This is∫

γ

1z

dz =∫ 2π

0

ieiθ

eiθ dθ = 2πi

So. Something is going on here.

Remark. There are some variations in notation if you are reading different books on the subject. Some-times when integrating over closed curves, people denote this operation by∮

.

I will not be doing this. All integrals will be denoted by∫

.19

Before we proceed, let’s gather some facts about these integrals which will be useful later.The first is that these contour integrals switch signs upon reversal of the path. First, we have to define

reversal of contour.

Definition. Let γ : [0, 1] → Ω be a curve. Then −γ is γ run backwards, i.e. −γ : [0, 1] → Ω is given byγ(1− t). In general for γ : [a, b]→ Ω, −γ is γ((b− t) + a).

Lemma. Let γ : [a, b]→ Ω and f a complex function∫−γ

f (z) dz = −∫

γf (z) dz

Proposition. We have ∣∣∣∣∫γ

f (z) dz∣∣∣∣ ≤ ‖ f ‖∞

∫ b

a|γ′(t)|dt := ‖ f ‖∞`(γ)

Proof. Just write it out

There is also a version of the Fundamental Theorem of Calculus for complex functions. We need adefinition first.

Definition. Let f : Ω→ C be a function. A primitive for f is an F : Ω→ C holomorphic and such that

F′ = f

Theorem (Fundamental Theorem of Calculus). If f is continuous and f has a primitive in Ω and γ begins atγ0 and ends at γ1 then ∫

γf (z) dz = F(γ1)− F(γ0)

Remark. Note the conditions on this: f has to be continuous, we have to assume f has a primitive.

Proof. It’s just by definitions. You can do it! It would be a good exericse.

Corollary. If γ is a closed curve in an open set Ω and f is continuous and f has a primitive F in Ω, then∫γ

f (z) dz = 0

Remark. Make sure you get the hypotheses on this corollary correct. This corollary is just for functionswith primitives. And we don’t even know when functions have primitives yet.

The corollary does, however, buy us something we maybe would ahve like to know sooner:

Corollary. If f is holomorphic on Ω and f ′ = 0 then f is constant.

Proof. Note that f ′ = 0 means that f ′ is a primitive for 0. Thus, for any curve γ : [0, 1]→ Ω∫γ

f ′ dz = f (γ(1))− f (γ(0)) = 0

so f (γ(1)) = f (γ(0)) no matter what. But, γ(1) and γ(0) could be anything. So f must be constant.

Remark. Note that since∫

γ1z 6= 0 where γ is the unit circle, it must be the case that 1/z does not have

a primitive on the unit circle. That is, there is no everywhere holomorphic function F such that F′ = 1z .

HOWEVER, (Log z)′ = 1z on D− [0,−1]. As soon as we remove a branch, we’re ok.

20

5.2. Cauchy-Goursat. We now need to prove Cauchy’s Theorem. This is the theorem stated in the intro-duction to this section and it is absurdly important.

The strategy for proving this will be successive approximation. We’ll prove it when(1) The path is a rectangle(2) The path is any path in a disc

Theorem. Let f be holomorphic. If Ω ⊂ C is open and R ⊂ Ω is a rectangle then∫R

f (z) dz = 0

Remark. This proof is great. Memorize it.

Proof. Just for a nice shorthand, let’s define

ϕ(R) =∫

∂Rf (z) dz

Here’s what we’re going to do. We divide up our rectangle evenly into 4 other rectangles: R(1), R(2), R(3), R(4).Note that we have

ϕ(R) = ϕ(R1) + ϕ(R2) + ϕ(R3) + ϕ(R4)

For one of these rectangles, which we’ll rename, R1, we have

|ϕ(R1)| ≥|ϕ(R)|

4.

We can continue this process. Split up R1 into R(1)1 , R(2)

1 , R(3)1 , R(4)

1 and pick the one such that

|ϕ(R(k)1 )| ≥ |ϕ(R1)|

4.

and we call this rectangle R2.We get a nested sequence of rectangles

R ⊃ R1 ⊃ R2 ⊃ · · ·where we have

|ϕ(Rn)| ≥ 4−n ϕ(R).Now, by point-set topology there is a point z∗ ∈ ⋂∞

i=1 Ri. We’re going to focus on this point.First, we can choose δ1 so that f is holomorphic on |z− z∗| < δ1 (i.e. the neighborhood |z− z∗| is fully

contained in R — this is where we’re using the hypothesis that R the interior of R is OPEN). Second, givenε > 0 we can choose δ2 so that ∣∣∣∣ f (z)− f (z∗)

z− z∗− f ′(z∗)

∣∣∣∣ < ε

when |z− z∗| < δ2.Now, choose δ = min(δ1, δ2) and assume that Rn is contained in the ball |z− z∗| < δ. Then, and this

is the wonderful part,

ϕ(Rn) =∫

∂Rnf (z)− f (z∗)− f ′(z∗)(z− z∗) dz

Why is this the case? I want to point out that the terms besides f (z) are all constants or multiples of z.Both constants and multiples of z integrate to 0 around a closed curve. Ok. But now

|ϕ(Rn)| ≤ ‖ f (z)− f (z∗)− f ′(z)∗(z− z∗)‖∞L(∂Rn)

≤ ε‖z− z∗‖∞L(∂Rn)

≤ ε diag(Rn)perimeter(Rn)

= ε2−n diag(R)2−n perimeter(R)

= ε4−n diag(R)perimeter(R)

But now we haveε4−n diag(R)perimeter(R) ≥ |ϕ(Rn)| ≥ 4−n ϕ(R)

21

and soε diag(R)perimeter(R) ≥ ϕ(R).

Since ε was arbitrary, ϕ(R) = 0 and we are done.

We also need a strengthening of this theorem.

Theorem. Let R′ = R− a1, . . . , an be a rectangle missing a few interior points. Let f (z) be holomorphic on R′.If we have

limz→ai

(z− ai) f (z) = 0

for all i, then ∫∂R

f (z) dz = 0.

Remark. So this says that integrating around a closed path is still zero, provided that the way that thesingularities are unbounded is controlled.

Proof. Might as well just work in the case where there is one point removed. We can draw a bunch ofrectangle and draw them so that there is a little one, R0 around the point a. However, choose R0 smallenough that the inequality

| f (z)| ≤ ε

|z− a|holds. Then ∣∣∣∣∫

∂R0

f (z) dz∣∣∣∣ ≤ ε‖ f ‖∞L(γ) < 8ε

But ε was arbitrary, so we’re done.

Example. Let’s cook up an example where that type of boundedness condition holds. Let

f (z) =e−

1(z−a)

z− a.

Then,limz→a

(z− a) f (z) = 0.

So, if we choose a rectangle R which encirlces a, we have∫∂R

e−1

(z−a)

z− adz = 0

We are now ready to prove one of the central results in this business.

Theorem (Cauchy’s Theorem). If f is holomorphic in any disc Dr and γ is a closed curve in the disk, then∫γ

f (z) dz = 0

Proof. When do integrals around closed curves equal 0? When they have primitives. The idea is toconstruct a primitive. To this end, consider the point (x, y) which we identify with x + iy. Define

F(z) =∫

βf (z) dz

where β is the path that goes from the center of the disk, call it (x0, y0) to (x, y0) and then goes up to(x, y). Note that

∂F∂y

= i f (z)

(WHY?). The path just described is one half of a rectangle. If we integrate around the other half of therectangle (let’s call the path β) we get the same function F and in this case

∂F∂x

= f (z).22

Thus, F(z) is holomorphic and F′(z) = f . So we’re done.

We now have enough material to start knocking down some pretty big theorems. The following is oneof the many proofs of the fundamental theorem of algebra.

Theorem (The Fundamental Theorem of Algebra). Every non-constant polynomial has a root in C.

Proof. Assume that P(z) doesn’t have a root. Write P(z) = P(0) + zQ(z). This is just saying to separateout the constant term and then factor the rest. Divide that equation by zP(z) to obtain

1z=

P(0)zP(z)

+Q(z)P(z)

.

So far, so good. We now note that since P(z) → ∞ as z → ∞, we can pick R so that when |z| > R, P(z)is arbitrarily large. Let us integrate the entire equation above around a circle of radius R centered at theorigin (and call that circle γ). Then we have∫

γ

1z=∫

γ

P(0)zP(z)

+∫

γ

Q(z)P(z)

Now, since we are assuming P(z) doesn’t have a root, then Q(z)P(z) is holomorphic and integrates to 0. Also,

we already know that∫

γ1z = 2πi for any circle γ around the origin. We thus have

2πi =∫

γ

P(0)zP(z)

.

However, ∣∣∣∣∫γ

P(0)zP(z)

∣∣∣∣ ≤ ‖ P(0)zP(z)

‖∞2πRi

and we can choose R so large that |P(z)| is arbitrarily large, and so the term on the right is arbitrarilysmall. This is a contradiction.

5.3. Interlude: Integration and Contours. Cauchy’s theorem has some surprising consequences. Well, ithas lots of them. But the one we’re going to discuss now is the evaluation of some integrals which youprobably didn’t think you could evaluate. This will be a consistent theme in the course.

The following example contains a lot of good stuff(1) Good use of Cauchy’s theorem(2) Good example of clever choice of contour(3) Good use of estimates

Example. Let’s evaluate ∫ ∞

0

1− cos xx2 dx

This is essentially the real part of integrating 1−eiz

z2 along the interval [0, ∞). However, it will be good forus to introduce some symmetry. The function cos x is even, after all, so we might as well evaluate alongthe interval (−∞, ∞).

We integrate around an indented semi-circle. The integral around the whole circle is 0. We look at theconstituent parts of the integral:

0 =

(∫ −ε

−R

1− eix

x2 dx +∫ R

ε

1− eix

x2 dx)+∫

γε

1− eiz

z2 dz +∫

γR

1− eiz

z2 dz

Now we note that ∣∣∣∣1− eiz

z2

∣∣∣∣ ≤ 2|z|2

so this term goes to 0 as R→ ∞. Thus, we are lef to evaluate

−∫

γε

1− eiz

z2 dz.

23

keyhole.pdf

We now use our knowledge of eiz. Recall that eiz = 1 + iz + (iz)2

2 + · · · . Thus

1− eiz

z2 =−izz2 +

1z2

((iz)2

2!+

(iz)3

3!+

(iz)4

4!+ · · ·

)As we integrate around γε, the term on the right will go to 0.

Various contours:[[[PICTURES]]]Here is a useful theorem for contour integrals. The proof is mostly an exercise, so I’ll leave it as such,

but it’s still useful.

Proposition (Jordan’s Lemma).

5.4. Cauchy Integral Formula and consequences. This subsection will establish a number of the mostuseful theorems in complex analysis. It is a good idea to list the theorems that we prove so you can keepin mind how they depend upon each other. Also note that the first one, the Cauchy integral formula, isproved using Cauchy’s theorem.

(1) The Cauchy Integral formula.(2) Regularity / Derivative form of Cauchy integral formula(3) The Cauchy Inequalities(4) Power series expansion(5) Liouville’s Theorem(6) Fundamental Theorem of Algebra (revisited)(7) Morera’s Theorem

We can now state and prove one of the most important theorems in this business.

Theorem (Cauchy Integral Formula). Let f be holomorphic in a region Ω and suppose this region contains aclosed disc D. For any z ∈ D we have

f (z) =1

2πi

∫∂D

f (ζ)ζ − z

dζ

Proof. We consider a keyhole contour Γδ,εWe have ∫

Γδ,ε

f (ζ)ζ − z

dζ = 0

by Cauchy’s theorem.Now, we let δ→ 0 and we see that we get left with two circles: a big circle ∂D and Cε. Write

F(ζ) =f (ζ)− f (z)

ζ − z+

f (z)ζ − z

.

24

Let’s look at the first term. Recall that the definition of holomorphic says that that for every ε there is a δsuch that if |ζ − z| < δ then ∣∣∣∣ f (ζ)− f (z)

ζ − z

∣∣∣∣ < | f ′(ζ)|So, in particular, this is bounded, and if we integrate around smaller and smaller cirlces, this term goes to0.

Then we evaluate the second term.∫Cε

f (z)ζ − z

dζ = f (z)∫

Cε

dζ

ζ − z= − f (z)2πi.

Thus, we have obtained

0 =∫

∂D

f (ζ)ζ − z

dζ − 2πi f (z)

The following theorem is known as a “regularity result.” These are types of results that tell you aboutthe existence of various derivatives. In our case it tells us that all derivatives of f holomorphic exist andtells us how to compute them!

Corollary. Let f satisfy the same hypotheses as above. Then(1) f has infinitely many complex derivatives(2) In particular, all derivatives of f are holomorphic(3) Let γ be a circle in Ω then

f (n)(z) =n!

2πi

∫γ

f (ζ)(ζ − z)n+1 dζ

for all z in the interior.

Proof. It’s not so surprising that the proof is by induction. The case n = 0 is the Cauchy integral formula.Suppose the n = 1 case holds, i.e.

f (n−1)(z) =(n− 1)!

2πi

∫C

f (ζ)(ζ − z)n dζ.

Let’s compute the derivative of this:

f (n−1)(z + h)− f (n−1)(z)h

=(n− 1)!

2πi

∫C

f (ζ)1h

[1

(ζ − z− h)n −1

(ζ − z)n

]dζ

=(n− 1)!

2πi

∫C

f (ζ)1h

hXY

[Xn−1 + Xn−2Y + · · ·+ XYn−2 + Yn−1] dζ

=(n− 1)!

2πi

∫C

f (ζ)1

XY[Xn−1 + Xn−2Y + · · ·+ XYn−2 + Yn−1]dζ

where

X =1

(ζ − z− h)Y =

1(ζ − z)

.

As h→ 0, X → 1(ζ−z) = Y.

n!2πi

∫C

f (ζ)(ζ − z)n+1 dζ

Great! So we’ve finally verified a statement I made before that when f is holomorphic all derivativesexist. Of course, the corollary also gave a way to compute them. Of course, computing things is sometimesdifficult and so it’s useful to have estimates. The next corollary gives us those.

25

Corollary. If f is holomorphic in Ω open and f contains DR(z0) then

| f (n)(z0)| ≤n!‖ f ‖C

Rn

where as usual we’ve define‖ f ‖C = sup

z∈C| f (z)|

Proof. It’s just a series of inequalities using what we already know

| f (n)(z0)| =∣∣∣∣ n!2πi

∫C

f (ζ)(ζ − z0)n+1 dζ

∣∣∣∣ = n!2π

∣∣∣∣∫ 2π

0

f (z0 + Reit)

(Reit)n+1 Rieitdt∣∣∣∣

≤ n!2π

‖ f ‖CRn 2π =

n!‖ f ‖CRn

Corollary (Louiville’s Theorem). If f is holomorphic on all of C and bounded, i.e. | f | < B for some B, then f isconstant.

Proof. Recall that if f ′ = 0 then f is constant. So, we just have to prove that f ′ = 0.However, the Cauchy inequality gives

| f ′(z0)| ≤BR

and as R→ ∞ this gives the result.

We can give another proof of the Fundamental Theorem of algebra (although, it’s really the same proof,we’ve just packaged it a bit differently — nevertheless it’s good to see this proof)

Theorem. Every non-constant polynomial p(z) has a root in C.

Proof. Suppose not. Then 1p(z) is a bounded holomorphic function (very this). Then, by Liouville, 1

p(z) isconstant. This contradicts the non-constant assumption.

After establishing these massive results, we’re in a position to prove a converse to Cauchy’s theorem.

Theorem (Morera’s Theorem). Suppose f is continuous in an open disc Dr(z0) and for any rectangle in Dr(z0),∫∂R

f (z) dz = 0.

Then f is holomorphic.

Proof. By the proof of Cauchy’s theorem, f has a primitive F so that F′ = f . But F is infinitely differen-tiable, so F′′ exists. But F′′ = f ′ and so f ′ exists. Thus f is holomorphic.

Now, we state a consequence that has to do with power series. We haven’t said it up to this point, butpower series are an important part of the holomorphic story. It’s because of the following theorem

Theorem. Suppose f is holomorphic in an open set Ω. Suppose Dr(z0) ⊂ Ω. Then f has a power series expansion

f (z) =∞

∑n=0

an(z− z0)n

for z ∈ Dr(z0) with coefficients

an =f (n)(z0)

n!=

12πi

∫C

f (ζ)(ζ − z0)n+1 dζ

26

Proof. We’ll use some trickery. The Cauchy integral formula says

f (z) =1

2πi

∫∂Dr(z0)

f (ζ)ζ − z

dζ

Now, we write1

ζ − z=

1ζ − z0 − (z− z0)

=1

ζ − z0

1

1−(

z−z0ζ−z0

)=

1ζ − z0

∞

∑n=0

(z− z0

ζ − z0

)n=

∞

∑n=0

(z− z0)n

(ζ − z0)n+1

This series converges uniformly (whatever that means), and so we can exchange it with an integral. Wethus get

f (z) =1

2πi

∫C

f (ζ)ζ − z

dζ

=∞

∑n=0

(1

2πi

∫C

f (ζ)(ζ − z0)n+1 dζ

)(z− z0)

n

Remark. Note that this gives another proof that holomorphic functions are infinitely differentiable.

Remark. THIS IS VERY DIFFERENT FROM THE REAL CASE. In the real case, if we know that a functionis differentiable, it does not automatically have a power series approximation! This is because having apower series approximation would imply smoothness. All differentiable real functions definitely aren’tsmooth (C∞). However, once functions are “complex differentiable” (i.e. holomorphic), they have ALLderivatives, and may be represented as power series.

6. Practical Matters

(add L’Hopital’s rule as something that exists in this context)

7. Meromorphic Functions / Residue Calculus

We’re going to expand the type of functions we can deal with. So far, we’ve only proved theorems aboutholomorphic functions. But other functions are nice too. For example, 1

z isn’t so bad — it’s derivative existseverywhere but at 0. It seems a shame to discard such a function just because its derivative doesn’t exist atone point (out of an uncountable infinity of them!). Before we define exactly how we are going to expandour notion of function, we need some preliminary definitions.

7.1. Zeroes and Poles. The first definition is no great surprise, but will nevertheless be useful.

Definition. A number z0 ∈ C is a zero for a holomorphic function f if f (z0) = 0.

The following theorem will be very useful for us. It tells us what holomorphic functions “look like”near a zero. This is reasonable similar to the C∞ real case.

Theorem. Let f be holomorphic on Ω which is connected and open. Suppose that f (z0) = 0. Then there is someopen U ⊂ Ω and some g holomorphic on U such that for all z ∈ U

f (z) = (z− z0)ng(z).

Proof. We know that in a disc centered at z0 there is a power-series expansion

f (z) =∞

∑k=0

ak(z− z0)k.

Since f is non-zero (by assumption) there is a smallest k, call it n, such that ak 6= 0. Then

f (z) = (z− z0)n(an + an+1(z− z0) + · · · ) = (z− z0)

ng(z)27

and g(z) is holomorphic since it’s defined by a converging power series.There is an argument to show that the choice of n is unique. But it is easy and left as an exercise.

Definition. f has a zero of order n in the case above.

Definition. A deleted neighborhood of z0 is defined to be

Dr(z0) \ z0 = z : 0 < |z− z0| < r.We now introduce a new type of point which will become important for us:

Definition. A function f on a deleted neighborhod Dr(z0) \ z0 has a pole at z0 if the function

1f

:=

0 z = z01f otherwise

is holomorphic in Dr(z0).

Let’s think about what this is saying. The point is that if a function is somehow infinite at z0, e.g.the function 1

z−2 at 2, it’s kind of hard to characterize exactly how it’s behaving. But! If we look at thereciporical of the function, it’s much, much easier to see what’s going on.

Example. Consider

f (z) =1

(z− 2)2 +1

z− 1Then at, for example, the point 2, we could expand this is a power series.

Theorem. If f has a pole at z0 ∈ Ω, then there is a neighborhood U 30, an n and a holomorphic h such that

f (z) = (z− z0)−nh(z).

Proof. Follows directly from above.

Definition. In the above, n is called the order of the pole.

Near a pole, a function has something like a power series expansion.

Theorem (and definition). If f ahs a pole order n at z0 then

f (z) =a−n

(z− z0)n +a−n+1

(z− z0)n−1 + · · ·+ a−1

(z− z0)︸ ︷︷ ︸principal part

+G(z)

Proof. It’s not so bad. Going to omit it.

Now, consider integrating something like 1(z−z0)3 in the deleted neighborhood Dr(z0) \ z0. Note that

in that neighborhood, it actually has a primitive, so would integrate to zero. However, 1(z−z0)

does NOThave a primitive. It will thus take on particular importance.

We come to the most important definition for us:

Definition. The coefficient a−1 above is called the residue of f at the pole z0. This situation is denoted

Resz=z0

f = a−1

How do we compute this thing, though? Finding a−1 involved finding some power series expansion,which is a somewhat arduous task. It turns out it’s easy though. The following theorem gives us theprocedure:

Theorem. If f has a pole of order n at z0 then

Resz=z0

f = limz→z0

1(n− 1)!

(ddz

)n−1(z− z0)

n f (z)

Proof. Just uses the relation

(z− z0)n f (z) = a−n + a−n+1(z− z0) + · · ·+ a−1(z− z0)

n−1 + G(z)(z− z0)n

28

7.2. Residue Formula. What good is any of the above? For evaluating integrals course — this is the wholereason for focusing on a−1 and the residue.

Theorem. Let f be holomomorphic in a disc, except at a point z0, where f has a pole. Let C be a cirlce in the discenclosing z0. Then ∫

Cf (z) dz = 2πi Res

z=z0f

Proof. We know that there is some small neighborhood U 3 z0 such that

f (z) =a−n

(z− z0)n +a−n+1

(z− z0)n−1 + · · ·+ a−1

(z− z0)+ G(z)

where G is holomorphic in U.Use a keyhole contour to reduce the size of the circle: that is, we can write∫

Cf (z) dz =

∫Cε

f (z) dz

where ε is as small as we want. Then apply∫

Cεto f as it’s been expanded above. The only non-zero term

is ∫Cε

a−1

z− z0dz = a−12πi

which of course gives our theorem.

The following is a trivial corollary, but very useful and usually the theorem that gets cited.

Corollary (THE RESIDUE FORMULA). Suppose f is holomorphic in an open set except at poles z1, . . . , zn.Suppose γ is a toy contour around the zi. Then∫

γf (z) dz = 2πi

n

∑k=1

Reszk

f (z)

This is all well and good, but let’s see some examples of computing residues and actually using theseto do things.

Example. How do we compute Resz=11

z4−1 . Note the bottom has a zero of multiplicity 1, so we cancompute

Resz=1

1z4 − 1

= limz→1

(z− 1)1

z4 − 1= lim

z→1

1z3 + z2 + z + 1

=14

Example. Compute Resz=3i1

z2+9 . The root has multiplicity 1 so

Resz=3i

1z2 + 9

= limz→3i

(z− 3i)1

z2 + 9= lim

z→3i

1z + 3i

=16i

Example. How about cot z at z0 = 0? Let’s consider

limz→0

zcos zsin z

= limz→0

cos zsin z

z= 1

Remark. In general, we have the following easy rule: Suppose z0 is such that g(z) has a simple zero at z0.Then

Resz=z0

f (z)g(z)

=f (z0)

g′(z0)

(this is also a consequence of our rule earlier and the quotient rule: think about it)

Here is an example of actually using the theorem.29

Example. Evaluate ∫|z−1|=1

1z8 − 1

dz.

The function 1z8−1 is meromorphic with a poles at e

ikπ4 where k ranges from 0 to 7 (note that these are all

simple poles, so the remark will apply). We need to figure out how many of these poles lie inside of thecircle |z − 1| = 1. A little geometry (DO IT) tells us that the roots are eiπ/4, 1, e7πi/4. Then the residuetheorem tells us that ∫

|z−1|=1

1z8 − 1

dz = Resz=e7πi/4

+Resz=1

+ Resz=eiπ/4

To compute the residues, we use the result above: The residues are

18z0

in each case.

7.3. Fun Integrals. Now, we can finally use residues to evaluate some integrals. We typically evaluateimproper integrals ∫ ∞

−∞f (x) dx

by turning f into a complex function and choosing a contour that runs along the x-axis, and hoping thebit that doesn’t run along the x-axis goes to 0. This is the same technique we used above.

Example. Let’s compute ∫ ∞

−∞

11 + x2 dx.

As one may guess, we want to compute some contour integral of f (z) = 11+z2 . Let’s use a half circle with

radius R. Let’s call this curve γR. Then∫γR

1z2 + 1

= (2πi)Resz=i

1z2 + 1

= π

and

π =∫

γR

1z2 + 1

=∫ ∞

−∞

11 + x2 dx +

∫arcR

11 + z2 dz

However by our usual sorts of estimates∣∣∣∣∫arcR

11 + z2 dz

∣∣∣∣ ≤ ∥∥∥∥ 11 + z2

∥∥∥∥∞

πR <BR2 πR

where B is some bound. And this term goes to 0 as R→ ∞. Thus∫ ∞

−∞

11 + x2 dx = π

In fact, we can do much better. If you think about it carefully, exactly the technique above prove thefollowing proposition

Proposition. Let p(x), q(x) be polynomials with deg p(x) ≤ deg q(x) + 2. Assume further that q(x) has no realroots. Then ∫ ∞

−∞

p(x)q(x)

dx = ∑i

Resz=zi

p(z)q(z)

where the sum is over the roots in the upper half plane.

We can do even more! The proof of the above can be used if we multiply by any bounded thing, really.Consider ∫ ∞

−∞

p(x)q(x)

cos axdx∫ ∞

−∞

p(x)q(x)

sin ax dx

30

These can be obtained as the real and complex parts (respectively) of∫ ∞

−∞

p(z)q(z)

eiz dz.

Let’s see an example of this

Example. Evaluate ∫ ∞

−∞

cos xx4 + 1

dx.

As one may guess, we’ll evaluate the following around γR, the half-circle of radius R∫γR

eiz

z4 + 1dz

The usual trick shows that the integral around the arc goes to 0 as R→ ∞. Now, the roots of the numeratorthe 4th roots of −1 = eiπ which is e

ikπ2 e

iπ4 with k = 0, 1, 2, 3. However, the only roots in the upper half

plane are eiπ4 and e

i3π4 . We must compute the residues. We can use one of the formulas above

Resz=eiπ/4

eiz

z4 + 1=

ei(cos π4 +i sin π

4 )

4eiπ/4 =ei cos π

4 e− sin π4

4eiπ/4

Resz=ei3π/4

eiz

z4 + 1=

ei(cos 3π4 +i sin 3π

4 )

4ei3π/4

Grind through some algebra.

We now evaluate integrals involving trig functions. Suppose we are trying to integrate∫ 2π

0R(cos θ, sin θ) dθ.

One thing we could potentially do is consider this as a funciton of z by noting that z = eiθ and so dθ = dziz

and

cos θ =12

(z +

1z

)sin θ =

12i

(z− 1

z

)With these substitutions∫ 2πi

0R(cos θ, sin θ) =

1i

∫|z|=1

R(

1p2

(z +

1z

),

12i

(z− 1

z

))dz

and this can be evaluated with residues.We present an absolutely classic example.

Example. Evaluate the integral ∫ 2π

0

dθ

a + cos θ.

We know what to do. Replace cos θ with 12 (z + z−1) and we get∫ 2π

0

dθ

a + cos θ=

1i

∫|z|=1

1a + 1

2 (z + z−1)

dzz

=2i

∫|z|=1

dzz2 + 2az + 1

31

Now compute using the residue theorem. The polynomial z2 + 2az + 1 has two roots, but only one ofthem is in the unit circle. Namely z = −a +

√a2 − 1. The residue formula now gives∫

|z|=1

dzz2 + 2az + 1

= 2πi Resz=−a+

√a2−1

1z2 + 2az + 1

=1

2√

a2 − 1

where the latter equality is by our nifty formula for residues. Thus, altogether, we have∫ 2π

0

dθ

a + cos θ=

2π√a2 − 1

7.4. Integrals invovling za and log z. So far, the functions that we’ve been working with don’t have any“pathologies.” Pathologies for us involving things like branching: the functions we worked with abovewere all holomorphic. However, there are functions we know and love (such as za and log z) that can onlybe holomorphic once we make a cut. In general, this will be solved by dealing appropriately with phaseissues.

Let’s consider ∫ ∞

0xaR(x)dx. − 1 < a < 1

We would like to evalute this by integrating in a circle around zaR(z) somehow. However,

za = exp(a log z)

and thus za is only defined on the cut C \ [0, ∞). Thus, we integrate a keyhole contour

PICTURE

The value of zaR(z) when θ = 0 is as is. When θ = 2πi, zaR(z) is rotated by a phase of e2πia. Integratingover the keyhole contour, our integral gets split into 4 terms∫ R

εxaR(x)dx +

∫γε

zaR(z) +∫

γR

zaR(z) +∫ ε

Re2πiaxaR(x)dx

and by the resiude theorem, the above is equal to the residues around the poles that lie in that contour. AsR → ∞ and ε → 0 the middle two terms can be estimated to go to 0. Changing the order of integrationfor the last integral we get

(1− e2πia)∫ R

εxaR(x)dx = 2πi ∑

zi poleResz=zi

R(z)

Thus, we’ve proved

Theorem. Let R(x) be a rational function as we’ve deal with above and let −1 < a < 1, then∫ ∞

0xaR(x) dx =

2πi(1− e2πia) ∑

zi poleResz=zi

R(z)

I’m not going to bother with an example. It’s clear enough what to do.

Remark. Exercise: Think about how this works for an integral involving log z.

7.5. Summary of Integrals. It is useful to collect our list of ways to evaluate integrals:(1) By Definition. We recall here that if we had a path γ(t) on an interval [a, b] we defined∫

γf (z) dz =

∫ b

af (γ(t))γ′(t) dt

Recall we were actually able to evaluate some integrals this way. For example, we saw that∫|z|=1

1z= 2πi

32

(2) Cauchy’s Theorem. Cauchy’s theorem told us that if we integrate a holomorphic function arounda closed curve, we get 0: ∫

γf (z) dz = 0

(there were more hypotheses, but this is a review). By using estimates and arduously avoidingplaces where the function is not holomorphic, we could evaluate some integrals. For example∫ ∞

0=

1− cos xx2

(3) Cauchy’s Integral formula. Recall that this said

f (n)(z) =n!

2πi

∫γ

f (ζ)(ζ − z)n+1 dζ

This could of course be used to compute integrals.(4) Residue Theorem. We’ve been working with the residue theorem a lot above. Recall that it says

(again, with appropriate hypotheses) if γ is a path that encloses poles z1, . . . , zn of f (z) then∫γ

f (z) dz = 2πi ∑i

Resz=zi

f (z)

(5) Rational functions(6) Rational functions mult by trig functions(7) Trigonometric functions(8) Rational functions multiplied by xa or log z

Of course, many of the techniques above invovled computing residues, so you had best know how todo that.

7.6. Argument Principle and a Grab-bag of theoretical results. we are going to talk a little bit about howbadly a complex function can behave at a singularity (a singularity, in particular, an isolated singularity, isa place where f is not defined at z0, but is defined everywhere nearby)

Theorem. Suppose f is holomorphic on Ω open except possibly at z0 ∈ Ω. If f is bounded on Ω− z0, then z0is a removable singularity.

Proof. Let z ∈ Ω. Choose a multiple keyhole contour. We get∫C

f (ζ)ζ − z

dζ +∫

γε

f (ζ)ζ − z

dζ +∫

γ′ε

f (ζ)ζ − z

dζ = 0.

The cauchy integral formula gives the middle term is 2πi f (z). Since f is bounded and ζ stays away fromz, the second inegral is bounded. We then get

f (z) =1

2πi

∫C

f (ζ)ζ − z

dζ.

Corollary. Suppose f has an isolated singularity at z0. Then z0 is a pole of f iff | f (z)| → ∞ as z→ z0.

Proof. If z0 is a pole, then 1/ f has a zero at z0, so | f (z)| → ∞ as z→ ∞If the condition holds, then 1/ f is bounded near z0. Thus 1/ f has a removeable singularity at z0 and

must vanish there.

There are 3 kinds of isolated singularities:(1) Removed(2) poles(3) essential

Essential singularities are just bonkers.

Example. Consider e1/z. This behaves very poorly near 0.33

I present the following without proof for those who know what “dense” is.

Theorem. Suppose f is holomorphic on Dr(z0)− z0 and ahs an essential singularity at z0. Then the image ofDr(z0)− z0 under f is dense in C.

Definition. f is meromorphic on Ω if there is a sequence of points zi that has no limit points in Ω suchthat

(1) f is holomorphic on Ω− zi(2) f has poles at zi.

What does it mean to be holomorphic on the extended complex plane, i.e. the Riemann sphere?[[[FILL THIS IN]]]Now, we discuss the argument principle.The issue here is that in complex analysis, log is not particularly well behaved. In particular, log( f1 f2) =

log( f1) + log( f2) fails in general, and log f (z) = log | f (z)|+ i arg f (z) depends very strongly on arg z.HOWEVER, the derivative of log is single valued:

(log f (z))′ =f ′(z)f (z)

and it is perfectly reasonable to consider such a thing.For derivatives, the additivity formula will hold.Consider f holomorphic with a zero of order n at z0 so that

f (z) = (z− z0)ng(z).

Thenf ′(z)f (z)

=n

z− z0+

g′(z)g(z)

Thus, if f has a zero of order n at z0, f ′/ f has a simple pole at z0 with reside n. The same thing works forpoles

Theorem. Suppose f is meromorphic on an open region Ω containing C and it’s interior. If f has no poles or zeroson C, then

12πi

∫C

f ′(z)f (z)

dz = #(zero inside C)− #(poles inside C)

Remark. This is another case of “probing” the interior of a circle by using information about fu

Of course, it’s perfectly reasonable to ask “why do I care”?Well, turns out we can prove some interesting things!

(1) Rouche’s theorem(2) The open mapping theorem(3) The maximum modulus principle

Theorem (Rouche). Suppose f and g are holomorphic in Ω containing a disc C. If

| f (z)| > |g(z)|for all z ∈ C then f and f + g have the same number of zeros inside the circle.

Proof. Define ft(z) = f (z) + tg(z). Let nt denote the number of zeros. The argument principle gives

nt =1

2πi

∫C

f ′t (z)ft(z)

dz

We now show that nt is continuous. But this is easy since f ′t (z)/ ft(z) is continuous in t and z. Done.

Definition. f is open if it takes open sets to open sets.

Example. f mapping a region to a point is not open.

Theorem (Open Mapping). If f is holomorphic and non-constant in a region Ω, then f is open.34

Proof. Let w0 be in the image, i.e. f (z0) = w0. We have to prove that there is an ε such that if |w−w0| < ε,then w is in the image as well.

Let g(z) = f (z)− w so that

g(z) = [ f (z)− w0] + (w0 − w) := F(z) + G(z)

Choose δ such that |z− z0| < δ is in Ω and f (z) 6= w0 on |z− z0| = δ. Choose ε so that | f (z)− w0| ≥ εon |z− z0| = δ. If |w−w0| < ε then |F(z)| > |G(z)| on |z− z0| = δ and by Rouche, we have that g = F + Ghas a zero inside the cirlce since F has one. So, w is hit by f .

Theorem. If f is a non-constant holomorphic function in a region ω, then f cannot attain a maximum in Ω.

Proof. Suppose it did, and at the point z0. Let D be a small disc centered at z0 so that f (D) is open by theopen mapping theorem. f (D) will also contain f (z0). But, since f (D) is open, there must be points z ∈ Dsuch that | f (z)| > | f (z0)| which is a contradiciton.

PICTURE

8. Sums and Products

8.1. Sums. We’ll do a bit of discussion of sums — this is a breezy review of the material. Much of it carriesover mutis mutandis from calculus. There is one area, uniform convergence, where we’ll focus because it’simportant, and not emphasized in calculus.

We’ll be concerned with sums of the form ∑ fn(z).

Definition. A partial sum for ∑ fn(z) is

SN =N

∑n=1

fn(z)

Definition. ∑ fn(z) converges at z if SN(z) as N → ∞ has a limit. If ∑ fn(z) converges for all z ∈ Ω, thenwe say ∑ fn(z) converges

Definition. The series ∑∞n=0 fn(z) converges absolutely if ∑ | fn(z)| does.

Definition. A sum ∑ fn(z) converges uniformly if for all z ∈ Ω there is an ε such that |Sn(z)− S(z)| < εfor all n > N.

I’ll gloss over the standard theorems on convege — they are mostly the same as in calculus. Here aresome things to remember:

(1) The ratio test(2) Root test(3) Alternating series test

For uniform convergence, there is another test:

Proposition (Weierstrass M-test). Suppose | fn(z)| ≤ Mn, with Mn independent of z ∈ Ω and ∑ Mn convergent,then ∑ fn(z) is uniformly convergent in Ω.

Here are some useful facts about uniform convergence:

Theorem. —(1) fn(z) continuous in Ω and ∑ fn(z) converges uniformly in Ω, then ∑ fn(z) is continuous in Ω.(2) If ∑ fn(z) converges uniformly in Ω and γ is a curve in Ω then∫

γ∑ fn(z) dz = ∑

∫fn(z) dz

(3) If f ′n(z) =ddz fn(z) exists in Ω and ∑ fn(z) converges uniformly and ∑ fn(z) converges, then

ddz ∑ fn(z) = ∑ f ′n(z).

Most of the series we’ll be dealing with are power series.

Definition. A power series centered at z0 is a series of the form ∑∞k=1 ak(z− z0)

k

35

Of course, we’ve written down some kind of expression, but how do we know where it converges? If itconverges?

Theorem. Let ∑ akzk be a power series. Then there is an R such that for |z| < R the series converges absolutelyand for |z| > R the series does not converge.

Definition. R is called the radius of convergence.

Remark. I’m not going to include a proof of this theorem. It relies on some stuff that will not be veryimportant to us. I just want everyone to know the definitions.

Theorem. The power series f (z) = ∑ anzn gives a holomorphic function in its disc of convergence. Also

f ′(z) = ∑ nanzn−1.

Definition. f on Ω is analytic at z0 ∈ Ω if there is an expansion ∑ an(z− z0)n with a positive radius of

convergence and such that f (z) = ∑ an(z− z0)n in some neighborhood of z0.

Definition. f is analytic if we have the above condition at every point of Ω.

The theorem above says that analytic implies holomorphic. We’ve already seen (by using Cauchy’stheorem and some tricks) that holomorphic implies analytic. Thus, they are equivalent notions in complexanalysis. But, it’s good to have different names for them, because they are useful at different times.

Recall that we had the Taylor series

f (z) = ∑f (n)

n!(z− a)n

which is valid when f is holomorphic on a disc Da(r) ⊂ Ω.There is also such a thing as a Laurent series. We won’t prove the following theorem, but it’s very, very

useful.

Theorem. Suppose f is holomorphic on the annulus z : R1 ≤ |z− z0| ≤ R2 then

f (z) =∞

∑n=−∞

an(z− z0)n

and the series converges absolutely on the annulus. Furthermore,

an =1

2πi

∫∂DR1 (z0)

f (z)(z− a)n+1 dz n ∈ Z.

We’ll now discuss the practical ways to find Laurent series for functions. In practice, can’t evaluateLaurent series just using the formula above. Instead, we use tricks and series that we already know.

Example. Find the Laurent series fore2z

(z− 1)3

about z = 1.For convenience, we set W = z− 1 so that z = w + 1 and the above is

e2w+2

w3 =e2

w3 e2w =e2

2w

(1 + 2w +

(2w)2

2!+

(2w)3

3!+ · · ·

)and substitute back in w = z− 1.

Example. Consider

(z− 3) sin1

z + 2.

let w = z + 2 so that z = w− 2. Then

(w− 5) sin12= (w− 5)

(1w− 1

3!1

w3 +15!

1w5 + · · ·

)36

Now multiply this out and substitue w = z + 2 back in to get

1− 5z + 2

+ · · ·

Note we can see from this that z = −2 is an essential singularity.

Remark. In fact, we can classify singularities in general by looking at Laurent series

Example. We’ll find Laurent series for

f (z) =1

z(z− 2)valid in two different annuli:

0 < |z| < 2 |z| > 2First, we use partial fractions to write

1z(z− 2)

=12z− 1

z− 2

Now, when 0 < |z| < 2 we can write

− 1z− 2

=12

11− z

2=

12

(1 +

z2+

z2

4+

z3

8+ · · ·

)Note that this sum convergnes when |z| < 2. This, together with the partial fraction decomposition givesus a Laurent series

12z

+12+

14

z +18

z2 + · · ·

For |z| > 21

z− 2=

1z

11− 2

z=

1z

(1 +

2z+

4z2 +

8z3 · · ·

)which converges when |z| > 2.

8.2. Evaluation of Sums. Suppose you want to understand a sum ∑∞n=−∞ f (n) (assuming it converges

and whatnot) using complex analysis. You might hope to get yourself into the following situation: thef (n) could arise as the residues of some function, and the sum would then correspond to a residuecomputation, or perhaps some integral. Let’s imagine we want to carry out this program. The first thingwe would want is a function with poles at the integers. Such a function is π cot πz. Furthermore, theresidue of π cot πz at z = n is easily computable:

limz→n

(z− n)π cot πz = π

(z− nsin πz

)cos πz = 1

by L’ Hopital. Thus, the residue of f (z)π cot πz is exactly f (n).Now, let’s try to relate this to some sort of integral. Let R be the (N + ε)× (N + ε) rectangle centered

at 0. Then by the residue theorem∫∂R

f (z)π cot πz dz = 2πi ∑ residues

= 2πi ∑poles of f (z)

Res+2πi ∑poles of cot πz

Res

Now, we’ll try to make the integral go away. Weneed the following fact:

Lemma. On ∂R, |π cot πz| < A for some A

Furthermore, assume that | f (z)| ≤ M|z|k (this is completely reasonable...our sum isn’t going to converge

otherwise). Then ∣∣∣∣∫∂R

f (z)π cot πz dz∣∣∣∣ ≤ πAM

Nk (8N + 4)→ 0

37

and so∞

∑n=−∞

f (n) = ∑z0

poles of f (z)

Resz=z0

f (z)π cot πz

Example.

Let’s evaluate∞

∑n=−∞

1n2 + a2 a > 0

The poles of f (z) = 1z2+a2 are at z = ±ia and

Resz=ia

1z2 + a2 π cot πz = lim

z→ia

(z− ia)z2 + a2 π cot πz =

π cot πai2ai

= − π

2acoth πa

We get the same pole at z = −ia and so algother we get∞

∑n=−∞

1n2 + a2 =

π

acoth πa.

Example. Usually, we don’t want to sum from −∞ to ∞. In cases where we want to sum from 1 to ∞, weusually have to do some manipulations. For example

∞

∑n=−∞

1n2 + a2 =

1a2 + 2

(∞

∑n=1

1n2 + a2

)and upon rearranging we see

∞

∑n=1

1n2 + a2 =

π

2acoth πa− 1

a2

Example. Take the limit a→ 0 in the above to get Euler’s famous identity∞

∑n=1

1n2 =

π2

6

There are a few other useful formulas, derivable in the same way as the boxed formula above.

∞

∑n=−∞

(−1)n f (n) = − ∑z0

poles of f (z)

Resz=z0

f (z)π csc πz

∞

∑n=−∞

f(

2n + 12

)= ∑

z0poles of f (z)

f (z)π tan πz

∞

∑n=−∞

(−1)n f(

2n + 12

)= ∑

z0poles of f (z)

f (z)π sec πz

Example. Compute

1− 133 +

153 −

173 + · · ·

We just use the formulas above. The only thing that requires any cleverness is the fact that this is a sumfrom n = 0 to ∞ (and so we want to use half of the value) and that the function is f (z) = 1

8z3 . Given thiswe have (after a residue computation)

1− 133 +

153 −

173 + · · · = π3

3238

9. Special Functions, Solving Equations

9.1. Gamma Function. We define a new function, the Gamma function which will be very useful to us.It should not be immediately clear why this is useful, exactly, but it will become clear(er) eventually.

The goal is to generalize the factor function, n!. We’d first like to define this for non-integers, and thencomplex numbers with a positive real part, and then all complex number (or almost all of them).

Definition. We defineΓ(z) =

∫ ∞

0e−ttz−1 dt Re(z) > 0

Lemma. If Re(z) > 0 thenΓ(s + 1) = sΓ(s)

and thus, Γ does indeed agree with the factorial function

Γ(n + 1) = n! n = 0, 1, 2, . . .

Proof. Do some integration by parts. We note that∫ 1/ε

ε

ddt(e−tts) ds = −

∫ 1/ε

εe−tts dt + s

∫ 1/ε

εe−tts−1 dt.

Now, the left side vanishes as ε→ 0, which gives Γ(s + 1) = sΓ(s).We just have to check a normalization. That is, we have to check Γ(1) = 1. This is not hard.

We have a function that is defined for complex numbers with the real part greater than 0. In fact, thiscan be extended to the entire complex plane.

Theorem. Γ(z) can be extended to a meromorphic function on C whose only singularities are simple poles atz = 0,−1, . . . . Furthermore,

Resz=−n

Γ(z) =(−1)n

n!.

Proof. We extend Γ(z) to be meromorphic on Re(z) > −m for each m ≥ 1. We define

Fm(s) =Γ(s + m)

(s + m− 1)(s + m− 2) · · · s .

Note that this agrees with Γ(s) for Re(s) > 0 by the functional equation. Furthermore, this function isdefinitely meromorphic on Re(z) > −m. We compute the residue

Ress=−n

Fm(s) =Γ(−n + m)

(m− 1− n)!(−1)(−2) · · · (−n)

=(m− n− 1)!

(m− 1− n)!(−1)(−2) · · · (−n)

∗ = (−1)n

n!

Lemma. The following will be useful: For Re(s) > 0

Γ(s) =∞

∑n=0

(−1)n

n!(n + s)+∫ ∞

1e−tts−1 dt

Proof. We split the integral

Γ(s) =∫ 1

0e−tts−1dt +

∫ ∞

1e−tts−1 dt.

We then note ∫ 1

0e−tts−1 dt =

∞

∑n=0

(−1)n

n!(n + s)

by using the power series for e−t and integrating term by term.

39

Theorem. For all z ∈ C we haveΓ(z)Γ(1− z) =

π

sin πzProof. Note that

Γ(1− s) =∫ ∞

0e−uu−s du = t

∫ ∞

0e−vt(vt)−s dv

where we made a change of variable u = vt. Then we get

Γ(1− s)Γ(s) =∫ ∞

0e−tts−1Γ(1− s) dt

=∫ ∞

e−tts−1(

t∫ ∞

0e−vt(vt)−sdv

)dt

=∫ ∞

0

∫ ∞

0e−t[1+v]v−sdvdt

We’ve hit a spot where we need the fact that for 0 < a < 1∫ ∞

0

va−1

1 + vdv =

π

sin πaand then using this fact we continue the above to∫ ∞

0

v−s

1 + vdv =

π

sin π(1− s)=

π

sin πs

Interestingly, we get the following bonkers fact:

Corollary. We have the identity

Γ(

12

)=√

π.

Now, we know that Γ is relatively well behaved. I mean, it’s meromorphic and we know the poles andthey are simple poles. In fact 1

Γ(z) is an entire function:

Theorem. The function 1Γ(s) is entire with simple zeros at s = 0,−1,−2, . . . and vanishes only there.

Proof. We can write1

Γ(s)= Γ(1− s)

sin πss

and the poles of Γ(1− s) and sin πs cancel.

10. Laplace Transforms

Definition. The Laplace transform is given by

[L( f )](s) =∫ ∞

0e−st f (t)dt := F(s)

Remark. Here s ∈ C.

We won’t talk about this for a little bit, but I’ll give away the game now and say that the inverse Laplacetransform is given as follows:

f (t) = L−1F(t) = 12πi

∫ γ+i∞

γ−i∞estF(s)ds

where γ is greater than the real part of all singularities of F(s).40

Remark. In fact, we can evaluate this using residues, assuming F(s) doesn’t grow too fast

LF(t) =n

∑k=0

Resz=zk

(F(s)est)

Remark. If the singularities lie in the left half plane, this is the same as the inverse Fourier transform.

Remark. In general, this inverse Laplace transform is why we’re interested in this as part of a complexanalysis class.

The goal is to use the Laplace transform to turn a differential equations problem into an algebraicproblem, solve the algebraic problem, and then transform back. The form our differential equation willusually take is

amy(m) + · · ·+ a1y′ + a0y = f (t) f (0) = a f ′(0) = b ...

We’ll begin by computing a few Laplace transforms to see how we get an algebraic expression out of adifferential equation.

Example. First, let’s compute the laplace transform of something. f (t) = 1 is a good place to start.

[L(1)](s) =∫ ∞

0e−stdt = −1

se−st

∣∣∣∣∞0= −1

s

Ok. So this is a thing we can get.

Example. We’d also like to compute the Laplace transform of derivatives. It involves integration by parts,so probably remember how to do that.

[L( f ′)](s) =∫ ∞

0e−st f ′(t)dt

= e−st f (t)∣∣∣∣∞0− (−s)

∫ ∞

0e−st f (t)dt

= sL( f )(s)− f (0)

Example. From the above example we can interatively work:

[L( f ′′)](s) = sL( f ′)(s)− f ′(0)

= s[sL( f )(s)− f (0)]− f ′(0)

= s2L( f )(s)− s f (0)− f ′(0)

and similarly[L( f ′′′)](s) = s3L( f )(s)− s2 f (0)− s f ′(0)− f ′′(0)

and you can see the pattern.

We’ll start trying to solve an easy equation with this

Example. Consider the equation

y′′ − 3y′ + 2y = 0 y(0) = 0, y′(0) = 2.

We compute, using the above

L(y′′ − 3y′ + 2y) = [s2Y(s)− sy(0)− y′(0)]− 3(sY(s)− y(0))− 2Y(s)

= s2Y(s) + 3− 2− 3sY(s)− 3 + 2Y(s)

= s2Y(s)− 3sY(s) + 2Y(s)− 2 = 0

So, solving, we have

Y(s) =2

s2 − 3s + 2.

41

So, we’ve got some algebraic equation. We’d now like to take an “inverse” of this. I already told youwhat it is, but let’s see that it actually works. We have

F(s) =∫ ∞

0est f (t) dt.

This is a complex function. Let’s assume that | f (t)| ≤ Meγt for some γ and for t > t0. Suppose alsothat | f ′(t)| ≤ Meγt. Then it’s not so hard to show that the Laplace transform exists for Re(s) > γ and|F(s)| < M′

|s| . Note that if it exists, it’ll be holomorphic. Since it’s holomorphic we can write

F(s) =1

2πi

∫C

F(z)z− s

dz

where the countor is the boundary of some half disk extending into the right-half plane. By order ofgrowth considerations, it’s the case that

F(s) =1

2πi

∫ γ+i∞

γ−i∞

F(z)z− s

dz

Now, we can apply L−1 to both sides of this.

f (t) = L−1[F(s)] =1

2πi

∫ γ+i∞

γ−i∞f (z)L−1

1

s− z

dz

=1

2πi

∫ γ+i∞

γ−i∞f (z)eszdz

which is what we wanted.

Remark. Note that this “proof” is telling is that the main property of the inverse Laplace transform thatwe need is that

L−1

1s− z

= esz

and once this is true, we’re good. In fact, this is a good way to remember what the inverse Laplacetransform should be — and you can remember it easily using the Cauchy integral formula.

Let’s now compute an inverse Laplace transform. We’ll see what’s involved.

Example. Let

Y(s) =2

s2 − 3s + 2.

The inverse Laplace transform is

L−1[Y(s)](t) =1

2πi

∫ γ+i∞

γ−i∞est 2

s2 − 3s + 2ds

This boils down to a residue computation: It’s clear that the poles are at 1 and 2. We take residues ats = 1 and s = 2 and the integral becomes

−2et + 2e2t

We can check that this satisfies the equations above.

It seems like a good idea that we should compute a bunch of Laplace transforms (and inverse Laplacetransforms). There are many tables of these things, but it seems nice to be able to have computations atour fingertips. What kind of things would we like?

• exponentials• sine/cosine• polynomials

42

Example. We’ll simultaneously compute the Laplace transform of cos ωt and sin ωt. We look at eiωt.

L[eiωt](s) =∫ ∞

0e−steiωt dt

=∫ ∞

0et(iω−s)dt

=1

iω− set(iω−s)

∣∣∣∣∞0=

1s− iω

Writing this in a more usual form

1s− iω

=s + iω

s2 + ω2 =s

s2 + ω2 + iω

s2 + ω2 .

Looking at real and complex parts we get

L[cos ωt](s) =s

s2 + ω2 L[sin ωt](s) =ω

s2 + ω2

Example. We’ll compute the Laplace transform of eat.

L[eat](s) =∫ ∞

0e−steatdt =

1s− a

Example. Let f (t) = tp with Re(p) > −1. Then, by definition∫ ∞

0e−sttpdt =

Γ(p)sp+1

Note that there is an interesting special case with f (t) =√

t:

L(√

t)(s) =√

π

2s3/2

Let’s do another example.

Example. We consider the differential equation the models a harmonic oscillator with a time-dependentexternal force f (t) = t. Also assume that y(0) = a and y′(0) = b. Take the Laplace transform of bothsides:

L[y′′(t) + ω2y(t)](s) = s2Y(s)− sy(0)− y′(0) + ω2Y(s)

= (s2 + ω2)Y(s)− sa− b

And the right side is just Γ(1)p2 = 1

p2 . So, the Laplace transform of the entire thing becomes

[s2 + ω2]Y(s)− sa− b =1s2

and solving, we get

Y(s) =s3a + s2b + 1s2(s2 + ω2)

The poles are at 0, ±ω. After we do a computation, we get

y(t) =t

ω2 + a cos ωt +(

bω− 1

ω3

)sin ωt

Which is pretty cool. This is obviously not something that we could have guessed at.43

10.1. More Complicated Laplace Transform. We are going to try to solve a differential equation

(a2 + b2t)y′′(t) + (a1 + b1t)y′(t) + (a0 + b0t)y(t) = 0.

Remark. For convenience later on we will denote the differential operator in the equation above by D.That is,

D = (a2 + b2t)d2

dt2 + (a1 + b1t)ddt

+ (a0 + b0t)

Suppose we can write

y(t) =∫

CestF(s)ds

for some contour. Differentiating we get

D[y(t)] =∫

CF(s)est(P(s) + Q(s)t)ds = 0

whereP(s) = a2s2 + a1s + a0 Q(s) = b2s2 + b1s + b0

Now, how do we pick F(s)? Well, we’d like that contour integral to be zero, so we’d like the thing underthe integral sign to be a derivative (that way it’s a primitive and we know the integral is 0). Consider

dds[F(s)estQ(s)

]=

dds

[F(s)Q(s)]est +dds

[est](F(s)Q(s)).

If we want this to be the integrand, we’d require

dds

[F(s)Q(s)] = P(s)F(s)

or in other words

F(s) =1

Q(s)

∫ s

s0

P(s)Q(s)

ds

which is determined up to a constnant. As long as the curve has the same value of F(s)estQ(s) at it’sendpints, we good.

So

y(t) =∫

Cest[

1Q(s)

∫ s

s0

P(s)Q(s)

ds]

Here is another derivation:First, let’s collect some Laplace transforms:

L[ty(t)](s) = −Y′(s)

L[ty′(t)](s) = −sY′(s)−Y(s)

L[ty′′(t)](s) = −s2Y′(s)− 2sY(s) + y(0)

and

L[y(t)](s) = Y(s)

L[y′(t)](s) = sY(s)− y(0)

L[y′′(t)](s) = s2Y(s)− sy(0)− y′(0)

Now, we have to compute the Laplace transform of the equation above. We get

a2(s2Y(s)− sy(0)− y′(0)) + b2(−2sY(s)− s2Y′(s) + y(0))

+ a1(sY(s)− y(0)) + b1(−Y(s)− sY′(s))

+ a0Y(s) + b0Y′(s)

= Y(s)(a2s2 + a1s + a0 − b22s− b1)−Y′(s)(b2s2 + b1s + b0)

+ (a2sy(0) + b2y(0)− a1y(0)− a2y′(0))

44

Some particularly nice things happen when we make certain assumptions. Assume that a2 = 0 andfurther assume that b2y(0)− a1y(0)− a2y′(0) = 0. This amounts to being agnostic about initial conditions.This is all well and good, we’re just trying to find some solution to an equation, and we’re not worryingtoo much about initial conditions. In that case, we have

Y′(s)Y(s)

=a1s2 + a1s + a0 − b22s− b1

b2s2 + b1s + b0

Let P(s) = a2s2 + a1s + a0 and Q(s) = b2s2 + b1s + b0 and then above becomes

dds

log Y(s) =P(s)Q(s)

− dds

log Q(s).

Integrating, we obtain

log Y(s) + log Q(s) =∫ P(s)

Q(s)ds + C1

and exponentiating

Y(s) = eC11Q

exp(∫ P(s)

Q(s)ds)

Now, we have to take the inverse laplace transform (let C2 = eC1 ) and we get

y(t) = C2

∫ γ+i∞

γ−i∞est[

1Q

exp(∫ P(s)

Q(s)ds)]

10.2. Some Equations from Physics. We are now in the business of solving some differential equations.I’ll tell you where they come from — but if you don’t care, that’s ok too. The point is, I’m showing youmethods for solving differential equations, and so I should pick some examples to work with. I might aswell pick ones that mean something.

Example (Hermite Polynomials). Consider the following equation:

y′′ − 2xy′ + 2ny = 0

Where does something like the above come from? It turns out this equation occurs when you try to solveSchrödinger’s equation for a quantum harmonic oscillator. We won’t worry too much about what thatmeans, but it’s an equation that comes up quite naturally in the real world.

We use the method above to tackle this guy. We note

a2 = 1, b2 = 0 a1 = 0, b1 = −2 a0 = 2n, b0 = 0

so thatP = s2 + 2n Q = −2s.

Then, using the expression we’ve derived, the Laplace transform is

− 12s

exp(∫ s2 + 2n

−2sds)= − 1

2sn+1 exp(− s2

4

).

(That last step requires a little bit of computation). We pick n > 1 so that this thing doesn’t have divergenceissues and note the singularity is obvious at 0. So, the inverse Laplace transform is (up to an absorbedconstant) ∫ γ+iγ

γ−iγ

12sn+1 exp

(− s2

4

)est ds

We could compute this by doing a really irritating residue calculation. What answer would we get?

1n!

lims→0

dn

dsn exp(− s2

4

)est

45

One could actually do things this way, but it would be really, really irritating. Instead, we make a changeof variables. Let u = s

2 − t. Note that u2 = s2

4 − st + t2. Also, du = ds2 . After these substitutions, we have

(up to a constant again)

et2∫

C

eu2

(u− t)n+1 du

This is a much nicer computation. This is because we can recognize this using Cauchy’s integral formulaas

et2(−1)n dn

dtn e−t2

Example (The Airy Function). The Airy function is the solution to the differential equation

y′′ − xy = 0

Where does this come up? It roughly comes up when we consider the Schrödinger equation with potentialV = x.

[[[FILL IN]]]

Example. Laguerre polynomials. We consider the differential equation

ty′′ + (1− t)y′ + ny = 0

We could either just take the straight-up Laplace transform, and this satisfies the conditions of ourinitial derivation. Or we could do it the fancier way. We opt for the fancier way. In this case

P(s) = s + n Q(s) = s2 − s

andP(s)Q(s)

=−n

s+

n + 1s− 1

so that ∫ P(s)Q(s)

ds = −n log s + (n + 1) log(s− 1)

= log s−n + log(s− 1)n+1.

We further have that1

Q(s)exp

((s− 1)n+1

sn

)=

1s(s− 1)

((s− 1)n+1

sn

)=

(s− 1)n

sn+1 .

So, the solution is a contour integral

12πi

∫ γ+i∞

γ−i∞

est(s− 1)n

sn+1 ds

where est (s−1)n+1

sn has to return to its original value on endpoints.We now go about the business of computing this integral. First, make the substitution u = s− 1, just to

make things look a bit cleaner. We know we probably want to use the Cauchy integral formula to evaluatethis, so we move the subtraction to the bottom. Here’s what it looks like after the subtituion:

et

2πi

∫ γ+i∞

γ−i∞

eutun

(u + 1)n+1 ddu

Now, we’d like to make it look even more like Cauchy’s integral formula. As is, it would be irritating toevaluate. Let v = −ut so that dv = −tdu. Then the integral becomes∫ (

−1t

)e−v(−v/t)n(− v

t + 1)n+1 =

∫ e−vvn

(v− t)n+1 dv = n!dn

dvn e−vvn∣∣∣∣v=t

and all together we get

etn!dn

dtn (e−ttn)

46

10.3. Confluent Hypergeometric Functions. We consider the differential equation

zu′′ + (γ− z)u′ − αu = 0

where α is arbitrary, γ 6= 0,−1,−2, . . . . Where does this differential equation arise? The answer is whenyou try to solve Schrödinger’s equation in a Coulomb potential. You can check by hand that this has asolution

F(α, γ, z) = 1 +α

γ

z1!

+α(α + 1)γ(γ + 1)

z2

2!+ · · ·

Well, so that’s a power series solution. But, we can solve it using our techniques above. You’ll certainlyrecognize it as something that you can represent using the techniques we’ve used thus far.

We getP(s) = γs− α Q(s) = s2 − s

We computeP(s)Q(s)

=γs− α

s2 − s=

α

s+

γ− α

s− 1and so

exp(∫ P(s)

Q(s)

)= exp

(α

s+

γ− α

s− 1

)ds = sα(s− 1)γ−α.

Altogether1Q

exp(∫ P

Q

)= sα−1(s− 1)γ−α−1.

Thus, we would seek a full solution of the form∫C

estsα−1(s− 1)γ−α−1ds

Suppose we’re in a case where Re(γ− α) > 0 where γ is not a positive integer. Then a path around the

unit cricle is such that est exp(∫ P

Q

)returns to its original value (0).

NOte that F(α, γ, 0) = 1. Now, it is a FACT (which I will prove later)

F(α, γ, z) = − 12πi

Γ(1− α)Γ(γ)Γ(γ− α)

∫C

esz(−s)α−1(1− s)γ−α−1ds

Let’s say that α = −n so that we have a polynomial (we can see this from the power series). Let’s try tofigure out what this polynomial is. First, let’s consider the leading constant:

Γ(n + 1)Γ(γ)Γ(γ + n)

= n!Γ(γ)

(γ + n− 1)(γ + n− 2) · · · Γ(γ) = n!1

(γ + n− 1)(γ + n− 2) · · · γNow we take care of the integral

− 12πi

∫C

esz(−s)α−1(1− s)γ−α−1

Make the substitution s 7→ 1− uz so that ds = − 1

z . Then the integral becomes

ez

2πi

∫C

e−uz1−α(u− z)α−1zα+1−γ 1z

uγ−α−1du

=z1−γez

2πi

∫C

e−uuγ+n−1

(u− z)n+1 du

= z1−γ ez

n!dn

dzn e−zzγ+n−1

Thus, we altogether get

F(−n, γ, z) =1

γ(γ + 1) · · · (γ + n− 1)z1−γez dn

dzn (e−zzγ+n−1).

47

Now, let’s get even more specific and assume that γ = m > n is a (positive) integer. This is often thecase in real-life examples. Then we get

F(−n, m, z) =(−1)m−1

m(m + 1) · · · (m + n− 1)ez dm+n−1

dzm+n−1 (e−zzn)

When m = 0 this is the Laguerre polynomial we discussed before. When m > 0 this is often calleda associated Laguerre polynomial (or perhaps just Laguerre polynomial, depending on where you’relooking).

11. Fourier Series

Note: almost all of this is taken from Wallace’s Mathematical Analysis of Physical Problems; an awe-some book (with a lot of irritating typos).

Here we cover the basics of Fourier series and Fourier integrals. We start with a basic definition.

Definition. A real function f is periodic of period a if f (x + a) = f (x) for all x.

Now, suppose we have a periodic function y(x) and it’s defined so that y(a/2) = y(−a/2). It seemsreasonable, based on the properties of sines and cosines, that this could be written in terms of sines andcosines

y =∞

∑n=0

an sin2πnx

a+ bn cos

2πnxa

for various an, bn. This can be re-written in a more compact form, and a form that is much more conduciveto our goals: using complex numbers to understand things. The alternative form is

y =∞

∑n=−∞

cne2πnx

a cn =12(bn − ian) c−n = cn =

12(bn + ian)

How would we compute the cn were they to exist? (Which, again we haven’t shown that functions CANbe written in this form, but I just asserting it). Well, here is a fact that will be useful∫ a/2

−a/2e

2πinxa e−

2πimxa dx = aδmn

so

cn =1a

∫ a/2

−a/2f (x) exp

(−2πinx

a

)dx

This number is telling us “how much” a2πn perioidc stuff is in f (x). So, given a perioidic function, we can

decompose it into its bits (often called harmonics).We’re going to soup this up a bit. In particular, not all functions are perioidc. Sometimes they’re

just...functions. So, we’ll take a limit as a → ∞ and consider normal functions as somehow being ∞-periodic.

Let’s start by naming yn = 2πna . From the stuff above, we have

a2π

cn = F(yn) =1

2π

∫ a/2

−a/2f (x)e−ixyn dx

f (x) =2π

a

∞

∑n=−∞

F(yn)eixyn

Let a→ ∞ so that the yns get closer and closer. By how we have defined everything,

f (x) = lim∆yn→0

∞

∑n=−∞

F(yn)eixyn ∆yn =∫ ∞

−∞F(y)eixydy

where in the last equality we have used the definition of Riemann sum. Note also that

F(y) =1

2π

∫ ∞

−∞f (x)e−ixydx

We are now ready to make our main definitions:48

Definition. Let f be a function. The Fourier transform of f , denoted f (k) is defined to be

f (k) =1

2π

∫ ∞

−∞f (x)e−ikxdx

and the inverse Fourier transform isf (x) =

∫ ∞

−∞f (k)eikxdk

Remark. It is often the case that these transforms are defined with different constants out front. That is,you will often see a definition

f (k) =1√2π

∫ ∞

−∞f (x)e−ikxdx

f (x) =1√2π

∫ ∞

−∞f (k)eikxdk

Remark. I also need so say something about WHEN Fourier transforms are defined. To cleanly state this,I need to define a set of functions called L2(R). The set L2(R) consists of those functions f such that∫ ∞−∞ | f (x)|2dx < 0 (another way of saying this is that f has finite L2-norm). That is

L2(R) =

f | f : R→ R and

∫ ∞

−∞| f (x)|2dx < ∞

Note that f need not be continuous or anything like that.

Anyhow, Fourier transforms are only defined for L2-functions and it is a fact that if f is L2, then f is L2.

Remark. One final remark about naming conventions. Much like when we did Laplace transforms wealways used the variable t for the initial function, and the variable s for the transformed function, wewill typically use x for the initial function and k for the transformed function when considering Fouriertransforms. The exception to this is that if we are thinking of a Fourier transform as a function of time, andaccordingly name the variable t, then the transform will have variable ω, in order to suggest frequency.

Now we can begin to derive some basic facts about Fourier transforms. Like Laplace transforms, Fouriertransforms turn an analytic problem into an algebraic one.

Example. We compute the Fourier transform of the derivative of f

d fdx

(k) =1

2πi

∫ ∞

−∞f ′(x)e−ikxdx

=1

2πf (x)e−ikx

∣∣∣∣∞−∞

+ ik1

2π

∫ ∞

−∞f (x)e−ikxdx

= ik1

2π

∫ ∞

−∞f (x)e−ikxdx

= ik f (k)

Example. A direct and easy generalization of the above example is that

dn fdx2 (k) = (ik)n f (k)

I want to at least state the following theorem, even though we will not use it much (or at all)

Theorem (Parseval’s theorem). Let f ∈ L2(R). Then∫| f (x)|2dx =

∫| f (k)|2dk

Remark. This is important for a number of reasons, but one of them is that in quantum mechanics, allfunctions have to be of L2-norm 1, and this is saying that a function remains normalized when we moveto momentum space (don’t worry if you don’t know what that means, it’s just interesting motivation).

49

We’ve proved some general properties, so let’s do some examples.

Example. Let

f (t) =

sin ω0t |t| < 2πN

ω0

0 otherwseWe want to take the Fourier transform of this. That is, we want to compute

12π

∫ 2πN/ω0

−2πN/ω0

sin ω0te−itωdt

How do we go about this? First, remember that

e−itω = − cos tω + i sin ωt.

Now, note that cos is an even function, while sin is an odd function, so the term involving cosine vanishes.This leaves us to evaluate

12π

∫ 2πN/ω0

−2πN/ω0

i sin ω0t sin ωt dt.

Now we recall some angle addition formulas. We have

sin ω0t sin ωt =12[cos(ω0t + ωt)− cos(ωt−ω0t)]

and so we can evaluate our Fourier transform∫ 2πN/ω0

−2πN/ω0

cos(ω0t + ωt)− cos(ωt−ω0t)dt

=1

ω0 + ωsin(ω0t + ωt)− 1

ω−ω0sin(ωt−ω0t)

∣∣∣∣2πN/ω0

−2πN/ω0

=1

ω0ωsin

2πNω0− 1

ω0 −ωsin

2πNω

ω0

= 2ω01

ω2 −ω20

sin2πNω

ω0

Throw in a factor of 12π and i and we get the result.

Example. The Fourier transform of e−Ax2. First, the following result will be useful:∫ ∞

−∞e−x2

dx =√

π.

Now, we proceed with the computation:1

2π

∫ ∞

−∞e−Ax2

e−ikxdx =1

2π

∫ ∞

−∞e−A(x2− ikx

A )dx

=1

2π

∫ ∞

−∞exp−A

((x− ik

2A

)2− k2

4A2

)dx

=1

2π

∫ ∞

−∞exp−A

((x− ik

2A

)2)

dx · e−k2/4A

= e−k2/4A 12π

∫ ∞

−∞e−Ax2

dx =1

2πe−k2/4A

√π

AWe come to an important construction:

Definition. Let f , g ∈ L2(R) then define the convolution to be

( f ∗ g)(x) =∫ ∞

−∞f (x)g(x− x′)dx′

The following fact is now incredibly important:50

Theorem. We havef ∗ g(k) = 2π f (k)g(k)

We can use the Fourier transform to solve some differential equation.

Example. Let’s consider the differential equation

d2 fdt2 + k

d fdt

+ ω20 f = ϕ(t).

We take the Fourier transform of both sides

[−ω2 + ikω + ω20 ]F = Φ

and solve

F(ω) = − Φ(ω)

ω2 −ω20 − ikω

We’d like an inverse Fourier transform of this. Either we could plug in a specific function Φ and thencompute an inverse Fourier transform using complex analytic techniques. Another thing we could do islet G(t) be such that

G(t) = − 12π

1ω2 −ω2

0 − ikω

Then

f (t) =∫ ∞

−∞ϕ(t′)G(t− t′)dt′

So, it’s enough to figure out G, which is obtained by an inverse Fourier transform:

G(t) = − 12π

∫ ∞

−∞

eiωtdω

ω2 −ω20 − ikω

This is something we can evaluate using contour integration. It is in fact explicitly one of the forms we’vedealt with before. The poles occur at

ω =12[±√

4ω20 − k2 + ik]

when 4ω20 − k2 > 0 these are nice complexn umbers. Call these ω+ and ω−. It’s pretty easy to compute

the residues and the final answer is

G(t) =1√

ω20 −

14 k2

e−12 kt sin

√ω2

0 −14

k2t

We come to another application of Fourier series. For an L2 function f we can define a sum

S(x) =∞

∑n=−∞

f (x + nx0).

This is periodic with period x0 (check this!). It thus has a Fourier series expansion

S(x) =∞

∑l=−∞

cl exp(

2πilxx0

).

We can compute the Fourier coefficients:

cl =1x0

∫ 12 x0

− 12 x0

S(x) exp(−2πilx

x0

)dx

=1x0

∫ 12 x0

− 12 x0

f (x + nx0) exp(−2πilx

x0

)51

We let y = x + nx0 in each term invovling n. If we do that, we get

cl =1x0

∫ ∞

−∞f (y) exp

(−2πily

x0

)dy

=2π

x0f(

2πlx0

).

Altogether we have the Poisson summation formula

S(x) =2π

x0

∞

∑l=−∞

F(

2πlx0

)exp

(2πilx

x0

).

So, we’ve re-written S(x) using Fourier coefficients.There is a convenient special case when x = 0 and x0 = 1. We get a more standard statement of the

Poisson summation formula∞

∑n=−∞

f (n) = 2π∞

∑k=−∞

f (k)

Example. We’ll consider the sum

S0 =∞

∑n=0

1a2 + n2 =

12a2 +

12

∞

∑n=−∞

1a2 + n2 .

f here is f (u) = 1a2+u2 and x0 = 1, x = 0. We have, by the Poisson summation formula that we have to

compute the Fourier transform:

f (k) =1

2π

∫ ∞

−∞

1a2 + u2 e−iukdu.

It is not hard to compute that the the transform is

12a

e−a|k|.

So the sum is

S0 =1

2a2 +2π

4a

∞

∑k=−∞

e−a|2πk|

=1

2a2 −π

2a+

π

a

∞

∑k=0

e2πka

=1

2a2 −π

2a+

π

a1

1− e−2πa

We give another example showing how to use Fourier (and Laplace!) transforms to solve a differentialequation.

Example. Consider an infinitely long vibrating string under tension (with small vibrations). Let y(x, t) bethe height of the string at point x and time t. It turns out that the differential equation for this is

∂2y∂x2 =

1c2

∂2y∂t2

and initial conditions

y = y0(x) t = 0

∂y∂t

= v0(x) t = 0

Take the Laplace transform in t to get

∂2Y∂x2 =

1c2 [−sy0(x)− v0(x)− s2Y(s, x)]

52

and take the Fourier transform in space, where we define ξ(s, k) = Y, and we get

−k2ξ = − sc2 y0(k)−

1c2 v0(k) +

s2

c2 ξ

and solve to get

ξ(k, s) =sy0(k) + v0(k)

s2 + c2k2

Now! We take the inverse laplace transform

12π

∫ γ+i∞

γ−∞

sy0(k) + v0(k)s2 + c2k2 eistds

The residues are s = ±ck and we get the the transform is

y0(k) cos kt + v0(k)sin kct

kc

Now the inverse Fourier transform! We’ll work on the first term, and leave the second term to you.Note that

cos kct =12(eikct + e−ikct).

Then12

∫ ∞

−∞y0(k)(eikct + e−ikct)eikxdk =

12[y0(x + ct)− y0(x− ct)].

If you take the inverse transform of the second term, we get, in total

y(x, t) =12[y0(x + ct)− y0(x− ct) +

1c

∫ x+ct

x−ctv0(x′)dx′]

Here is yet another example of an interesting thing that can be done with Fourier series. The moral ofthe following example is that you can sometimes just write down a Fourier series and learn somethingnew and interesting.

Example. Consider the function given by cosh h(x) when |x| ≤ π and is otherwise periodic. We wouldlike to write this as a Fourier series. To this end, write

cn =1

2π

∫ π

−π

(ex + e−x

2

)exp(−inx)dx

=12

12π

∫ π

−πex(1−in)dx +

12

12π

∫ π

−πex(1+in)dx

=1

4π

11− in

ex(1−in)∣∣∣∣π−π

+1

4π

11 + in

ex(1+in)∣∣∣∣π−π

=1

4π

11− in

(eπ(1−in) − e−π(1−in)) +1

4π

11 + in

(eπ(1+in) − e−π(1+in))

Now, eiπn = (−1)n and the above becomes

14π

11− in

(−1)n(eπ − e−π) +1

4π

11 + in

(−1)n(eπ − e−π)

=1π

11 + n2 (−1)n sinh(π)

And so the series is thensinh(π)

π

∞

∑n=−∞

(−1)n

n2 + 1einx.

53

Well, we haven’t learned anything new yet. But look at x = π. We get

cosh(π) =sinh(π)

π

∞

∑n=−∞

(−1)n

n2 + 1(−1)n

=sinh(π)

π

∞

∑n=−∞

1n2 + 1

We then finally get

π coth(π) = 2∞

∑n=0

1n2 + 1

− 1

and we get the identity:∞

∑n=0

1n2 + 1

=π coth(π)

2+

12

We present another example of a partial differential equation that one can solve using Fourier trans-forms.

Example. The following is the (one dimensional) heat equation which describe how heat would dispersealong, for example, a metal wire.

∂u∂t

= c∂2u∂x2 t > 0, −∞ < x < ∞

u(0, x) = g(x)

Solving this will predictably invovle taking a Fourier transform of both sides of the equation. This gives

∂u∂t

= −ck2u u(0, k) = g(k)

But the equation above is easy to solve, it’s a very standard initial value problem from ODEs:

u(t, k) = g(k)e−ck2t.

So, we know that the solution is a convolution of g and the inverse Fourier transform for e−ck2t, which itremains to compute.

We recall that the Fourier transform of e−Ax2is 1

2π e−k2/4A√

πA . In this case it would appear that A = 1

4ct

and so the inverse Fourier transform is 12π e−

14ct x2√

4πct.

54