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Complex Analysis
T.K. Subrahmonian Moothathu
Contents
1 Basic properties of C 2
2 Holomorphic functions, and the branches of logarithm 7
3 Power series 12
4 Mobius maps 17
5 Integration of a continuous function along a path 23
6 Cauchy’s integral formula and power series representation 28
7 Liouville’s theorem and Zeroes theorem 34
8 Some more fundamental theorems about holomorphic functions 38
9 Winding number and Cauchy’s general theory 43
10 Singularities 49
11 Residues 54
12 More on zeroes and poles 56
13 The automorphism group 60
14 Harmonic functions 66
This is just a first course in Complex Analysis; to master some of the deeper theorems in the
subject, the student should study an advanced course in Complex Analysis also. We will study
1
differentiable functions f : U → C, where U ⊂ C is a connected open set. Such functions are called
holomorphic or analytic functions. The notion of differentiability in Complex Analysis is so strong
that it imposes many rigid properties on the function f . Some such properties are the following.
Assume f : U → C is holomorphic, where U ⊂ C is a connected open set. Then:
(i) f is differentiable infinitely often.
(ii) Locally f is a combination of a stretching/shrinking and a rotation, and locally f has a power
series expansion. Moreover, f preserves angles provided f ′ is non-vanishing in U .
(iii) Any holomorphic g : U → C agreeing with f on an uncountable subset of U must agree with
f on the whole of U .
(iv) If f is non-constant, then f is an open map, and moreover, the maximum of |f | on any
nonempty compact set K ⊂ U is attained only on the boundary of K.
(v) If U has no ‘holes’, then for any two points z1, z2 ∈ U and any two smooth paths α, β in U
from z1 to z2, we have that∫α f =
∫β f .
(vi) If U = C, then f behaves somewhat like a polynomial. For instance, if f : C → C is holomorphic
and non-constant, then f(C) must contain all complex numbers with one possible exception.
Another aspect of the theory is the study of singularities. Let U ⊂ C be a connected open set
a ∈ U and f : U \ {a} → C be holomorphic. Then a is called a singularity of f . In some cases, f
or (z − a)kf (for some k ∈ N) will have a holomorphic extension to U . Otherwise, a is called an
essential singularity of f , and in this case the behavior of f near a will be quite wild: the f -image
of any deleted neighborhood of a will be dense in C. We also remark that many proofs in Complex
Analysis makes use of the following fact:∫|z−a|=r
dz
z − a= 2πi = 0 for a ∈ C and r > 0.
Some books for further reading:
1. D. Alpay, A Complex Analysis Problem Book.
2. J.B. Conway, Functions of One Complex Variable.
3. T.W. Gamelin, Complex Analysis.
4. T. Needham, Visual Complex Analysis.
5. R. Remmert, Theory of Complex Functions.
1 Basic properties of C
Let us recall the preliminary notions. If z = x + iy ∈ C, where x, y ∈ R, then x and y are called
respectively the real part and imaginary part of z, and we write Re(z) = x and Im(z) = y. The
2
conjugate of z is z = x− iy and the absolute value of z is |z| =√x2 + y2. If z = 0, we may write
z in polar coordinates as z = reiθ, where r = |z| > 0 and θ ∈ [0, 2π) (or θ ∈ [−π, π)); here θ is
called the argument of z and we write arg(z) = θ; note that x = r cos θ and y = r sin θ. Also recall
how the addition and multiplication of two complex numbers are defined: for a + ib, c + id ∈ C,
we have that (a+ ib) + (c + id) = (a+ c) + i(b + d) and (a+ ib)(c + id) = (ac− bd) + i(ad+ bc).
The geometric meaning of complex multiplication is better understood in polar coordinates: if
z = r1eiθ1 and w = r2e
iθ2 , then zw = r1r2ei(θ1+θ2). There is scaling as well as rotation involved in
complex multiplication. In particular, we have that zn = rneinθ if z = reiθ, and iz is the complex
number obtained by rotating z by an angle π/2 in the anticlockwise direction since i = eiπ/2.
Exercise-1: For z, w ∈ C, we have that:
(i) z = z, |z| = |z|, z + w = z + w, and zw = z w.
(ii) Re(z) = (z + z)/2, Im(z) = (z − z)/2i, and |z|2 = zz.
(iii) z ∈ R ⇔ z = z; and z ∈ iR ⇔ z = −z.
(iv) |zw| = |z||w|, |z + w| ≤ |z|+ |w| and |z ± w| ≥ ||z| − |w||.
Exercise-2: (i) [Cauchy’s inequality] If n ∈ N, then Cn is a Hilbert space with respect to the
inner product ⟨z, w⟩ :=∑n
k=1 zkwk for z = (z1, . . . , zn) and w = (w1, . . . , wn). Consequently,
|∑n
k=1 zkwk|2 ≤ (∑n
k=1 |zk|2)(∑n
k=1 |wk|2) for z1, . . . , zn, w1, . . . , wn ∈ C.
(ii) [Parallelogram law] |z + w|2 + |z − w|2 = 2(|z|2 + |w|2) for every z, w ∈ C.
(iii) If z = x+ iy ∈ C, then |z| ≤ |x|+ |y| ≤√2|z|.
[Hint : (iii) By (ii), we have (|x|+ |y|)2 + (|x| − |y|)2 = 2(|x|2 + |y|2) = 2|z|2.]
Algebraically, C is a field for which R is a subfield. Therefore, C is a vector space of dimension
one over itself, and of dimension two over R. Accordingly, we may talk about C-linearity and
R-linearity of maps T : C → C. As a metric space, we may identify C with R2, where a natural
metric on C is given by d(z, w) = |z − w|. Hence C is a complete separable metric space which is
path connected, locally path connected, and locally compact. Moreover, K ⊂ C is compact iff K
is closed and bounded in C (here, bounded means there is M > 0 such that |z| ≤M ∀ z ∈ K).
Notation: If z ∈ C and r > 0, then B(z, r) := {w ∈ C : |z−w| < r} denotes the open ball centered
at z with radius r. Also let B(z, r) := {w ∈ C : |z − w| ≤ r}. We will denote the open unit disc
B(0, 1) = {z ∈ C : |z| < 1} by D. Therefore, ∂D = {z ∈ C : |z| = 1} is the unit circle. The exercises
below describe various properties of C.
Exercise-3: (i) The usual order of R cannot be extended to C to make C a totally ordered field.
3
(ii) Describe all field automorphisms f : C → C such that f(x) = x for every x ∈ R.
[Hint : (i) One of the axioms of a totally ordered field is the following: xy > 0 whenever x > 0 and
y > 0. Now, if 0 < i, then 0 = 02 ≤ i2 ≤ −1, a contradiction. If 0 > i, then 0 < −i, leading to the
same contradiction. (ii) Since f(i)2 = f(i2) = f(−1) = −1, either f(i) = i or f(i) = −i. So f is
either the identity map or the conjugation map z 7→ z.]
Exercise-4: For a map T : C → C, the following are equivalent: (i) T is R-linear.
(ii) T (x+ iy) = xT (1) + yT (i) for every x+ iy ∈ C.
(iii) ∃ a real matrix
a b
c d
such that T (x+ iy) = (ax+ by) + i(cx+ dy) ∀ x+ iy ∈ C.
(iv) There exist p, q ∈ C such that T (z) = pz + qz for every z ∈ C.
Moreover, if statements (i)-(iv) hold, then T (1) = a + ic, T (i) = b + id, p =T (1)
2+T (i)
2i, and
q =T (1)
2− T (i)
2i. [Hint : (iii) ⇒ (iv): Use the fact that x = (z + z)/2 and y = (z − z)/2i.]
Exercise-5: Let T : C → C be R-linear. Then the following are equivalent: (i) T is C-linear.
(ii) T (i) = iT (1).
(iii) If T is given by the real matrix
a b
c d
, then a = d and b = −c.
(iv) There is p ∈ C such that T (z) = pz for every z ∈ C.
[Hint : Use Exercise-4 as needed. (ii) ⇒ (iii): b + id = T (i) = iT (1) = i(a + ic). (iii) ⇒ (iv): If
T (z) = pz + qz, then q =a+ ic
2− b+ id
2i= 0.]
Take a special note of (i) ⇔ (iii) in Exercise-5, as this will be needed later.
Exercise-6: Let U ⊂ C be open.
(i) If A ⊂ U is compact, then there is δ > 0 such that {z ∈ C : dist(z,A) < δ} =∪
a∈AB(a, δ) ⊂ U .
(ii) If B(a, r) ⊂ U , then there is s > 0 such that B(a, s) ⊂ U .
Exercise-7: Let z ∈ C. (i) If |z| < 1, then limn→∞ zn = 0.
(ii) If |z| > 1, then limz→∞ zn = ∞ (this means, for every C > 0, there isM > 0 such that |zn| ≥ C
whenever |z| ≥M).
(iii) If |z| = 1 and z = e2πiθ = cos 2πiθ + i sin 2πiθ for some θ ∈ Q, then zn = 1 for some n ∈ N
(complex numbers with this property are called nth roots of unity).
(iv) If |z| = 1 and z = e2πiθ for some θ ∈ R \ Q, then {zn : n ∈ N} is dense in the unit circle ∂D
(this is a bit non-trivial to prove).
Exercise-8: Let n ≥ 3 and z1, . . . , zn be the vertices of an n-sided regular polygon inscribed in the
4
unit circle ∂D. Then∑n
k=1 zk = 0. [Hint : If we put z = e2πi/n, then there is w ∈ ∂D such that
zk = wzk for 1 ≤ k ≤ n. Also 0 = zn − 1 = (z − 1)∑n
k=1 zk, and z = 1.]
Exercise-9: (i) The unit circle ∂D is an abelian group with respect to complex multiplication.
(ii) ϕ : (R,+) → (∂D, ·) given by ϕ(x) = eix is a continuous surjective group homomorphism.
(iii) A continuous group homomorphism ψ : (R,+) → (∂D, ·) can never be injective.
[Hint : (iii) ψ(R) must be connected and hence must contain a neighborhood of 1 ∈ ∂D. Hence
there is an nth root of unity in ψ(R) \ {1} for some n ≥ 2.]
Exercise-10: Let zn = xn + iyn ∈ C for n ∈ N. (i) If both (xn) and (yn) have convergent subse-
quences, should (zn) have a convergent subsequence?
(ii) If (|zn|) has a convergent subsequence, should (zn) have a convergent subsequence?
Exercise-11: If p ∈ C and q ∈ C \ {0}, recognize the sets L = {z ∈ C : Im( z−pq ) = 0} and
H = {z ∈ C : Im( z−pq ) > 0}. [Hint : L = {p + tq : t ∈ R} is a line and H is one of the two open
hyperplanes determined by L.]
Exercise-12: [Upper bound] (i) If z1, z2 ∈ D = {z ∈ C : |z| ≤ 1} and |z1−z2| ≥ 1, then |z1+z2| ≤√3.
(ii) If z1, z2, z3 ∈ D are distinct, then there exist j = k in {1, 2, 3} with min{|zj − zk|, |zj + zk|} ≤ 1.
(iii) If z1, . . . , zn ∈ D, then there exist ε1, . . . , εn ∈ {−1, 1} such that |∑n
j=1 εjzj | ≤√3.
[Hint : (i) Use parallelogram law. (ii) Assume zj ’s are non-zero. Let {w1, . . . , w6} = {±z1,±z2,±z3}.
Choose j = k such that the anticlockwise angle θ from wj to wk is ≤ π/3. Then cos θ ≥ 1/2 and
hence |wj−wk|2 = |wj |2+ |wk|2−2|wj ||wk| cos θ ≤ 1. (iii) Use induction on n, where the case n = 2
is covered in (i). Assume the result for values up to n, and consider z1, . . . , zn+1 ∈ D, where n ≥ 2.
We may suppose zj ’s are distinct, for otherwise the conclusion is easy by induction. Applying (ii)
to zn−1, zn, zn+1 (and relabelling if necessary), find ε ∈ {−1, 1} such that |zn + εzn+1| ≤ 1. Now
use the induction hypothesis for z1, . . . , zn−1, zn + εzn+1.]
Exercise-13: [Lower bound] If z1, . . . , zn ∈ C\{0}, then there is a nonempty set J ⊂ {1, . . . , n} such
that |∑
j∈J zj | > (1/6)∑n
k=1 |zk|. [Hint : After a rotation, assume arg(zj) /∈ {π/4, 3π/4, 5π/4, 7π/4}
for 1 ≤ j ≤ n. Let U = {z ∈ C \ {0} : π/4 < arg(z) < 3π/4}. After a rotation, we
may assume∑
zj∈U |zj | ≥ (1/4)∑n
k=1 |zk|. Since |zj | <√2Im(zj) for zj ∈ U , we get that
|∑
zj∈U zj | ≥∑
zj∈U Im(zj) >1√2
∑zj∈U |zj | ≥ 1
4√2
∑nk=1 |zk| > (1/6)
∑nk=1 |zk|.]
Exercise-14: Let a0, a1, . . . , an ∈ C. Then there is t ∈ [0, 1] such that |∑n
k=0 ake2πikt|2 ≥
∑nk=0 |ak|2.
[Hint : Let fk(t) = e2πikt. Then {f0, f1, . . . , fn} is an orthonormal set in the Hilbert space L2C[0, 1].
Hence by Parseval’s identity,∫ 10 |∑n
k=0 akfk(t)|2dt = ∥∑n
k=0 akfk∥22 =∑n
k=0 |ak|2.]
5
Exercise-15: Let a1, . . . , an, b1, . . . , bn ∈ C be such that∑n
k=1 ask =
∑nk=1 b
sk for every s ∈ {1, . . . , n}.
Then there is a permutation σ ∈ Sn such that bk = aσ(k) for 1 ≤ k ≤ n. [Hint : Let p(z) =∏n
k=1(z−
ak) and q(z) =∏n
n=1(z − bk). Enough to show p = q. Write p(z) = zn − c1zn−1 + · · · + (−1)ncn
and q(z) = zn − d1zn−1 + · · ·+(−1)ndn, where cs =
∑k1<···<ks
ak1 · · · aks and ds =∑
k1<···<ks
bk1 · · · bks .
So it suffices to show cs = ds for 1 ≤ s ≤ n. Use induction on s. By hypothesis, c1 = d1. Next, for
s < n, compute (∑n
k=1 ak)s+1 and (
∑nk=1 bk)
s+1 to deduce that cs+1 = ds+1.]
Exercise-16: If a1, . . . , an ∈ C are such that |∑n
k=1 ak| =∑n
k=1 |ak|, then there exist b ∈ C and
positive reals tk such that ak = tkb for 1 ≤ k ≤ n. [Hint : Use induction on n. If n = 2, then
squaring the given statement |a1+ a2| = |a1|+ |a2|, we may deduce Re(⟨a1, a2⟩) = |a1||a2|, and the
required conclusion follows. If n ≥ 2, then from∑n+1
k=1 |ak| = |∑n+1
k=1 ak| ≤ |∑n
k=1 ak| + |an+1|, we
get∑n
k=1 |ak| ≤ |∑n
k=1 ak|, and hence∑n
k=1 |ak| = |∑n
k=1 ak|. Similarly,∑n+1
k=2 |ak| ≤ |∑n+1
k=2 ak|.
Now apply twice the induction assumption for n.]
Exercise-17: Let a0 > a1 > · · · > an > 0 be reals, where n ≥ 2, and p(z) =∑n
k=0 akzk. If w ∈ C is
with p(w) = 0, then |w| > 1. [Hint : (1−w)p(w) = 0 and hence a0 =∑n
k=1(ak−1−ak)wk+anwn+1.
If |w| < 1, then a0 <∑n
k=1(ak−1 − ak) + an = a0, a contradiction. If |w| = 1, then a0 =
|∑n
k=1(ak−1 − ak)wk + anw
n+1| ≤∑n
k=1(ak−1 − ak) + an = a0 (and hence equality throughout) so
that we may deduce w = 1 using Exercise-16. But then p(w) =∑n
k=0 ak = 0, a contradiction.]
Exercise-18: (i) If θ ∈ (0, 2π), then limn→∞(1/n)∑n−1
k=0 eikθ = 0.
(ii) If θ ∈ (0, 2π), then∑n
k=1 cos kθ =cos(n+1
2 θ) sin(nθ2 )
sin( θ2)and
∑nk=1 sin kθ =
sin(n+12 θ) sin(nθ2 )
sin( θ2).
(iii) If r ∈ (0, 1) and θ ∈ R, then∑∞
n=0 rn cosnθ =
1− r cos θ
1 + r2 − 2r cos θand
∑∞n=0 r
n sinnθ =
r sin θ
1 + r2 − 2r cos θ.
[Hint : (i) (1/n)|∑n−1
k=0 eikθ| = |1− einθ|
n|1− eiθ|≤ 2
n|1− eiθ|→ 0.
(ii)∑n
k=1 eikθ =
eiθ(einθ − 1)
eiθ − 1=ei(n+1)θ/2(einθ/2 − e−inθ/2)
eiθ/2(eiθ/2 − e−iθ/2). Now expand and separate real and
imaginary parts. (iii)∑∞
n=0 rneinθ =
1
1− reiθ=
1− re−iθ
|1− reiθ|2=
1− r cos θ + ir sin θ
1 + r2 − 2r cos θ.]
Seminar topic: Consider the one-point compactification C∞ = C∪ {∞} of C (we may identify C∞
with a sphere) and discuss stereographic projection.
6
2 Holomorphic functions, and the branches of logarithm
Definition: Let U ⊂ C be open and f : U → C be a function. We say f is complex differentiable or
holomorphic at a ∈ U if limw→0f(a+ w)− f(a)
wexists (here w varies in C\{0} with a+w ∈ U). If
the limit exists, it is denoted as f ′(a). We say f is holomorphic (or analytic) in U if f is holomorphic
at each point of U . Let H(U) = {f : U → C : f is holomorphic in U}.
[101] (i) Let U ⊂ C be a nonempty open set. Then H(U) is a complex vector space with respect
to pointwise operations. Moreover, the pointwise product of two members of H(U) also belongs
to H(U). We have that (af + bf)′ = af ′ + bg′ and (fg)′ = fg′ + f ′g for every f, g ∈ H(U) and
a, b ∈ C. Also, (f/h)′ = (f ′h− fh′)/h2 for f, h ∈ H(U) provided h is non-vanishing in U .
(ii) If U ⊂ C is open, then every f ∈ H(U) is continuous in U .
(iii) Let U, V ⊂ C be open, f ∈ H(U), g ∈ H(V ) and suppose f(U) ⊂ V . Then g ◦ f ∈ H(U) and
(g ◦ f)′(z) = g′(f(z))f ′(z) for every z ∈ U .
Proof. Proofs are as in Real Analysis.
Example: The identity map and every constant function belong to H(C). Being a finite product
of the identity map, the function z 7→ zn belongs to H(C) for every n ∈ N. Being the linear
combination of such functions, every complex polynomial belongs to H(C). If p, q are two complex
polynomials and U = {z ∈ C : q(z) = 0} (which is open), then the rational function p/q ∈ H(U).
In Real Analysis, producing a continuous function which is nowhere differentiable is not trivial.
But in Complex Analysis, the notion of differentiability is so strong that many natural continuous
functions are nowhere differentiable:
Exercise-19: The following continuous functions are nowhere differentiable in C: z 7→ z, z 7→
Re(z), z 7→ Im(z), z 7→ |z|. [Hint : Consider the quotientf(a+ w)− f(a)
win the definition of
differentiability and compare the limits when w is real and when w is purely imaginary.]
Definition: Let f : U → C be a function, where U ⊂ C is open. (i) Identifying C with R2, we can
find functions u, v : U ⊂ R2 → R such that f = u+ iv. Here u is called the real part and v is called
the imaginary part of f . We write u = Re(f) and v = Im(f). The partial derivatives of u and v
with respect to the two real variables x and y (when they exist) will be denoted as ux, uy, vx, vy.
(ii) Since x =z + z
2and y =
z − z
2i, we may consider the partial derivative of f = u + iv with
respect to z. Formally, we have that∂f
∂z=∂u
∂z+ i
∂v
∂z= [
∂u
∂x
∂x
∂z+∂u
∂y
∂y
∂z] + i[
∂v
∂x
∂x
∂z+∂v
∂y
∂y
∂z] =
7
(1/2)[(ux − vy) + i(uy + vx)]. Therefore we define the action of the differential operator∂
∂zon
f = u+ iv as∂f
∂z= (1/2)[(ux − vy) + i(uy + vx)] when the partial derivatives exist.
Exercise-20: Let f = u + iv : U → C be holomorphic and a ∈ U . Then f ′(a) = ux(a) + ivx(a) =
vy(a)− iuy(a). [Hint : Considerf(a+ w)− f(a)
w, and let w → 0 through R and through iR.]
[102] Let f : U → C be a function, where U ⊂ C is open, let u = Re(f), v = Im(f), and let a ∈ U .
Then the following are equivalent: (i) f is complex differentiable at a.
(ii) f is real differentiable at a, and the differential of f at a is C-linear.
(iii) f is real differentiable at a, and ux(a) = vy(a), uy(a) = −vx(a).
(iv) f is real differentiable at a and∂f
∂z(a) = 0.
Proof. The result follows from the two observations below about f = u+ iv (also see Exercise-5).
Observation-1 : f is real differentiable at a iff there is a linear map Dfa : R2 → R2 such that
limR2∋w→0|f(a+ w)− f(a)−Dfa(w)|
|w|= 0; and the matrix of Dfa is
ux(a) uy(a)
vx(a) vy(a)
.
Observation-2 : f is complex differentiable at a iff limC∋w→0|f(a+ w)− f(a)− f ′(a)w|
|w|= 0; and
the differential of f at a is the map z 7→ f ′(a)z from C to C. As as a map from R2 to R2,
it becomes (x, y) 7→ (Re(f ′(a))x − Im(f ′(a))y, Im(f ′(a))x + Re(f ′(a))y) so that its matrix is Re(f ′(a)) −Im(f ′(a))
Im(f ′(a)) Re(f ′(a))
=
ux(a) −vx(a)
vx(a) ux(a)
=
vy(a) uy(a)
−uy(a) vy(a)
by Exercise-20.
Remark: The equations ux = vy, uy = −vx are known as Cauchy-Riemann equations (C-R equa-
tions for short). Just because Cauchy-Riemann equations are true for f : U → C at a ∈ U , we
cannot conclude that f is complex differentiable at a (since f may not be real differentiable at
a). For example, consider f : C → C given by f(0) = 0 and f(z) = z5/|z|4 for z = 0; then f is
continuous on C, the Cauchy-Riemann equations hold at 0, but f is not (complex) differentiable
at 0. However, we have two positive results:
[103] Let U ⊂ C be open and f = u+ iv : U → C be a function.
(i) [Looman-Menchoff] If f is continuous on U and the Cauchy-Riemann equations hold at every
a ∈ U , then f ∈ H(U).
(ii) If the Cauchy-Riemann equations hold at every a ∈ U and the partial derivatives ux, uy, vx, vy
are continuous on U , then f ∈ H(U).
8
Proof. The proof of (i) is left to the student as a reading assignment. Statement (ii) follows from
[102] because the continuity of the partial derivatives implies that f is real differentiable on U .
[104] Let U ⊂ C be a connected open set and f ∈ H(U).
(i) If f ′ ≡ 0, then f is a constant function.
(ii) If for some n ∈ N, the nth derivative f (n) ≡ 0, then f is a polynomial of degree ≤ n− 1.
Proof. (i) Write f = u+ iv. Then ux, uy, vx, vy are identically zero by Exercise-20. Hence u, v are
constants as a standard fact in Real Analysis (since U is connected). Hence f is constant.
(ii) Apply (i) repeatedly.
Exercise-21: For each f below, determine {z ∈ C : f is holomorphic at z}: (i) f(x + iy) = (x3 +
y) + i(2xy − x2), (ii) f(x+ iy) = (x2 − 5xy) + i(5xy − y2), (iii) f(x+ iy) = 3x3y2 + i3x2y3. [Hint :
Since f is real differentiable, it suffices to check Cauchy-Riemann equations. (i) {−i/2, 2/3 + i/6},
(ii) {0}, (iii) {x+ iy : x = 0 or y = 0}.]
Exercise-22: Let U ⊂ C be a connected open set and f ∈ H(U). If either f(U) ⊂ R or f(U) ⊂ iR,
then f is a constant. [Hint : If f = u+ iv, then v ≡ 0 or u ≡ 0. Use Cauchy-Riemann equations.]
Exercise-23: [A geometric insight] Let U ⊂ C be open and f ∈ H(U). Think of f as a map from
U ⊂ R2 to R2, and let Jf (z) be the Jacobian of f at z. Then,
(i) det(Jf (z)) = |f ′(z)|2.
(ii) If U is bounded, and f is bounded and injective, then Area(f(U)) =∫ ∫
U |f ′(z)|2dxdy.
[Hint : (i) Use Exercise-20 after noting that Jf (z) =
ux(z) uy(z)
vx(z) vy(z)
=
ux(z) −vx(z)
vx(z) ux(z)
by
Cauchy-Riemann equations. (ii) Area(f(U)) =∫ ∫
U det(Jf (z))dxdy by Multivariabele Calculus.]
[105] The (complex) exponential function e : C → C is defined as e(x + iy) = ex(cos y + i sin y).
Then, (i) e ∈ H(C), ez = ez, anddez
dz= ez.
(ii) e is a surjective group homomorphism from (C,+) onto (C \ {0}, ·) with ker(e) = 2πiZ. In
particular, ez+w = ezew and e−z = 1/ez.
(iii) For each b ∈ R, the exponential function is injective on {z ∈ C : b ≤ Im(z) < b+ 2π}.
Proof. (i) Let u = Re(e) and v = Im(e). Check that the partial derivatives of u and v are
continuous and Cauchy-Riemann equations hold. Use [103](ii) to conclude e is holomorphic. It is
easy to see ez = ez. To showdez
dz= ez, use Exercise-20.
9
Statement (ii) is left to the student. For (iii), note that if |Im(z)− Im(w)| < 2π and ez = ew,
then z − w ∈ ker(e) = 2πiZ and |Im(z − w)| < 2π, and hence z = w.
Remark: (i) For c ∈ R, the vertical line x = c in C is wound around the circle |z| = ec infinitely
many times with period 2π by the exponential function. Hence, the left half-plane {x+ iy : x < 0}
is mapped onto the bounded set D \ {0} by e. (ii) The real exponential function x 7→ ex from R to
R is injective with range (0,∞). So for every y ∈ (0,∞), we have defined log y (in Real Analysis)
as the unique x ∈ R satisfying ex = y. Since the complex exponential function is not injective, it is
not possible to define complex log function in a unique way. Instead, we have a countably infinite
collection of complex log functions; they are called the branches of logarithm (defined below).
Definition: Let U ⊂ C be a connected open set. A continuous function f : U → C is called a branch
of logarithm on U if ef(z) = z for every z ∈ U . Note that a branch of logarithm is always injective.
It may happen that a connected open set U ⊂ C does not admit any branch of logarithm. For
example, if 0 ∈ U , then U cannot admit a branch of logarithm because the exponential function
never takes the value 0. Thus there is no branch of logarithm on C and D. It is of interest to find
necessary and sufficient conditions for a connected open set to admit a branch of logarithm.
[106] Let U ⊂ C be a connected open set and f : U → C be a branch of logarithm. Then,
(i) f ∈ H(U) and f ′(a) = 1/a for every a ∈ U .
(ii) A continuous function g : U → C is a branch of logarithm iff there is k ∈ Z such that
g(z) = f(z) + 2πik for every z ∈ U .
Proof. (i) limw→0ef(a+w) − ef(a)
f(a+ w)− f(a)=dez
dz|z=f(a) = ef(a) = a = 0 since f is continuous and injective,
and since 0 /∈ U . Hence 1/a = limw→0f(a+ w)− f(a)
ef(a+w) − ef(a)= limw→0
f(a+ w)− f(a)
w.
(ii) ⇒: (g − f)(U) ⊂ ker(e) = 2πiZ. Moreover, (g − f)(U) must be connected since f, g are
continuous and U is connected. Therefore g − f ≡ 2πik for some k ∈ Z.
[107] Let U ⊂ C be a connected open set containing the unit circle. Then U does not admit a
branch of logarithm. In particular, C \ {0} does not admit a branch of logarithm.
Proof. Suppose f : U → C is a branch of logarithm, i.e., f is continous and ef(z) = z for every z ∈ U .
For each eiθ ∈ ∂D ⊂ U , we have ef(eiθ) = eiθ and hence there is kθ ∈ Z with f(eiθ) = i(θ + 2πkθ).
Since f is uniformly continuous on the compact set ∂D, there is δ ∈ (0, π) such that |f(z)−f(w)| < π
for every z, w ∈ ∂D with |z−w| < δ. This implies kθ1 = kθ2 if |eiθ1 − eiθ2 | < δ. Consequently, there
10
is a unique k ∈ Z such that f(eiθ) = i(θ + 2πk) for every θ ∈ [0, 2π). Let θ ∈ (π, 2π) be near to 2π
so that |1− eiθ| < δ. Then π > |f(1)− f(eiθ)| = |i(2πk)− i(θ+2πk)| = θ > π, a contradiction.
[108] (i) C \ [0,∞) and C \ (−∞, 0] admit branches of logarithm.
(ii) If B ⊂ C is an open ball with 0 /∈ B, then B admits a branch of logarithm.
Proof. Any ball B ⊂ C \ {0} is included in either C \ [0,∞) or C \ (−∞, 0]. Hence it suffices to
prove (i). If z ∈ C \ [0,∞), then |z| > 0 and there is a unique θ ∈ (0, 2π) with z = |z|eiθ. Define
f : C \ [0,∞) → C as f(z) = log |z|+ iθ. It is easy to verify that f is continuous and ef(z) = z for
every z ∈ C\ [0,∞). Similarly, if z ∈ C\(−∞, 0], then there is a unique θ ∈ (−π, π) with z = |z|eiθ,
and therefore f : C \ (−∞, 0] → C defined as f(z) = log |z|+ iθ is a branch of logarithm.
Remark: Let f : C \ (−∞, 0] → C be f(z) = log |z|+ i arg(z), where arg(z) ∈ (−π, π). We know f
is a branch of logarithm. Moreover, f agrees with the real logarithm on (0,∞). Hence f is often
called the principal branch of logarithm, and is sometimes denoted by the symbol ‘log’. That is,
log z := log |z| + i arg(z), z ∈ C \ (−∞, 0]. It must be kept in mind that this is only a convenient
notation, and there are infinitely many branches of logarithm.
If a, b ∈ (0,∞), then ab = eb log a. We may generalize this using branches of logarithm to define
ab, when a, b are complex numbers with a = 0. It must be noted that since we use branches of
logarithm, in general there in no unique value for ab for a ∈ C \ {0} and b ∈ C.
Definition: Let U ⊂ C be a connected open set and f : U → C be a branch of logarithm. If a ∈ U
and b ∈ C, then ebf(a) is called a branch of ab.
Example: Let U = C \ (−∞, 0] and n ≥ 2 be an integer. Then z1/n is not a unique-valued
function in U ; on the other hand, z1/n has exactly n distinct holomorphic branches in U ; they are
ef0/n, ef1/n, . . . , efn−1/n, where fk(z) = log |z|+ i(arg(z) + 2πk) for z ∈ U for 0 ≤ k ≤ n− 1.
Remark: If a ∈ C \ [0,∞) and b ∈ C \ Z, then we may see from the definition that ab will have
at least two branches and therefore ab is not defined as a unique complex number. In particular,
the identity (ab)c = abc is not true in general for complex numbers a, b, c. If this is not understood
properly, one may get into absurdities. Here is a false proof showing that every integer is zero: if
k ∈ Z, then e2πk = (e2πik)−i = 1−i = 1, and hence 2πk = 0, which implies k = 0.
Definition: Motivated by the identities cos y = (eiy + e−iy)/2 and sin y = (eiy − e−iy)/2i, we
define cos z = (eiz + e−iz)/2 and sin z = (eiz − e−iz)/2i for z ∈ C. Since the exponential function
is holomorphic, it follows that cos z and sin z are holomorphic. It may be verified that (cos z)′ =
11
− sin z, (sin z)′ = cos z, and cos(z+2πk) = cos z and sin(z+2πk) = sin z for every k ∈ Z. Moreover,
standard trigonometric identities such as cos2 z + sin2 z = 1 are true.
Exercise-24: The functions cos z and sin z are surjective on C. [Hint : Consider w ∈ C. We have
cos z = w iff u2 − 2wu+1 = 0, where u = eiz. Since the constant term is non-zero, the roots of the
quadratic polynomial are non-zero and hence are of the form eiz.]
Remark: (i) By Exercise-24, cos z and sin z are unbounded (unlike the real functions cosx and
sinx); this can be seen directly also: | cos(in)| → ∞ and | sin(in)| → ∞ as n → ∞. (ii) Let c > 0,
L = {z ∈ C : Im(z) = c}, a = (ec + e−c)/2, and b = (ec − e−c)/2. It may be seen that cos z
maps the line L onto the ellipse {w ∈ C :[Re(w)]2
a2+
[Im(w)]2
b2= 1}, winding L around the ellipse
infinitely many times in a periodic manner.
Seminar topic: Further analysis about complex trigonometric functions.
3 Power series
We will see a little later that all holomorphic functions can be locally expressed as a power series.
Because of this, here we will look at some of the basic properties of a complex power series.
Definition: Let X be a set, fn : X → C be functions for n ∈ N and W ⊂ X. We say that
(i) the series∑∞
n=1 fn converges absolutely on W if∑∞
n=1 |fn(w)| <∞ for each w ∈W ,
(ii) the sequence (fn) converges uniformly on W to a function f :W → C if for every ε > 0, there
is n0 such that sup{|f(w)− fn(w)| : w ∈W} < ε for every n ≥ n0,
(iii) the series∑∞
n=1 fn converges uniformly on W to a function f :W → C if for sn := f1+ · · ·+fn,
the sequence (sn) converges uniformly on W to f .
Exercise-25: (i) [Weierstrass’ M-test] Let X be a set, fn : X → C be functions for n ∈ N and
W ⊂ X. Suppose there are Mn > 0 such that∑∞
n=1Mn <∞ and sup{|fn(w)| : w ∈W} ≤Mn for
each n ∈ N. Then∑∞
n=1 fn converges absolutely and uniformly on W , where the uniform limit is
the function f : W → C given by f(w) =∑∞
n=1 fn(w) for w ∈ W . Moreover, for any permutation
σ of N, the series∑∞
n=1 fσ(n) also converges uniformly on W to the original limit function.
(ii) [Special case] Let∑∞
n=0 an(z− a)n be a formal power series in C, where a, an ∈ C. Let W ⊂ C,
and Mn > 0 be such that∑∞
n=1Mn < ∞ and sup{|an||z − a|n : z ∈ W} ≤ Mn for each n ≥ 0.
Then the series∑∞
n=0 an(z − a)n converges absolutely and uniformly on W .
[Hint : (i) Recall from Real Analysis. To prove uniform convergence, let sn = f1 + · · · + fn, and
12
note that sup{|sn+1(w)−sn(w)| : w ∈W} ≤Mn+1, which implies (sn) is uniformly Cauchy on W .]
Definition: The radius of convergence R of a formal power series∑∞
n=0 an(z − a)n in C is defined
as R = sup{r ≥ 0 : (anrn)∞n=0 is bounded} ∈ [0,∞]. For any R ∈ (0,∞), the series
∑∞n=0
zn
Rnhas
radius of convergence equal to R.
Let B(a, r) = {z ∈ C : |z − a| ≤ r} for r ∈ [0,∞].
[109] Let∑∞
n=0 an(z − a)n be a formal power series in C with radius of convergence R. Then,
(i) [Cauchy-Hadamard] R =1
lim supn→∞ |an|1/n(where 1/0 = ∞ and 1/∞ = 0 by convention).
(ii) R = limn→∞ | anan+1
|, provided the limit exists.
Proof. (i) Let L =1
lim supn→∞ |an|1/n. If 0 < r < L, then 1/r > lim supn→∞ |an|1/n, and hence
there is n0 ∈ N such that 1/r > |an|1/n, or equivalently |an|rn < 1 for every n ≥ n0. This means
r ≤ R, and hence L ≤ R. If L < r < ∞, choose t ∈ (L, r). Then 1/t < lim supn→∞ |an|1/n so
that 1/tn < |an| for infinitely many n ∈ N. Then (r/t)n < |an|rn for infinitely many n ∈ N, which
implies R < r and hence R ≤ L.
(ii) Suppose that the limit L := limn→∞ | anan+1
| exists. If 0 < r < L, then there is n0 ∈ N such that
|an/an+1| > r, or equivalently |an+1| < |an|/r for every n ≥ n0. Hence |an|rn ≤ |an0 |rn/rn−n0 =
|an0 |rn0 for every n ≥ n0, which shows r ≤ R and hence L ≤ R. If L < r < ∞, choose t ∈ (L, r).
Then there is n0 ∈ N such that |an/an+1| < t, or equivalently |an+1| > |an|/t for every n ≥ n0.
Hence |an|rn > |an0 |rn/tn−n0 = |an0 |tn0(r/t)n for every n ≥ n0. Since t < r, this means (anrn) is
unbounded and hence R < r. Therefore R ≤ L.
Exercise-26: Find the radius of convergence for the following series:
(i)∑∞
n=1
zn
nk, where k ∈ N ∪ {0}.
(ii)∑∞
n=1 zkn , where (kn) is a strictly increasing sequence in N ∪ {0}.
(iii)∑∞
n=0
zn
n!.
(iv)∑∞
n=0 n!zn.
(v)∑∞
n=0(sinn)zn.
[Hint : (i) R = limn→∞ | anan+1
| = limn→∞(1 + 1/n)k = 1. (ii) lim supn→∞ |an|1/n = 1 and hence
R ≥ 1. By considering z = 1, we see R = 1. (iii) R = limn→∞ |an/an+1| = limn→∞ |n + 1| = ∞.
(iv) R = limn→∞ |an/an+1| = 0. (v) Clearly, lim supn→∞ |an|1/n ≤ 1. Since the interval [π/4, 3π/4]
has length > 1, n (mod 2π) ∈ [π/4, 3π/4] for infinitely many n ∈ N and hence sinn ≥ 1/√2 for
infinitely many n ∈ N. Hence lim supn→∞ |an|1/n ≥ 1 also. So R = 1.]
13
Remark: Since R = ∞ for Exercise-26(iii), we deduce limn→∞(1/n!)1/n = 0 by Cauchy-Hadamard.
[110] Let∑∞
n=0 an(z − a)n be a formal power series in C with radius of convergence R. Then,
(i) If z ∈ C and |z − a| > R, then∑∞
n=0 an(z − a)n does not converge.
(ii) If 0 ≤ r < R, then∑∞
n=0 an(z − a)n converges absolutely and uniformly on B(a, r).
(iii) We cannot say anything about the case where |z−a| = R (it depends on the particular series).
(iv) The radius of convergence of the series∑∞
n=1 nan(z − a)n−1 is also equal to R.
Proof. (i) Let sn(z) =∑n
k=0 ak(z − a)k. Assume |z − a| > R. Then (an|z − a|n) is unbounded
by the definition of R, and hence (sn(z)) should also be unbounded since |an+1(z − a)n+1| ≤
|sn+1(z)|+ |sn(z)|. Therefore (sn(z)) cannot converge in C.
(ii) If r = 0, there is nothing to prove. So assume 0 < r < R, and choose t ∈ (r,R). By the
definition of R, there is M > 0 such that supn≥0 |an|tn ≤ M . Then for Mn := M(r/t)n, we have
that∑∞
n=0Mn <∞ and |an|rn ≤Mn for every n ≥ 0. So Weierstrass’ M-test applies.
(iii) Check that the radius of convergence is equal to 1 for the following three series:∑∞
n=1
zn
n2,∑∞
n=0 zn, and
∑∞n=1
(−1)n−1zn
n. By Weierstrass’ M-test,
∑∞n=1
zn
n2converges for every z ∈ ∂D. The
series∑∞
n=0 zn does not converge for any z ∈ ∂D since |sn+1(z)−sn(z)| = 1, where sn(z) =
∑nk=0 z
k.
The series∑∞
n=1
(−1)n−1zn
nconverges for z = 1 but does not converge for z = −1.
(iv) Let R′ be the radius of convergence of∑∞
n=1 nan(z − a)n−1. If 0 < r < R′, then (nanrn−1)
is bounded, which implies (anrn) is bounded, and hence R′ ≤ R. If 0 < r < R, choose t ∈ (r,R).
Then M := supn≥0 |an|tn < ∞. Now n|an|rn−1 ≤ nMrn
rtn→ 0 as n → ∞ since r < t, and hence
(nanrn−1) is bounded. This shows r ≤ R′ and hence R ≤ R′.
Remark: If∑∞
n=0 an(z − a)n is a power series in C with radius of convergence R = ∞, then the
series converges uniformly in the compact ball B(a, r) for every r ∈ (0,∞), but the series may not
converge uniformly on the whole of C. Read [110](ii) carefully.
[111] Let∑∞
n=0 an(z − a)n be a power series in C with radius of convergence R > 0, and let
f(z) =∑∞
n=0 an(z − a)n be the limit function for |z − a| < R. Then, for |z − a| < R, we have:
(i) f is holomorphic and f ′(z) =∑∞
n=1 nan(z − a)n−1.
(ii) f is infinitely often differentiable and f (k)(z) =∑∞
n=k
n!
(n− k)!an(z − a)n−k for every k ∈ N.
(iii) ak =f (k)(a)
k!for every k ≥ 0 so that f(z) =
∑∞n=0
f (n)(a)
n!(z − a)n .
Proof. A repeated application of (i) and [110](iv) will give (ii); and (iii) can be deduced from (ii)
14
by putting z = a in the expression for f (k)(z). Therefore it suffices to establish (i). Let g(z) =∑∞n=1 nan(z − a)n−1 for z ∈ B(a,R), where g is well-defined by [110](iv). Consider z0 ∈ B(a,R)
and ε > 0. We need to find δ > 0 such that |f(z)− f(z0)
z − z0− g(z0)| < ε for every z ∈ B(a,R) with
0 < |z − z0| < δ. Choose r > 0 such that |z0 − a| < r < R. Let sn(z) =∑n
k=0 ak(z − a)k and
Rn(z) = f(z)− sn(z) =∑∞
k=n+1 ak(z − a)k for z ∈ B(a,R). Since (s′n) → g uniformly on B(a, r),
there is n1 ∈ N such that |g(z)− s′n(z)| < ε/3 for every n ≥ n1 and z ∈ B(a, r).
Next note that for z ∈ B(a, r) \ {z0}, |Rn(z) − Rn(z0)| ≤∑∞
k=n+1 |ak||(z − a)k − (z0 − a)k| =∑∞k=n+1 |ak||z − z0||
∑k−1j=0(z − a)k−1−j(z0 − a)|, and hence |Rn(z)−Rn(z0)
z − z0| ≤
∑∞k=n+1 |ak|krk−1.
Since∑∞
n=1 nan(z− a)n−1 converges absolutely and uniformly in B(a, r), there is n2 ∈ N such that
|Rn(z)−Rn(z0)
z − z0| < ε/3 for every n ≥ n2 and every z ∈ B(a, r) \ {z0}. Let n0 = max{n1, n2}.
Since the polynomial sn0 is differentiable, we may choose δ > 0 such that B(z0, δ) ⊂ B(a, r) and
such that |sn0(z)− sn0(z0)
z − z0− s′n0
(z0)| < ε/3 for every z ∈ B(z0, δ)\{z0}. Then for 0 < |z− z0| < δ,
we have |f(z)− f(z0)
z − z0−g(z0)| ≤ |sn0(z)− sn0(z0)
z − z0−s′n0
(z0)|+|Rn(z)−Rn(z0)
z − z0|+|s′n0
(z0)−g(z0)| <
ε/3 + ε/3 + ε/3 = ε.
Exercise-27: A = {all complex power series with radius of convergence = ∞}. Define addition and
multiplication on A as follows: (∑∞
n=0 anzn) + (
∑∞n=0 bnz
n) :=∑∞
n=0(an + bn)zn, and
(∑∞
n=0 anzn)(∑∞
n=0 bnzn) :=
∑∞n=0 cnz
n, where cn :=∑
j+k=n ajbk. Then,
(i) A is closed under addition and multiplication.
(ii) A is a commutative ring with unity. Moreover, A is an integral domain.
(iii)∑∞
n=0 anzn ∈ A has a multiplicative inverse iff a0 = 0.
[112] (i) Let f ∈ H(C) be such that f(0) = 1 and f ′ = f . Then f(z) = ez for z ∈ C.
(ii) ez =∑∞
n=0
zn
n!, cos z =
∑∞n=0
(−1)nz2n
(2n)!, and sin z =
∑∞n=0
(−1)nz2n+1
(2n+ 1)!for z ∈ C.
(iii) Let U ⊂ C be a connected open set and f ∈ H(U) be such that f ′(z) = 1/z for every z ∈ U
and suppose there is z0 ∈ U with ef(z0) = z0. Then f is a branch of logarithm on U .
(iv) If f is the principal branch of logarithm on C \ (−∞, 0], then f(z) =∑∞
n=1
(−1)n−1(z − 1)n
nfor |z − 1| < 1.
Proof. (i) Let g : C → C be g(z) = f(z)e−z. Then g ∈ H(C), g(0) = 1, and g′(z) = f ′(z)e−z +
f(z)(−e−z) = 0 for every z ∈ C. Hence g ≡ 1 by [104](i).
(ii) By Exercise-26(iii) and [111](i), f : C → C defined as f(z) =∑∞
n=0
zn
n!is holomorphic. More-
over, f(0) = 1 and f ′ = f . Hence by part (i), f(z) = ez for z ∈ C. Now the power series
15
expressions for cos z and sin z can be derived from the definition that cos z = (eiz + e−iz)/2 and
sin z = (eiz − e−iz)/2i.
(iii) Let g : U → C be g(z) = ze−f(z). Then g ∈ H(U), g(z0) = 1 and g′(z) = e−f(z) +
z(−1/z)e−f(z) = 0 for every z ∈ U . Hence g ≡ 1 by [104](i).
(iv) Recall that f(z) = log |z|+ iarg(z) for z ∈ C \ (−∞, 0]. The function g : B(1, 1) → C defined
as g(z) =∑∞
n=1
(−1)n−1(z − 1)n
nis holomorphic by [111](i) since the radius of convergence of
the power series is 1. Moreover, term by term differentiation yields (as in [111](i)) that g′(z) =∑∞n=1(−1)n−1(z − 1)n−1 =
1
1 + (z − 1)= 1/z for z ∈ B(1, 1). Also, eg(1) = e0 = 1. Hence g must
be a branch of logarithm on B(1, 1) by part (iii). So g − f ≡ 2πik on B(1, 1) for some k ∈ Z by
[106](ii). Since g(0) = 1 = f(0), we deduce that k = 0, and thus g = f on B(1, 1).
Exercise-28: [Real Analysis!] (i) log(1+x) =∑∞
n=1
(−1)n−1xn
nfor |x| < 1, and hence we have that
log(1− x) = −∑∞
n=1 xn/n for 0 < x < 1. Consequently, log(1− x) + x < 0 for 0 < x < 1.
(ii)x
1 + x< log(1 + x) < x for x > 0.
(iii) Let an = (∑n
k=1 1/k)− logn. Then an > 0 for every n ∈ N and (an) is a decreasing sequence.
Hence (an) converges (the limit is called the Euler’s constant).
[Hint : (ii) Let f(x) = log(1+x). By Mean value theorem, there is c ∈ (0, x) such that log(1+x) =
f(x) − f(0) = f ′(c)x = x/(1 + c) < x. Next, let y = 1/(1 + x) and note that 1 − y = x/(1 + x).
By (i), we have 0 > y + log(1 − y) = x/(1 + x) + log(1/(1 + x)) = x/(1 + x) − log(1 + x). (iii)
Since 1/x ≥ 1/k for x ∈ [k, k + 1], we have an ≥ (∫ n+11
dx
x)− log n = log(n+ 1)− log n > 0. Also,
an − an−1 = log(n− 1)− log n+ 1/n = log(1− 1/n) + 1/n < 0 by (i).]
Exercise-29: Let f(z) =∑∞
n=0 anzn, where the series converges for every z ∈ C.
(i) If f(R) ⊂ R, then an ∈ R for every n ≥ 0.
(ii) If f(R) ⊂ R and f(iR) ⊂ iR, then f(−z) = −f(z) for every z ∈ C.
[Hint : Write an = bn + icn. (i) Since |bn|, |cn| are ≤ |an|, the functions g(z) =∑∞
n=0 bnzn and
h(z) =∑∞
n=0 cnzn are holomorphic on C. Also f(z) = g(z)+ ih(z), g(R) ⊂ R and h(R) ⊂ R. Hence
by hypothesis h ≡ 0 so that f = g. (ii) In addition, b2n = 0 for every n ≥ 0.]
Exercise-30:∑∞
n=1
n
2n= 2. [Hint : Differentiate the identity
∑∞n=0 z
n = (1 − z)−1 (for |z| < 1) to
obtain∑∞
n=1 nzn−1 = (1− z)−2 for |z| < 1. Next put z = 1/2.]
Exercise-31: (i) If z ∈ C is with Re(z) > 0, then∫ 0−∞
dt
(t− z)2=∫∞0 e−tzdt.
(ii) If z1, z2 ∈ {z ∈ C : Re(z) ≤ 0}, then |ez1 − ez2 | ≤ |z1 − z2|.
(iii) |1− (1− z)ez| ≤ |z|2 whenever |z| ≤ 1.
16
[Hint : (i)∫ 0s
dt
(t− z)2=
1
z − t|0t=s → 1/z as s → −∞. Similarly,
∫ s0 e
−tzdt =e−tz
−z|st=0 → 1/z as
s→ ∞. (ii) Note that ez − 1 = z∫ 10 e
tzdt and hence |ez − 1| ≤ |z|∫ 10 e
Re(tz)dt. Assuming Re(z1) ≤
Re(z2), we obtain |ez1 − ez2 | = |ez2 ||ez1−z2 − 1| ≤ |ez1−z2 − 1| ≤ |z1 − z2|∫ 10 e
Re(tz1−tz2)dt ≤ |z1 − z2|
using the fact that ew ≤ 1 when Re(w) ≤ 0. (iii) 1 − (1 − z)ez =∑∞
n=2 zn(
1
(n− 1)!− 1
n!). Hence
for |z| ≤ 1, we see |1− (1− z)ez| ≤ |z|2∑∞
n=2(1
(n− 1)!− 1
n!) = |z|2.]
Exercise-32: (i) Let an ∈ C be such that∑∞
n=0 |an+1 − an| < ∞. Then the radius of convergence
of the power series∑∞
n=0 anzn is ≥ 1.
(ii) Let∑∞
n=0 anzn be a complex power series with radius of convergence > 1, and also assume
|∑∞
n=0 anzn| ≤ 1 whenever |z| ≤ 1. Then
∑∞n=0 |an|2 ≤ 1.
[Hint : (i) Let M =∑∞
n=0 |an+1 − an| < ∞. Then |an| ≤ |a1| +∑n−1
k=1 |ak+1 − ak| ≤ |a1| +M ,
and hence (an) is bounded. (ii) Let f(z) =∑∞
n=0 anzn, where the series converges absolutely
and uniformly for |z| ≤ 1, and M :=∫ 10 |f(e2πit)|2dt ≤ 1 by hypothesis. Moreover, |f(e2πit)|2 =
f(e2πit)f(e2πit) =∑∞
n=0 aname2πi(n−m)t, and
∫ 10 e
2πiktdt = 0 if k ∈ Z \ {0} and = 1 if k = 0.
Therefore 1 ≥M =∑∞
n=1 anan =∑∞
n=0 |an|2.]
4 Mobius maps
Definition: Let C∞ = C ∪ {∞} be the one-point compactification of C, and note that C∞ may be
identified with a sphere. If a, b, c, d ∈ C are with ad− bc = 0, then the map T : C∞ → C∞ given by
T (z) =az + b
cz + dis called a Mobius map. Here, 1/0 = ∞ and 1/∞ = 0 by convention; and moreover,
T (∞) =a+ b/∞c+ d/∞
= a/c when c = 0. If p ∈ C \ {0}, then paz + pb
pcz + pd=az + b
cz + d, and therefore we
may assume that ad − bc = 1 in the definition of a Mobius map. When this assumption holds, a b
c d
is called the matrix of the Mobius map. If T, S are Mobius maps, then T ◦ S is also a
Mobius map, and the matrix of T ◦ S is the product of the matrix of T with the product of the
matrix of S. Consequently, every Mobius map T : C∞ → C∞ is invertible, and the matrix of T−1
is the inverse of the matrix of T up to multiplication by a non-zero constant, i.e., if T (z) =az + b
cz + d,
then T−1(z) =dz − b
−cz + a.
Remark: The similarity of Mobius maps to linear maps can be explained as follows. Let C2∗ =
C2 \ {(0, 0)}. Define an equivalence relation ∼ on C2∗ by the condition that (z1, w1) ∼ (z2, w2) iff
z1w2 = z2w1. Let C2∗/ ∼ be the space of equivalence classes. If L : C2 → C2 is an invertible linear
17
map given by the matrix
a b
c d
, then L induces a map L : C2∗/ ∼→ C2
∗/ ∼ by the rule that
[(z, w)] 7→ [L((z, w))]. It may be noted that for any p ∈ C\{0}, both L and pL induce the same map
on C2∗/ ∼. Hence we may assume ad− bc = 1, as far as the induced map L on C2
∗/ ∼ is concerned.
Next observe that C2∗/ ∼ can be identified with C∞ via the correspondence [(z, 1)] ↔ z ∈ C and
[(1, 0)] ↔ ∞ ∈ C∞. Through this correspondence, L induces a map from C∞ to C∞, which is
precisely the Mobius map z 7→ az + b
cz + d.
[113] [Some properties of Mobius maps] (i) If T (z) =az + b
cz + dis a Mobius map on C∞, then T is
holomorphic on C \ {−d/c}, and T has at least one fixed point in C∞.
(ii) If T is a Mobius map with at least three fixed points in C∞, then T = I.
(iii) Any Mobius map can be written as a composition of the following types of Mobius maps:
translations (z 7→ z + b for some fixed b ∈ C), dilations (z 7→ az for some a ∈ C \ {0}), and the
inversion (z 7→ 1/z).
(iv) For any three distinct points z1, z2, z3 ∈ C∞, there is a unique Mobius map T such that
T (z1) = 0, T (z2) = 1, and T (z3) = ∞.
(v) Given two sets {z1, z2, z3} and {w1, w2, w3} of three distinct points each in C∞, there is a unique
Mobius map T with T (zj) = wj for j = 1, 2, 3.
Proof. (i) Since T is a rational function and {z ∈ C : cz + d = 0} = {−d/c}, we see T is
holomorphic on C \ {−d/c}. If T (∞) = a/c = ∞, then c = 0, and therefore the quadratic
equation cz2 + (d− a)z + b = 0, which is equivalent to “T (z) = z”, has a solution in C.
(ii) Let T (z) =az + b
cz + d. If T (∞) = a/c = ∞, then c = 0, and by hypothesis, there are at least three
solutions in C for T (z) = z, i.e., for cz2 + (d− a)z − b = 0, which is impossible since c = 0. So we
must have T (∞) = ∞ and c = 0. Then the equation (d− a)z − b = 0 has at least two solutions in
C by hypothesis, which implies d = a and b = 0. Thus c = 0 = b and a = d so that T = I.
(iii) Let T (z) =az + b
cz + d. If c = 0, then T (z) = (a/d)z + b/d. If c = 0, then T (z) =
c(az + b)
c(cz + d)=
a(cz + d) + bc− ad
c(cz + d)=a
c+
bc− ad
c2z + cd.
(iv) Uniqueness can be deduced using (ii) since if T, S are two Mobius maps with the required
property, then T ◦ S−1 has at least three fixed points. For the existence part, we give two proofs.
First proof : After applying a translation and then the inversion z 7→ 1/z if necessary, suppose
z1, z2, z3 ∈ C \ {0}. Let T (z) =(z3 − z2)(z − z1)
(z1 − z2)(z − z3). Check that T is indeed a Mobius map and
18
T (zj)’s are as required. Another Proof : Let T1(z) = z or T1(z) = 1/(z − z3) depending upon
whether z3 = ∞ or z3 = ∞. Then T1 is a Mobius map with T1(z3) = ∞. Since Mobius maps are
bijective, w1 := T1(z1) and w2 := T1(z2) are two distinct elements of C. Let T2(z) = w1+z(w2−w1).
Then T2 is a Mobius map with T2(0) = w1, T2(1) = w2 and T2(∞) = ∞. Take T = T−12 ◦ T1.
(v) Let T1, T2 be Mobius maps with (T1(z1), T1(z2), T1(z3)) = (0, 1,∞) = (T2(w1), T2(w2), T2(w3)).
Take T = T−12 ◦ T1. Uniqueness is argued as before using (ii).
Definition and notation: (i) By a circle in C∞ we mean either a circle in C or a set of the form
L ∪ {∞}, where L is a line in C. In particular, R ∪ {∞} is a circle in C∞. (ii) For z1, z2, z3 ∈ C,
let [z1, z2] denote the line segment with end points z1 and z2, let ∆(z1, z2, z3) denote the triangle
with vertices z1, z2, z3, and let ](z1, z2, z3) denote the angle at z2 determined by z1 and z3.
Exercise-33: (i) Let a, b ∈ C be distinct and C be a circle in C for which [a, b] is a diameter. Then
for c ∈ C \ {a, b}, we have that c ∈ C iff ](a, c, b) = π/2.
(ii) Let a, b ∈ C \ {0} be distinct. Then the triangles ∆(a, 0, b) and ∆(1/b, 0, 1/a) are similar.
[Hint : (i) This is a standard fact in elementary geometry. (ii) Writing a, b in polar coordinates, we
may see ](a, 0, b) = ](1/b, 0, 1/a). Moreover,|a||b|
=1/|b|1/|a|
.]
[114] If T is a Mobius map and C is a circle in C∞, then T (C) is also a circle in C∞.
Proof. By [113](iii), T is a composition of translations, dilations and the inversion map. It is easy
to see that translations and dilations map circles in C∞ to circles in C∞. Therefore we may assume
that T is the inversion map z 7→ 1/z.
Case-1 : C = L ∪ {∞}, where L is a line in C with 0 ∈ L. Then L = {ta : t ∈ R} for some
a ∈ C \ {0}. Since T (0) = ∞, and T (ta) = 1/(ta) =a
t|a|2, we see T (C) = {sa : s ∈ R} ∪ {∞}.
Case-2 : C = L ∪ {∞}, where L is a line in C with 0 /∈ L. Then there is a unique point a ∈ L
such that [0, a] ⊥ L. If b ∈ L \ {a}, then the triangles ∆(a, 0, b) and ∆(T (b), 0, T (a)) are similar by
Exercise-33(ii), and hence ](0, T (b), T (a)) = ](0, a, b) = π/2, which implies by Exercise-33(i) that
T (C) is the circle in C for which [0, T (a)] is a diameter.
Case-3 : C is a circle in C with 0 ∈ C. Since T = T−1, we may retrace the argument in case-2 to
see that T (C) = L ∪ {∞} for some line L in C for which 0 /∈ L.
Case-4 : C is a circle in C with 0 /∈ C. Choose a1, a2 ∈ C such that [a1, a2] is a diameter of
C and 0, a1, a2 are collinear. Then for any b ∈ C \ {a1, a2}, the triangle ∆(0, aj , b) is similar
to ∆(T (b), 0, T (aj)) for each j ∈ {1, 2} by Exercise-33(ii); and therefore, ](T (a1), T (b), T (a2)) =
19
](T (a1), T (b), 0) − ](T (a2), T (b), 0) = ](b, a1, 0) − ](b, a2, 0) = π − ](b, a1, a2)) − ](b, a2, a1) =
](a2, b, a1) = π/2. Hence by Exercise-33(i), T (C) is a circle in C for which [T (a1), T (a2)] is a
diameter (draw a picture to understand the argument clearly).
Exercise-34: [Convince yourself that the following is true, at least intuitively] If z1, z2, z3 ∈ C∞
are three distinct points, then there is a unique circle C in C∞ passing through z1, z2, z3. The
complement of C in C∞ consists of two connected open sets. We can call them the left side and
the right side of C if we fix an orientation [z1, z2, z3] for C (i.e., as we travel from z1 to z3 along
the arc in C containing z2). If T is any Mobius map, then the left side (respectively, the right side)
of C is mapped onto the left side (respectively, the right side) of T (C) in a bijective manner with
respect to the orientation [T (z1), T (z2), T (z3)] of the circle T (C). [In practice, we just evaluate T
at a suitable point in one side of C to deduce where this region will be mapped onto by T .]
Example: Suppose C1, C2 are two circles in C∞. Let z1, z2, z3 be three distinct points on C1, and
w1, w2, w3 be three distinct points of C2. By [113](v), there is a unique Mobius map T with T (zj) =
wj for j = 1, 2, 3. Then necessarily T (C1) = C2. By choosing zj ’s and wj ’s appropriately, we may
control which side of C1 is mapped to which side of C2 by T . A specific example is the following.
Let T (z) =(1 + i)(z + 1)
(−1 + i)(z − 1)=
−i(z + 1)
z − 1. Then T is a Mobius map with T (−1) = 0, T (−i) = 1
and T (1) = ∞. Hence T maps the unit circle ∂D onto the circle R ∪ {∞} in C∞. Since T (0) = i,
we deduce that T maps the open unit disc D onto the upper half-plane {w ∈ C : Im(w) > 0}.
Exercise-35: Let T = I be a Mobius map. Then,
(i) Fix(T ) := {z ∈ C∞ : T (z) = z} = {∞} iff T is a translation.
(ii) Fix(T ) = {0,∞} iff T is a dilation.
(iii) T (0) = ∞ and T (∞) = 0 iff T (z) = a/z for some a ∈ C \ {0}.
(iv) If S is a Mobius map, then Fix(S ◦ T ◦ S−1) = S(Fix(T )).
(v) |Fix(T )| = 1 iff T is conjugate to a translation.
(vi) |Fix(T )| = 2 iff T is conjugate to a dilation.
Exercise-36: Let S, T be Mobius maps different from aI, a ∈ C \ {0}. Then S ◦ T = T ◦ S iff
Fix(S) = Fix(T ). [Hint : First suppose Fix(T ) = {z1, z2}, where z1 = z2. If J is a Mobius map
with J(z1) = 0 and J(z2) = ∞, then replacing T with J ◦ T ◦ J−1 and S with J ◦ S ◦ J−1, we
may assume Fix(T ) = {0,∞}. Then Tz = az for some a ∈ C \ {0}. Now if Fix(S) = Fix(T ),
then S(z) = bz for some b ∈ C \ {0}, and hence S ◦ T = T ◦ S. Conversely, if S ◦ T = T ◦ S,
then S({0,∞}) = {0,∞} so that either S(z) = bz or S(z) = b/z; eliminate the second possibility
20
using S ◦ T = T ◦ S. Next suppose Fix(T ) is a singleton. After a conjugation, we may assume
Fix(T ) = {∞}, and then T (z) = z + b for some b ∈ C \ {0}. If Fix(S) = Fix(T ), then S is also
a translation and hence S ◦ T = T ◦ S. Conversely, if S ◦ T = T ◦ S, then S(∞) = ∞. Apply first
case to ensure that S has no other fixed points.]
Exercise-37: Let T (z) =az + b
cz + dbe a Mobius map. Then T (R ∪ {∞}) = R ∪ {∞} iff there is
p ∈ C \ {0} such that pa, pb, pc, pd ∈ R. [Hint : ⇒: First suppose T (∞) = ∞. Then c = 0 and
T (z) = (a/d)z + b/d. Since T (1), T (−1) ∈ R, we have a/d + b/d,−a/d + b/d ∈ R and therefore
a/d, b/d ∈ R. Take p = 1/d. If x = T (∞) ∈ R, then consider S(z) = 1/(T (z)− x), which fixes ∞,
and apply the first case.]
Exercise-38: Let U ⊂ C be a connected open set and f ∈ H(U). If f(U) ⊂ ∂D, then f is a constant.
[Hint : Each z ∈ U has a neighborhood B such that f(B) is a proper subset of ∂D. Since C is
second countable, we may write U =∪∞
n=1Bn, a countable union of open balls, such that f(Bn)
is a proper subset of ∂D for each n ∈ N. Choose Mobius maps Tn such that Tn(∂D) = R ∪ {∞}
and Tn(f(Bn)) ⊂ R. By Exercise-22, Tn ◦ f is a constant on Bn, and hence f is a constant on Bn
(since Tn is bijective). Thus f(U) =∪∞
n=1 f(Bn) must be a countable and connected subset of C,
and hence must be a singleton. Remark: The above reasoning depends on some topological facts.
After learning some more Complex Analysis, we can give an easier argument.]
Exercise-39: Let M be the collection of all Mobius maps. Then,
(i) (M, ◦) is a non-abelian group.
(ii) ϕ : SL(2,C) → M given by
a b
c d
7→ (z 7→ az + b
cz + d) is a surjective group homomorphism
with ker(ϕ) = {±I}, where SL(2,C) = {all 2× 2 complex matrices of determinant 1}.
(iii) Describe all abelian subgroups of M.
[Hint : (ii) From earlier theory, we know ϕ is a surjective group homomorphism. Consider T (z) =az + b
cz + d. If T (z) = z for every z ∈ C∞, then in particular 0, 1,∞ ∈ Fix(∞), and moreover ad− bc =
1. Deduce from this that b = c = 0, a = d and ad = 1; and conclude ker(ϕ) = {±I}. (iii) Let
Hw1,w2 = {T ∈ M : Fix(T ) = {w1, w2}} ∪ {I} for w1, w2 ∈ C∞ (where w1 = w2 is allowed). Then
H is an abelian subgroup of M; and moreover, any abelian subgroup H of M must be a subgroup
of some Hw1,w2 since Fix(S) = Fix(T ) for every S, T ∈ H \ {I} by Exercise-36. Also, Hw1,w2 is
conjugate to either H0,∞ = {all translations} ∼= (C,+) or H∞,∞ = {all dilations} ∼= (C \ {0}, ·).]
Exercise-40: Let N = {I} be a normal subgroup of M := {all Mobius maps}. Then,
(i) N must contain all translations and at least one dilation different from I.
21
(ii) z 7→ −1/z belongs to N .
(iii) We must have N = M.
[Hint : (i) Consider T ∈ N \ {I}. Then |Fix(T )| = 1 or = 2 by [113]. As N is normal, after a
conjugation we may assume that either Fix(T ) = {∞} or Fix(T ) = {0,∞}. In the first case, T
is a translation by Exercise-35, say T (z) = z + b. Conjugating with z 7→ az, we see z 7→ z + ab
belongs to N . Thus all translations belong to N . Conjugating z 7→ z + 1 with z 7→ 1/z, and then
composing with z 7→ z− 1, we see z 7→ −1/(z+1) belongs to N . This Mobius map has exactly two
fixed points, and hence is conjugate to a dilation by Exercise-35. Thus N contains a dilation. If
Fix(T ) = {0,∞}, then T is a dilation, say T (z) = az, a ∈ C\{0, 1}. Conjugating T with z 7→ z+1,
and then composing with T−1, we see the translation z 7→ z+(1− a)/a belongs to N , and then all
translations belong to N as above. (ii) z 7→ −1/(z+1) belongs to N and −1/z = −1/((z− 1)+1).
(iii) Consider T ∈ N \ {I}, T (z) = az, a = 0, 1, and consider b ∈ C \ {0}. Composing T with
z 7→ z + b, conjugating with z 7→ 1/z and then composing with z 7→ z − 1/b, we see z 7→ 1
b2z/a+ bbelongs to N . Composing this with z 7→ −1/z and z 7→ z + b, we get z 7→ −b2z/a belongs to
N . Thus all dilations belong to N . The inversion z 7→ 1/z has exactly two fixed points, hence is
conjugate to a dilation by Exercise-35, and therefore belongs to N . Finally apply [113](iii).]
Definition: Two open sets U, V ⊂ C are said to be holomorphically equivalent if there is a bijection
f : U → V such that both f and f−1 are holomorphic; and in this case, f is called a biholomorphism
from U to V . Note that being holomorphically equivalent is an equivalence relation on the collection
of all open subsets of C.
[115] Any two of the following open sets are holomorphically equivalent:
U1 = D (open unit ball),
U2 = {z ∈ C : Im(z) > 0} (upper half-plane),
U3 = {z ∈ C : Re(z) > 0} (right half-plane),
U4 = {z ∈ C : Im(z − a
b) > 0}, where a ∈ C and b ∈ C \ {0} (an open half-plane),
U5 = {z ∈ C : 0 < Im(z) < π} (a particular horizontal strip),
U6 = {z ∈ C : a < Im(z) < b}, where a < b are real (a horizontal strip),
U7 = {z ∈ C : a < Re(z) < b}, where a < b are real (a vertical strip),
U8 = C \ [0,∞),
U9 = C \ (−∞, 0].
Proof. Verify that each Ti,j below is a biholomorphism from Ui to Uj :
22
T1,2(z) =(1 + i)(z + 1)
(−1 + i)(z − 1)=
−i(z + 1)
z − 1, and
T2,1(z) = T−11,2 (z) =
dz − b
−cz + a=
−z + i
−z − i=z − i
z + i;
T2,3(z) = −iz and T3,2(z) = iz;
T2,4(z) = bz + a and T4,2(z) = (z − a)/b;
T2,5(z) = log |z|+ i arg(z) and T5,2(z) = ez;
T5,6(z) =(b− a)z
π+ ia and T6,5(z) =
π(z − ia)
b− a;
T6,7(z) = −iz and T7,6(z) = iz;
T2,8(z) = z2 and T8,2(z) = e(1/2)(log |z|+i arg(z));
T8,9(z) = −z and T9,8(z) = −z.
Remark: After learning more theory, we can show that no two of the following are holomorphically
equivalent: D, D \ {0}, {z ∈ C : 1 < |z| < 2}, C \ {0}, and C (keep this exercise in mind to be
solved later). However, we have the following:
Exercise-41: There is a surjective holomorphic function from D to each of the following: D \ {0},
{z ∈ C : 1 < |z| < 2}, C \ {0}, and C.
[Hint : D ∼= {z ∈ C : Re(z) < 0} ez−→ D \ {0}; D ∼= {z ∈ C : Im(z) > −1} z2−→ C ez−→ C \ {0};
and D ∼= {z ∈ C : 0 < Re(z) < log 2} ez−→ {z ∈ C : 1 < |z| < 2}.]
5 Integration of a continuous function along a path
Here, as a preparation for Cauchy’s integral theory, we will discuss the integration of complex-valued
continuous functions along paths. The general theory deals with the integration along ‘rectifiable
paths’. For simplicity, we will consider only piecewise smooth paths (defined below).
Definition: Let U ⊂ C be an open set. By a path (or curve) in U , we mean a continuous map
α : [a, b] → U , where a < b are reals. The image α([a, b]) of the path will be denoted as [α], which
is a compact subset of U . A path α : [a, b] → U is said to be (i) closed if α(a) = α(b), (ii) smooth
if α is continuously differentiable (i.e., if α′ is continuous), and (iii) piecewise smooth if there is a
partition a = a0 < a1 < · · · < ak = b of the domain [a, b] of α such that αj := α|[aj−1,aj ] is smooth
for 1 ≤ j ≤ k; and in this case we will use the formal expression α = α1 + · · ·+ αk.
Example: (i) For any k ∈ Z \ {0}, α : [0, 1] → C defined as α(t) = e2πikt is a smooth path and
its image [α] is the unit circle. (ii) A piecewise smooth path parametrizing the boundary of the
23
unit square in C in the anticlockwise direction is α : [0, 4] → C defined as α(t) = t for 0 ≤ t ≤ 1,
α(t) = 1 + i(t− 1) for 1 ≤ t ≤ 2, α(t) = (3− t) + i for 2 ≤ t ≤ 3, and α(t) = i(4− t) for 3 ≤ t ≤ 4.
Definition: (i) If g : [a, b] → C is continuous, we define∫ ba g(t)dt =
∫ ba Re(g(t))dt+ i
∫ ba Im(g(t))dt.
(ii) Let U ⊂ C be open and f : U → C be continuous. If α : [a, b] → U is a smooth path in U ,
then we define the integral of f along α (or over α) as∫α f =
∫α f(z)dz :=
∫ ba f(α(t))α
′(t)dt. This
definition is intuitively justified because if z = α(t), then dz = α′(t)dt. Moreover, the function
t 7→ f(α(t))α′(t) is continuous (and hence integrable as specified by (i)) because α is smooth. If
α is a piecewise smooth path in U , say α =∑k
j=1 αj , where each αj is smooth, then we define∫α f =
∑kj=1
∫αjf . (iii) If U ⊂ C is open, f : U → C is continuous, and α : [a, b] → U is a smooth
path, then∫α |f ||dz| :=
∫ ba |f(α(t))α′(t)|dt. Clearly, this definition implies |
∫α f | ≤
∫α |f ||dz|. If
f ≡ 1, then∫α |f ||dz| =
∫ ba |α′(t)|dt, which is by definition the length of α.
Examples: (i) Let U ⊂ C be open, f : U → C be continuous, and α : [a, b] → U be a smooth path.
If α is constant, then α′ ≡ 0; so∫α f = 0. If f is constant, say f ≡ c, then
∫α f = c(α(b) − α(a)).
(ii) Let α : [0, 1] → C \ {0} be α(t) = e2πit. Then∫α z
kdz = 2πi∫ 10 e
2πi(k+1)tdt = 0 ∀ k ∈ Z \ {−1}.
On the other hand,∫α
dz
z= 2πi
∫ 10 1 dt = 2πi. Similarly, if β : [0, 2π] → C \ {0} is β(t) = eit, then∫
β zkdz = 0 ∀ k ∈ Z \ {−1} and
∫β
dz
z= 2πi. More generally, if k,m ∈ Z, and α : [0, 1] → C \ {0}
is α(t) = e2πimt, then∫α z
kdz = 2πi if k = −1 and m = 0, and∫α z
kdz = 0 otherwise.
Remark: From the examples above, we observe that: (i) the integral of a holomorphic function
over a closed path need not be zero, even though a closed path has the same end points as that
of a constant path, and (ii) the integral of a holomorphic function over two paths with the same
image need not be equal.
Definition and remark: (i) Two smooth paths α : [a, b] → C and β : [c, d] → C are said to be
equivalent if there is an increasing bijection ϕ : [a, b] → [c, d] such that ϕ′ is continuous and α = β◦ϕ.
Note that if α, β are equivalent as defined above, then their images [α] and [β] are the same, and
moreover for any complex-valued f continuous in a neighborhood of [α] = [β], we have that∫α f =∫
β f because∫α f =
∫ ba f(α(t))α
′(t)dt =∫ ba f(β(ϕ(t)))β
′(ϕ(t))ϕ′(t)dt =∫ dc f(β(s))β
′(s)ds =∫β f .
(ii) The natural parametrization of the circle |z− a| = r is given by α : [0, 1] → C, α(t) = a+ re2πit
(or by β : [0, 2π] → C, β(t) = a + reit, which is equivalent to α). When we write∫|z−a|=r f , it
is understood that the integral is to be considered with respect to the natural parametrization of
the circle. Similarly, if w1, w2 ∈ C, then∫[w1,w2]
f is to be considered with respect to the natural
parametrization of the line segment, i.e., w.r.to α : [0, 1] → C given by α(t) = w1 + t(w2 − w1).
24
However, at this stage, it is meaningless to write∫ w2
w1f since there are many paths from w1 to w2
and the integrals over these paths need not be the same.
(iii) If α : [a, b] → C is a path, we denote by −α the path going in the ‘opposite direction’, i.e.,
−α : [a, b] → C is defined −α(a+ t(b−a)) = α(b+ t(a− b)) for t ∈ [0, 1]. If α is a piecewise smooth
path in an open set U ⊂ C, and f : U → C is continuous, verify that∫−α f = −
∫α f .
(iv) If α : [a, b] → C and β : [b, c] → C are paths with α(b) = β(b), then we define the path
α + β : [a, c] → C as (α + β)(t) = α(t) for a ≤ t ≤ b, and (α + β)(t) = β(t) for b ≤ t ≤ c. If α, β
are piecewise smooth paths in an open set U ⊂ C such that α + β is defined, and if f : U → C is
continuous, then clearly∫α+β f =
∫α f +
∫β f .
[116] Let a ∈ C, r > 0. Then∫|z−a|=r(z−a)
kdz = 0 for every k ∈ Z\{−1}, and∫|z−a|=r
dz
z − a= 2πi.
Proof. We have already noted this for the unit circle. Argue similarly.
Definition: Let U ⊂ C be open, and f : U → C be continuous. A function g ∈ H(U) is said to
be a primitive of f if g′ = f . The main advantage of having a primitive is the path independence
property stated in Exercise-42(ii) below.
Exercise-42: Let U ⊂ C be a connected open set, and f : U → C be continuous.
(i) If g1, g2 ∈ H(U) are primitives of f , then (g1 − g2)′ ≡ 0 and hence g1 − g2 is a constant.
(ii) [Path independence] Suppose f has a primitive g ∈ H(U) and let z1, z2 ∈ U . Then for any
piecewise smooth path α in U from z1 to z2, we have that∫α f = g(z2)− g(z1), and thus this value
is independent of the particular path α. This value is also independent of the particular choice of
a primitive g of f because of (i).
(iii) Let α be a piecewise smooth path in U . Suppose fn : U → C are continuous functions
converging to f uniformly on the image [α] of α. Then limn→∞∫α fn =
∫α f .
[Hint : (ii) Assume α : [a, b] → U is smooth. Then∫α f =
∫ ba f(α(t))α
′(t)dt =∫ ba g
′(α(t))α′(t)dt =∫ ba (g ◦α)
′(t)dt = g(α(b))− g(α(a)) = g(z2)− g(z1). (iii) We may assume α is not a constant so that
its length l(α) > 0. Given ε > 0, choose n0 ∈ N such that |f(z)− fn(z)| < ε/l(α) for every n ≥ n0
and z ∈ [α]. Then∫α |f − fn| ≤
∫ ba
ε|α′(t)|l(α)
dt = ε for every n ≥ n0.]
Definition: Let U ⊂ C be open and f : U → C be continuous. We say f is integrable if for
any two points z1, z2 ∈ U and any two piecewise smooth paths α, β in U from z1 to z2, we have
that∫α f =
∫β f (so that
∫ z2z1f(z)dz has a well-defined meaning). Note that f is integrable iff the
restriction of f to each connected component of U is integrable (since a path in U should be entirely
25
contained in a connected component of U).
Exercise-43: [Topological fact] If U ⊂ C is a nonempty connected open set, then U is polygonally
connected, i.e., for any two w, z ∈ U , there is a polygonal path in U from w to z. Here, a polygonal
path means a path whose image consists of finitely many line segments. [Hint : Fix w ∈ U , and let
A be the collection of all z ∈ U such that there is a polygonal path in U from w to z. Show that
A is both closed and open in U .]
[117] [Integrability criterion] Let U ⊂ C be open and f : U → C be continuous. Then the following
are equivalent: (i) f has a primitive g ∈ H(U).
(ii) f is integrable.
(iii)∫α f = 0 for every piecewise smooth closed path α in U .
Proof. (i) ⇒ (ii): We may suppose U is connected. For any two points z1, z2 ∈ U and any two
piecewise smooth paths α, β in U from z1 to z2, we have that∫α f = g(z2) − g(z1) =
∫β f by
Exercise-42(ii).
(ii) ⇒ (iii): Let α : [a, b] → U be a piecewise smooth closed path and β : [a, b] → U be the constant
path β ≡ α(a) = α(b). The integrability of f implies that∫α f =
∫β f = 0.
(iii) ⇒ (i): We may assume U is connected. Fix c ∈ U , and define g : U → C as follows: if p ∈ U ,
then g(p) :=∫α f , where α is any piecewise smooth path in U from c to p. Note that the existence
of such a path α is guaranteed by Exercise-43; and the value∫α f is independent of the particular
choice of a path because if β is another such path, then α− β is a piecewise smooth closed path in
U and hence 0 =∫α−β f =
∫α f −
∫β f by (iii). We claim that g is a primitive of f in U .
Consider p ∈ U and let ε > 0. Since U is open and f is continuous, we may choose δ > 0 such
that B(p, δ) ⊂ U and such that |f(p)− f(q)| < ε for every q ∈ B(p, δ). If h ∈ C is with 0 < |h| < δ,
let α, β be piecewise smooth paths in U from c to p and from c to p+h respectively. Then g(p) =∫α f
and g(p+h) =∫β f . The integral of f over the piecewise smooth closed path α+[p, p+h]−β is zero
by (iii), and hence g(p + h) − g(p) =∫β−α f =
∫[p,p+h] f(z)dz. Moreover, hf(p) =
∫[p,p+h] f(p)dz.
Therefore, for 0 < |h| < δ, we have that |g(p+ h)− g(p)− hf(p)| ≤∫[p,p+h] |f(z)− f(p)||dz| < ε|h|.
This shows that g is holomorphic at p with g′(p) = f(p).
Remark: In Real Analysis, every continuous function f : [a, b] → R is Riemann integrable, and has
a ‘primitive’; indeed, g : [a, b] → R defined as g(x) =∫ xa f(t)dt satisfies g
′ = f . Contrast this with
the following example in Complex Analysis. Let U = {z ∈ C : 1/2 < |z| < 2} and f : U → C be
26
f(z) = 1/z. Then, U is a bounded open set and f ∈ H(U) is bounded, but we note by [117] that
f is not integrable and f does not have a primitive in U because∫|z|=1 f = 2πi = 0.
Definition: Let U ⊂ C be open and f : U → C be continuous. We say f is locally integrable in U if
for each z ∈ U , there is δ > 0 such that B(z, δ) ⊂ U and the restriction of f to B(z, δ) is integrable.
Exercise-44: The holomorphic function f(z) = 1/z is locally integrable in any open subset U of
C \ {0}. Hence local integrability does not imply (global) integrability. [Hint : Given z ∈ U , we can
find δ > 0 with B(z, δ) ⊂ U ⊂ C \ {0}. There is a branch of logarithm g : B(z, δ) → C by [108](ii).
Since g′(z) = 1/z, g is a primitive of f in B(z, δ). Now apply [117].]
Remark: Let U ⊂ C be open and f : U → C be continuous. With more theory, it can be shown
that (i) f is locally integrable ⇔ f ∈ H(U), and (ii) if U is connected and has no ‘holes’, then f is
locally integrable ⇔ f is integrable.
Definition: An open set U ⊂ C is called a star region if there is c ∈ U such that [c, z] ⊂ U for every
z ∈ U (where [c, z] is the line segment joining c and z). Here c is called a center of U .
Example: Every convex open subset U of C is a star region, and in this case, every point of U is a
center of U . The sets C \ [0,∞) and C \ (−∞, 0] are not convex, but they are star regions: every
c ∈ (−∞, 0) is a center of C \ [0,∞), and every c ∈ (0,∞) is a center of C \ (−∞, 0]. Another
example of a star regions which is non-convex is {x+ iy : |x| < 1 and |xy| < 1}. On the other hand,
the following connected open sets are not star regions: D \ {0}, C \ {0}, {z ∈ C : 1 < |z| < 2}, and
{z ∈ C : Im(z) > 0 and |z| > 1}.
Remark: Being a star region is a geometric property. In general this property is not preserved
under topological or holomorphic equivalence. Let δ > 0 be very small, U be the open rectangle
with vertices 0, δ, δ+ iπ, iπ, and V be the image of U under ez. Then it can be verified that the star
region U is holomorphically equivalent to V via the exponential map, but V is not a star region
since V = {z ∈ C : 1 < |z| < eδ and Im(z) > 0} is the upper-half of a thin open annulus.
[118] Let U ⊂ C be a star region and f : U → C be continuous. If∫∂∆ f = 0 for every solid triangle
∆ ⊂ U , then f has a primitive g ∈ H(U) and consequently f is integrable (here the integral∫∂∆ f
is taken with respect to the natural parametrization of ∂∆ in the anticlockwise direction).
Proof. Let c ∈ U be a center of U . Define g : U → C as g(p) =∫[c,p] f(z)dz. Consider p ∈ U and
ε > 0. Choose δ > 0 such that B(p, δ) ⊂ U and such that |f(p) − f(q)| < ε for every q ∈ B(p, δ).
For h ∈ C with 0 < |h| < δ, note that ∆ := [c, p, p + h] ⊂ U and hence∫∂∆ f = 0 by hypothesis,
27
which implies that g(p+ h)− g(p) =∫[p,p+h] f(z)dz; moreover, hf(p) =
∫[p,p+h] f(p)dz. Therefore,
for h ∈ C with 0 < |h| < δ, we see that |g(p+ h)− g(p)− hf(p)| ≤∫[p,p+h] |f(z)− f(p)||dz| < ε|h|,
and this shows g′(p) = f(p). Thus g is a primitive of f , and then f is integrable by [117].
6 Cauchy’s integral formula and power series representation
The relevance of a star region in the context of holomorphic functions can be first observed by
noting that the hypothesis of [118] is satisfied by holomorphic functions:
[119] Let U ⊂ C be open and f ∈ H(U). Then,
(i) [Goursat’s integral lemma]∫∂∆ f = 0 for every solid triangle ∆ contained in U .
(ii) [Cauchy’s integral theorem for star regions] If U is a star region, then f has a primitive in U ,
f is integrable, and∫α f = 0 for every piecewise smooth closed path α in U .
Proof. Since (ii) is a direct consequence of part (i), [118] and [117], it suffices to prove (i). Let
l(∂∆) denote the length of the boundary of a solid triangle ∆. Note that if ∆′ is one of the four
solid triangles obtained by joining the midpoints of ∂∆, then l(∂∆′) = l(∂∆)/2. Below, all integrals
over triangles are considered with respect to the anticlockwise orientation.
Consider a solid triangle ∆ ⊂ U and ε > 0. We will show (and it is enough to show) that
|∫∂∆ f | ≤ ε. Divide ∆ into four solid triangles ∆1,∆2,∆3,∆4 of equal size by joining the midpoints
of the sides of ∂∆. Observe that∑4
j=1
∫∂∆j
f =∫∂∆ f since the integrals along common edges will
get cancelled due to opposite orientations. Hence one of the ∆j ’s, call it ∆(1), satisfies |
∫∂∆(1) f | ≥
|∫∂∆ f |/4. Also, l(∂∆
(1)) = l(∂∆)/2, as noted above.
At the n the step, divide ∆(n) into four solid triangles ∆(n)1 ,∆
(n)2 ,∆
(n)3 ,∆
(n)4 of equal size by
joining the midpoints of the sides of ∂∆(n). Since∑4
j=1
∫∂∆
(n)j
f =∫∂∆(n) f , there exists ∆(n+1) ∈
{∆(n)1 ,∆
(n)2 ,∆
(n)3 ,∆
(n)4 } with |
∫∂∆(n+1) f | ≥ |
∫∂∆(n) f |/4 = |
∫∂∆ f |/4
n+1. Moreover, l(∂∆(n+1)) =
l(∂∆(n))/2 = l(∂∆)/2n+1.
Since (∆(n)) is a decreasing sequence of nonempty compact sets with diam(∆(n)) → 0, there is
p ∈ U such that∩∞
n=1∆(n) = {p}. Since f is holomorphic, there is an open ball B ⊂ U centered at
p such that |f(z)− f(p)− (z − p)f ′(p)| < ε|z − p|l(∂∆)2
for every z ∈ B.
Choose n large enough with ∆(n) ⊂ B. Note that z 7→ f(p) − (z − p)f ′(p) is a poly-
nomial and hence has a primitive. Consequently,∫∂∆(n)(f(p) − (z − p)f ′(p))dz = 0 by [117],
28
and therefore |∫∂∆(n) f(z)dz| ≤
∫∂∆(n) |f(z) − f(p) − (z − p)f ′(p)||dz| ≤
∫∂∆(n)
ε|z − p|l(∂∆)2
|dz| ≤∫∂∆(n)
εl(∂∆(n))
l(∂∆)2|dz| ≤ εl(∂∆(n))2
l(∂∆)2= ε/4n. Hence |
∫∂∆ f | ≤ 4n|
∫∂∆(n) f | ≤ ε.
Remark: [119](ii) implies that if f is holomorphic in a star region U , then ‘∫ z2z1f(z)dz’ has a
well-defined meaning (independent of the path) for any two z1, z2 ∈ U .
Exercise-45: Let f : D → C be continuous and suppose f is holomorphic on D. Then∫|z|=1 f = 0.
[Hint : Define fn : B(0, n+1n ) → C as fn(z) = f( nz
n+1). Then fn is holomorphic on B(0, n+1n ) and
hence∫|z|=1 fn = 0 by [119](ii). The uniform continuity of f on the compact set ∂D implies that
(fn) → f uniformly on ∂D. Therefore,∫|z|=1 f = limn→∞
∫|z|=1 fn = 0 by Exercise-42(iii).]
[120] (i) Let a ∈ C and r > 0. Then∫|z−a|=r
dz
z − b= 2πi for every b ∈ B(a, r).
(ii) Let V ⊂ C be an open set such that ∂V is a polygon. Then∫∂V
dz
z − b= 2πi for every b ∈ V .
Proof. (i) Let f(z) = 1/(z − b), which is holomorphic on C \ {b}. Choose ε > 0 small enough with
B(b, ε) ⊂ B(a, r). See the figure with piecewise smooth closed paths β1 and β2. Since β1 is contained
in the star region C\{b+t : t ≤ 0} where f is holomorphic,∫β1f = 0 by [119](ii). Similarly,
∫β2f = 0
by [119](ii) since β2 is contained in the star region C \ {b+ t : t ≥ 0} where f is holomorphic. Thus
0 =∫β1f +
∫β2f =
∫|z−a|=r
dz
z − b−∫|z−b|=ε
dz
z − b. Also we know
∫|z−b|=ε
dz
z − b= 2πi by [116].
(ii) Similar to the proof of (i).
[121] [An improvement of Cauchy’s integral theorem for star regions] Let U ⊂ C be a star region,
p ∈ U be a center of U , let f : U → C be continuous, and suppose f is holomorphic on U \ {p}.
Then f has a primitive in U , f is integrable in U , and∫α f = 0 for every piecewise smooth closed
path α in U .
Proof. By [118], it suffices to show that∫∂∆ f = 0 for an arbitrary solid triangle ∆ ⊂ U .
29
Case-1 : p /∈ ∆. Then∫∂∆ f = 0 by Goursat’s integral lemma applied to U \ {p}.
Case-2 : p is a vertex of ∆. Let q1, q2 be the other two vertices of ∆. Choose points p1 ∈ (p, q1)
and p2 ∈ (p, q2) very near to p. Now,∫∂∆ f =
∫[p,p1,p2]
f +∫[p1,q1,p2]
f +∫[p2,q1,q2]
f , where the last
two integrals are zero by case-1. Moreover, |∫[p,p1,p2]
f | can be made arbitrarily small by taking p1
and p2 sufficiently close to p. Hence∫∂∆ f = 0.
Case-3 : p is an interior point of an edge of ∆. Divide ∆ into two solid triangles ∆1,∆2 by
joining p with the opposite vertex of ∆. Then∫∂∆j
f = 0 for j = 1, 2 by case-2, and hence∫∂∆ f =
∫∂∆1
f +∫∂∆2
f = 0.
Case-4 : p ∈ int(∆). Divide ∆ into three solid triangles ∆1,∆2,∆3 by joining p with the vertices
of ∆. Then∫∂∆j
f = 0 for j = 1, 2, 3 by case-2, and hence∫∂∆ f =
∑3j=1
∫∂∆j
f = 0.
Now we are ready to prove one of the most important results about holomorphic functions.
The result says in particular that a holomorphic function f is completely determined in an open
ball B(a, r) by the values of f on the boundary circle |z − a| = r provided f is holomorphic in a
neighborhood the closed ball B(a, r)
[122] [Cauchy’s integral formula for discs] Let U ⊂ C be open, f : U → C be holomorphic and
suppose B(a, r) ⊂ U . Then f(b) =1
2πi
∫|z−a|=r
f(z)
z − bdz for every b ∈ B(a, r).
Proof. Let s > r be with B(a, s) ⊂ U (such a choice is possible since the compact set B(a, r) is
contained in U). Fix b ∈ B(a, r) and define g : B(a, s) → C as g(b) = f ′(b) and g(z) =f(z)− f(b)
z − bfor z = b. Then g is continuous on B(a, s) and holomorphic on B(a, s) \ {b}. By [121] and [120](i),
we get 0 =∫|z−a|=r g(z)dz =
∫|z−a|=r
f(z)
z − bdz −
∫|z−a|=r
f(b)
z − bdz =
∫|z−a|=r
f(z)
z − bdz − 2πif(b).
[123] [Power series representation] Let U ⊂ C be open, f : U → C be holomorphic and let
B(a,R) ⊂ U . Then there is a complex power series∑∞
n=0 an(z − a)n with radius of convergence
≥ R such that f(z) =∑∞
n=0 an(z − a)n for every z ∈ B(a,R). The coefficients an are given by
an =f (n)(a)
n!=
1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz for any r ∈ (0, R).
Proof. Let 0 < r < R. Then B(a, r) ⊂ B(a,R) ⊂ U . Let b ∈ B(a, r). By [122], we have
f(b) =1
2πi
∫|z−a|=r
f(z)
(z − b)dz. Since
1
z − b=
1
(z − a)− (b− a)=
1
z − a
∑∞n=0
(b− a
z − a
)n
, we have
that∫|z−a|=r
f(z)
(z − b)dz =
∫|z−a|=r
∑∞n=0
(f(z)(b− a)n
(z − a)n+1
)dz. Here, we would like to interchange the
integration and summation. Define g, gn : ∂B(a, r) → C as g(z) =f(z)
z − b=∑∞
n=0
f(z)(b− a)n
(z − a)n+1and
30
gn(z) =∑n
k=0
f(z)(b− a)k
(z − a)k+1, which are continuous functions. Let M = sup{|f(z)| : |z − a| = r}.
For |z − a| = r, we see that |g(z) − gn(z)| ≤∑∞
k=n+1
|f(z)||b− a|k
|z − a|k+1≤∑∞
k=n+1
M |b− a|k
rk+1→ 0
independent of z as k → ∞. That is, (gn) → g uniformly on ∂B(a, r). Hence by Exercise-42(iii),∫|z−a|=r
f(z)
(z − b)dz =
∫|z−a|=r g = limn→∞
∫|z−a|=r gn = limn→∞
∫|z−a|=r
∑nk=0
f(z)(b− a)k
(z − a)k+1dz =∑∞
n=0
(∫|z−a|=r
f(z)
(z − a)n+1dz
)(b − a)n. So, f(b) =
1
2πi
∫|z−a|=r
f(z)
(z − b)dz =
∑∞n=0 an(b − a)n,
where an =1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz. Note that an is independent of the point b ∈ B(a, r). Thus
f(z) =∑∞
n=0 an(z − a)n for every z ∈ B(a, r). Hence the radius of convergence of the power series∑∞n=0 an(z − a)n is ≥ r. Now from the theory of power series, it follows that f is infinitely often
(complex) differentiable in B(a, r) and an =f (n)(a)
n!. Thus an’s are independent of the particular
choice of r ∈ (0, R) also. Since r ∈ (0, R) was arbitrary, the proof is complete.
[124] [Corollary] Let U ⊂ C be open and f ∈ H(U). Then,
(i) f is infinitely often complex differentiable in U .
(ii) If B(a, r) ⊂ U , then f (n)(a) =n!
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz for every n ∈ N.
(iii) [Cauchy’s estimate] If B(a,R) ⊂ U and M := sup{|f(z)| : |z − a| < R}, then |f (n)(a)| ≤ n!M
Rn
for every n ∈ N.
(iv) If B(a, r) ⊂ U , then f (n)(b) =n!
2πi
∫|z−a|=r
f(z)
(z − b)n+1dz for every n ∈ N and b ∈ B(a, r).
(v) [Cauchy’s estimate generalized] If B(a,R) ⊂ U and M := sup{|f(z)| : |z − a| < R}, then
|f (n)(b)| ≤ n!MR
(R− |b− a|)n+1for every n ∈ N and b ∈ B(a,R).
Proof. (i) This is clear from [123].
(ii) By [123], f(z) =∑∞
n=0 an(z − a)n in B(a, r) and an =f (n)(a)
n!=
1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz.
(iii) Let r ∈ (0, R). By (ii), |f (n)(a)| ≤ n!
2π
∫|z−a|=r
M
rn+1|dz| = n!M
2πrn+1× 2πr =
n!M
rn. Now let
r → R.
(iv) Fix b ∈ B(a, r). Let ε > 0 be small enough with B(b, ε) ⊂ B(a, r). We have that f (n)(b) =n!
2πi
∫|z−b|=ε
f(z)
(z − b)n+1dz by (ii). Let g : U \ {b} → C be g(z) =
f(z)
(z − b)n+1, which is holomorphic.
Consider piecewise smooth paths β1, β2 in U \ {b} as in the proof of [120](i), and establish the
following:∫βjg = 0 for j = 1, 2 and
∫β1g +
∫β2g =
∫|z−a|=r g −
∫|z−b|=ε g.
(v) Fix b ∈ B(a,R). Let r ∈ (0, R) be such that b ∈ B(a, r). By (iii), we get |f (n)(b)| ≤n!
2π
∫|z−a|=r
M
(r − |b− a|)n+1|dz| = n!M
2π(r − |b− a|)n+1×2πr =
n!Mr
(r − |b− a|)n+1. Now let r → R.
31
Exercise-46: Let f(z) =∑∞
n=0 an(z − a)n be holomorphic on B(a,R).
(i) If M = sup{|f(z)| : |z − a| < R} <∞, then |an| ≤M/Rn for every n ≥ 0.
(ii) If M ′ = sup{|f ′(z)| : |z − a| < R} <∞, then n|an| ≤M/Rn−1 for every n ∈ N.
[Hint : (i) For r ∈ (0, R), we get using [123] that |an| ≤1
2π
∫|z−a|=r
M
rn+1|dz| = M
rn. Now let r → R.
(ii) Since f ′(z) =∑∞
n=1 nan(z − a)n−1, we may apply (i).]
Exercise-47: Evaluate the following: (i)∫|z|=1
cos z
zdz, (ii)
∫|z|=1
sin z
z4dz, (iii)
∫|z|=1
sin z
(z − 3)5dz,
(iv)∫|z|=2
1
z2 + 1dz, (v)
∫|z|=2
9z + 1
(z − 1)2(z + 1)dz.
[Hint : (i) Let f(z) = cos z. Then∫|z|=1
cos z
zdz = 2πif(0) = 2πi. (ii) Let f(z) = sin z. Then∫
|z|=1
sin z
z4dz = 2πif (3)(0)/3! = −πi/3. (iii) f(z) :=
sin z
(z − 3)5is holomorphic in the star region
{z ∈ C : Re(z) < 2}. Hence∫|z|=1 f = 0. (iv)
1
z2 + 1=
i/2
z + i− i/2
z − i. Hence
∫|z|=2
1
z2 + 1dz =∫
|z|=2
i/2
z + idz −
∫|z|=2
i/2
z − idz = (i/2)2πi − (i/2)2πi = 0. (v)
9z + 1
(z − 1)2(z + 1)=
2z + 3
(z − 1)2− 2
z + 1.
Let f(z) = 2z + 3 and g ≡ 2. Then∫|z|=2
9z + 1
(z − 1)2(z + 1)dz =
2πi
1!f ′(1)− 2πig(−1) = 0.]
Exercise-48: (i) If b ∈ D is such that∫|z|=1
ez
(z − b)4dz =
2πi
3, determine b.
(ii) If c ∈ C is such that∫|z|=1
c
z3 − 4z2 + 4zdz = 2πi, determine c.
[Hint : (i) Let f(z) = ez. By hypothesis,2πi
3=
2πif (3)(b)
3!, and hence 2 = eb, which im-
plies b = log 2. (ii) Let f(z) = c/(z − 2)2, which is holomorphic in B(0, 2). Since 2πi =∫|z|=1
c
z3 − 4z2 + 4zdz =
∫|z|=1
f(z)
zdz = 2πif(0) = 2πic/4, we get c = 4.]
Exercise-49: (i) If f(z) =∑∞
n=0 anzn is holomorphic in a neighborhood of 0, then limn→∞
ann!
= 0.
(ii) There do not exist any R > 0 and f ∈ H(B(0, R)) with the property that |f (n)(0)| ≥ (n!)2 for
infinitely many n ∈ N.
(iii) Let k ∈ N and f : C \ {0} → C be f(z) =sin z
z2k. Then f does not have a primitive in C \ {0}.
[Hint : (i) Otherwise, there is δ > 0 such that lim supn→∞|an|n!
≥ δ > 0. Then lim sup |an|1/n ≥
lim δ1/n(n!)1/n = ∞ (since lim δ1/n = 1 and lim(n!)1/n = ∞), implying that the radius of con-
vergence of f is zero, a contradiction. (ii) This follows from (i) since an = f (n)(0)/n!. A slightly
different argument is as follows. Suppose f is holomorphic in a neighborhood B(0, R), and let
M = sup{|f(z)| : |z| ≤ R} < ∞. By Cauchy’s estimate, |f (n)| ≤ n!M
Rnfor every n ∈ N, and hence
|f (n)| ≥ (n!)2 for at most finitely many n ∈ N. (iii) If f has a primitive in C \ {0}, then we must
have∫|z|=1 f = 0 by [117]. But
∫|z|=1
sin z
z2kdz =
±2πi cos 0
(2k − 1)!= 0.]
Exercise-50: Let α1, α2 be smooth paths in C with α1(t1) = α2(t2) =: w, and α′1(t1) = 0, α′
2(t2) = 0.
32
Then the angle between α1 and α2 at w is defined as arg(α′1(t1))− arg(α′
2(t2)). If U ⊂ C is open,
f ∈ H(U) and w ∈ U is with f ′(w) = 0, then f preserves angles at w in the following sense: if
α1, α2 are smooth paths in U such that the angle between α1 and α2 at w is defined, then this angle
is same as the angle between the smooth paths f ◦ α1 and f ◦ α2 at f(w). Here f ◦ αj is smooth
since f is infinitely often differentiable. [Hint : (f ◦ αj)′(tj) = f ′(αj(tj))α
′j(tj) = f ′(w)α′
j(tj) and
hence arg((f ◦ αj)′(tj)) = arg(f ′(w)) + arg(α′
j(tj)).]
Exercise-51: [Prove this using power series and not using [119](ii)] Let U = D or U = C. Then,
(i) If f ∈ H(U), then f has a primitive g ∈ H(U). Consequently, f is integrable and∫α f = 0 for
every piecewise smooth closed path α in C.
(ii) If f, g ∈ H(U) are such that f (n)(0) = gn(0) for every n ≥ 0, then f = g.
[Hint : Write f(z) =∑∞
n=0 anzn. (i) Take g(z) =
∑∞n=0
ann+ 1
zn+1. (ii) an =f (n)(0)
n!by [123].]
A few more interesting corollaries of [123] and [124] are the following:
[125] [Morera’s theorem - converse of Goursat’s integral lemma] Let U ⊂ C be open and f : U → C
be continuous. If∫∂∆ f = 0 for every solid triangle ∆ ⊂ U , then f ∈ H(U).
Proof. Fix a ∈ U , and let B ⊂ U be an open ball centered at a. Since B is a star region, f |B is
integrable by [118], and then by [117], f |B has a primitive g ∈ H(B). Since g is infinitely often
differentiable by [124], it follows that f |B = g′ is also infinitely often differentiable. In particular f
is holomorphic at a.
[126] [Comparing integrability and differentiability] Let U ⊂ C be open and f : U → C be
continuous. Then (i) f ∈ H(U) ⇔ f is locally integrable in U .
(ii) If U is a star region, then f ∈ H(U) ⇔ f is integrable in U .
Proof. Since being holomorphic is a local property, the implication ‘⇐’ follows by [125] in both (i)
and (ii). Moreover, the implication ‘⇒’ in (ii) is nothing but Cauchy’s integral theorem for star
regions. Therefore, it remains to show ‘⇒’ in (i). Assume f ∈ H(U), and write U =∪∞
n=1Bn,
a countable union of open balls. Since Bn is a star region, f |Bn is integrable as noted above, by
Cauchy’s integral theorem for star regions. Hence f is locally integrable in U .
[127] [Riemann continuation theorem] Let U ⊂ C be open, A ⊂ U be discrete and closed in U and
f : U \A→ C be holomorphic. Then the following are equivalent:
(i) f has a holomorphic extension to the whole of U .
(ii) f has a continuous extension to the whole of U .
33
(iii) For each a ∈ A, there is an open ball B ⊂ U centered at a such that f |B\{a} is bounded.
(iv) For each a ∈ A, limz→a(z − a)f(z) = 0.
Proof. The implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear. To prove (iv) ⇒ (i), we may assume
A = {a} and U = B := B(a, r) for some r > 0 since being holomorphic is a local property. Define
g : B → C as g(a) = 0 and g(z) = (z − a)f(z) for z = a. Then g is continuous in the star region
B by (iv), and g is holomorphic in B \ {a}. By [121], g is integrable in B and hence∫∂∆ g = 0
for every solid triangle ∆ ⊂ B, which implies by Morera’s theorem [125] that g ∈ H(B). By [123],
we may write g(z) =∑∞
n=0 an(z − a)n for z ∈ B. Since g(a) = 0, we have a0 = 0, and hence
g(z) = (z − a)[∑∞
n=1 an(z − a)n−1]. Define f : B → C as f(z) =∑∞
n=1 an(z − a)n−1. By [110](v)
and [111](i), f ∈ H(B). Moreover, f(z) = g(z)/(z − a) = f(z) for every z ∈ B \ {a}.
Usually, we will use the following special case of [127]:
[127′] Let U ⊂ C be open, a ∈ U , f : U → C be continuous, and suppose f is holomorphic in
U \ {a}. Then f ∈ H(U).
Remark: Result [127′] indicates the strong nature of complex differentiability. In contrast note that
in Real Analysis, f : R → R given by f(x) = |x| is continuous on R and differentiable on R \ {0},
but not differentiable at 0.
7 Liouville’s theorem and Zeroes theorem
Definition: Any f ∈ H(C) is called an entire function. If f is an entire function, then we may write
f(z) =∑∞
n=0 anzn, where the power series converges for every z ∈ C. We will see that an entire
function behaves somewhat like a polynomial.
[128] (i) [Liouville’s theorem] A bounded entire function is a constant.
(ii) Let f ∈ H(C). If there is c ∈ C such that limz→∞ f(z) = c, then f ≡ c.
(iii) If f ∈ H(C) is non-constant, then f(C) is dense in C.
Proof. (i) Let f ∈ H(C) be such that M := sup{|f(z)| : z ∈ C} < ∞. For any z ∈ C and r > 0,
applying Cauchy’s estimate to B(z, r) we get |f ′(z)| ≤M/r, and consequently f ′(z) = 0 since r > 0
is arbitrary. Thus f ′ ≡ 0, and therefore f is a constant by [104](i).
(ii) There is r > 0 such that |f(z)− c| < 1 for |z| > r, and f is bounded on the compact set B(0, r).
So f is bounded on C and hence constant by (i). The constant has to be c since limz→∞ f(z) = c.
34
(iii) If f(C) is not dense in C, there exist a ∈ C and r > 0 such that |f(z)− a| ≥ r for every z ∈ C.
Then g : C → C defined as g(z) = 1/(f(z)− a) is holomorphic and |g| ≤ 1/r. By (i), g is constant,
and then f should also be constant, a contradiction.
Exercise-52: (i) If f : C → D is holomorphic, then f is constant.
(ii) Let W be equal to one of the following: {z ∈ C : Re(z) > 0}, {z ∈ C : Im(z) > 0}, C \ (−∞, 0],
or C \ [0,∞). If f : C →W is holomorphic, then f is constant.
(iii) If f : C \ {0} → C is bounded and holomorphic, then f is constant.
(iv) If f ∈ H(C) and Re(f) is bounded, then f is constant.
[Hint : (ii) By [115], there is a Mobius map T taking W biholomorphically onto D. Then T ◦ f :
C → D is holomorphic and hence constant by (i). Since T is injective, f is also constant. (iii) Since
f is bounded on C \ {0}, f extends to an entire function by [127] so that [128] can be applied. (iv)
Note that f(C) cannot be dense and apply [128](iii). Another argument is: g := ef is bounded
and hence constant by Liouville’s theorem, which implies f(C) is discrete, and consequently f is
constant since f(C) must be connected.]
Exercise-53: Let p(z) =∑n
j=0 ajzj , where n ∈ N, aj ∈ C, and an = 0. Then,
(i) There is r > 0 such that 12 |anz
n| ≤ |p(z)| ≤ 32 |anz
n| for every z ∈ C with |z| ≥ r.
(ii) limz→∞ |p(z)| = ∞.
(iii) [Fundamental theorem of algebra] There is a ∈ C with p(a) = 0.
[Hint : (i) Let M =∑n−1
j=0 |aj |. Then for |z| ≥ 1, |anzn| −M |zn−1| ≤ |p(z)| ≤ |anzn| +M |zn−1|.
Let r = max{1, 2M/|an|}. Then for |z| ≥ r, we have M |zn−1| ≤ |anzn−1|r/2 ≤ |anzn|/2. Use this
in the earlier inequalities. (iii) Assume p is non-vanishing. Then f(z) := 1/p(z) defines an entire
function, and limz→∞ f(z) = 0 by (ii). So f (= 1/p) must be constant by [128](ii), a contradiction.]
Exercise-54: Let f ∈ H(C). (i) If there is k ∈ N such that limz→∞|f(z)||z|k
= 0, then f a polynomial
of degree ≤ k − 1. In particular, if limz→∞|f(z)||z|
= 0, then f is constant.
(ii) If there is g ∈ H(C) such that g is non-vanishing and |f(z)| ≤ |g(z)| for every z ∈ C, then there
is c ∈ C such that f = cg. In particular, if there is C > 0 such that |f(z)| ≤ C|ez| for every z ∈ C,
then there is c ∈ C such that f(z) = cez for every z ∈ C.
[Hint : (i) Write f(z) =∑k−1
n=0 anzn + zk
∑∞n=k anz
n−k = g(z) + zkh(z). By Exercise-53(i), we have
limz→∞|g(z)||z|k
= 0. Hence by hypothesis, 0 = limz→∞|f(z)||z|k
= 0 + limz→∞ |h(z)|. By [128](ii),
h ≡ 0. Thus f = g. (ii) h := f/g ∈ H(C) and is bounded.]
Exercise-55: Let f ∈ H(C) be doubly periodic in the following sense: there exist R-linearly inde-
35
pendent vectors a, b ∈ C such that f(z+ a) = f(z) = f(z+ b) for every z ∈ C. Then f is constant.
[Hint : Let K be the parallelogram with vertices 0, a, b, a + b. Then f(C) = f(K) because f is
doubly periodic. Also f(K) is bounded since K is compact. By Liouville’s theorem, f is constant.]
[129] [Zeroes theorem] Let U ⊂ C be a connected open set and f ∈ H(U). Then the following are
equivalent: (i) f ≡ 0, (ii) {z ∈ U : f(z) = 0} has a limit point in U , (iii) There is a ∈ U such that
f (n)(a) = 0 for every n ≥ 0.
Proof. (ii) ⇒ (iii): Let a ∈ U be a limit point of {z ∈ U : f(z) = 0}. By continuity, f(a) = 0. Let
B(a, r) ⊂ U , and write f(z) =∑∞
n=0 an(z − a)n for z ∈ B(a, r) by [123]. Since f (n)(a) = n!an by
[123], it suffices to show an = 0 for every n ≥ 0. Note that a0 = f(a) = 0. If possible, let an = 0
for some n ∈ N, and choose the smallest such n. Then f(z) = (z − a)ng(z) for z ∈ B(a, r), where
g(z) =∑∞
k=n ak(z − a)k−n. Since g(a) = an = 0, there is 0 < δ < r such that f(z) = 0 for every
z ∈ B(a, δ) \ {a}, a contradiction since a is a limit point of {z ∈ U : f(z) = 0}.
(iii) ⇒ (i): Let A = {z ∈ U : f (n)(z) = 0 for every n ≥ 0}. Then A = ∅ by (iii), and A is closed in
U since each f (n) is continuous. If a ∈ A and B(a, r) ⊂ U , then f(z) =∑∞
n=0
f (n)(a)
n!(z − a)n = 0
for every z ∈ B(a, r), which means B(a, r) ⊂ A. Thus A is also open in U . Since U is connected,
it follows that A = U , and in particular f ≡ 0.
Remark: Let U = B(1, 1) = {z ∈ C : |z − 1| < 1} and f : U → C be f(z) = sin(1/z). Then
f ∈ H(U), the set A := {z ∈ U : f(z) = 0} is infinite (consider the points 12πn) and 0 ∈ C \ U is a
limit point of A, but f is not identically zero.
Definition: Let U ⊂ C be open and connected, and suppose f ∈ H(U) is not identically zero. If
a ∈ U is with f(a) = 0, then m := min{n ∈ N : f (n)(a) = 0} (which exists by [129]) is called the
order of the zero a for f .
[130] [Corollaries of [129]] Let U ⊂ C be open and connected. Then,
(i) If f, g ∈ H(U) are such that {z ∈ C : f(z) = g(z)} has a limit point in U , then f = g.
(ii) If f ∈ H(U) is not identically zero, then |{z ∈ K : f(z) = 0}| <∞ for any compact set K ⊂ U .
(iii) [Factorization] Assume f ∈ H(U) is not identically zero. If a ∈ U is a zero of f of order m ∈ N,
then there is g ∈ H(U) such that f(z) = (z− a)mg(z) for every z ∈ U and g(a) = 0. Consequently,
we can find δ > 0 such that B(a, δ) ⊂ U and f(z) = 0 for every z ∈ B(a, δ) \ {a}.
(iv) [Structure of the set of zeroes] Assume f ∈ H(U) is not identically zero. Then f−1(0) = {z ∈
U : f(z) = 0} is discrete and closed in U (possibly empty), and consequently f−1(0) is countable.
36
(v) [Structure of the pre-image set] Assume f ∈ H(U) is non-constant. Then for every b ∈ C, the
set f−1(b) is discrete and closed in U (possibly empty), and consequently f−1(b) is countable.
(vi) H(U) is an integral domain.
Proof. (i) Apply [129] to f − g.
(ii) Note that any infinite subset of a compact K set must have a limit point in K.
(iii) Let r > 0 be with B(a, r) ⊂ U , and write f(z) =∑∞
n=0 an(z − a)n for z ∈ B(a, r). Since a is a
zero of order m, we have that an = f (n)(a)/n! = 0 for 0 ≤ n < m and am = f (m)(a)/m! = 0. Hence
f(z) = (z − a)mg(z) for z ∈ B(a, r), where g(z) :=∑∞
k=m ak(z − a)k−m. Then g ∈ H(B(a, r)) and
g(a) = am = 0. Extend g holomorphically to U by putting g(z) = f(z)/(z−a)m for z ∈ U \B(a, r).
(iv) f−1(0) is closed in U by the continuity of f , and f−1(0) is discrete in U either by part (iii) or
by [129]. To see f−1(0) is countable, argue as follows. For each z ∈ U , we can find an open ball Bz
centered at z such that Bz contains at most one zero of f (since the zeroes of f are isolated). We
may cover U with countably many such balls since U is second countable.
(v) Fix b ∈ C and apply part (iv) to the function g(z) := f(z)− b.
(vi) Verify that H(U) is a commutative ring with unity under pointwise operations. If f, g ∈ H(U)
are such that f(z)g(z) = 0 for every z ∈ U , then either f−1(0) or g−1(0) should be uncountable,
and consequently either f ≡ 0 or g ≡ 0 by (iv).
Exercise-56: (i) There does not exist any f ∈ H(D) such that f(1/n) = (−1)n/n for every n ≥ 2.
(ii) If f ∈ H(C) is such that f(f(z)) = f(z) for every z ∈ ∂D, then either f = I or f is constant.
(iii) Let U ⊂ C be open and connected, and f ∈ H(U). If there exist a ∈ U and k ∈ N such that
f (n)(a) = 0 for every n ≥ k, then f is a polynomial of degree ≤ k − 1.
[Hint : (i) f( 12n) = 1
2n for every n ∈ N and hence f = I by [130], a contradiction. (ii) f(∂D) is
compact and connected, and hence f(∂D) is either uncountable or a singleton. In the first case,
f = I by [130] since f(w) = w for every w ∈ f(∂D). In the second case, f is constant by [130] since
∂D has a limit point in C. (iii) f (k) ≡ 0 by Zeroes theorem.]
Exercise-57: Let U ⊂ C be open and connected. (i) Let f : U → C be continuous, and define
g : U → C as g(z) = f(z)2. If g ∈ H(U), then f ∈ H(U).
(ii) If f ∈ H(U), g ∈ H(U) \ {0} are such that fg ∈ H(U), then f is constant.
(iii) Let B ⊂ U be an open ball, and Φ : H(U) → H(B) be defined as Φ(f) = f |B. When is Φ
injective? When is Φ surjective?
37
[Hint : (i) Fix a ∈ U . Assume f is not identically zero, and choose an open ball B ⊂ U centered
at a by [130](iv) such that (g and hence) f is non-vanishing in B \ {a}. Since g(z) − g(w) =
(f(z) + f(w))(f(z) − f(w)), we get that limz→wf(z)− f(w)
z − w=
g′(w)
2f(w)for every w ∈ B \ {a}.
Hence f is holomorphic in B \ {a}, and hence by [127′] holomorphic in B. (ii) Choose an open ball
B ⊂ U such that g does not vanish in B. Since 1/g ∈ H(B), we get f = (fg)/g ∈ H(B). Then
Re(f) = (f + f)/2 and Im(f) = (f − f)/2i are holomorphic in B, and hence constants. Thus f |Bis constant. Now [130] may be used. (iii) Φ is always injective by [130]. If there is a ∈ U \ B, put
g(z) = 1/(z − a). If there is f ∈ H(U) with f |B = g|B, then f = g on U \ {a} by [130]; but this is
not possible since g is unbounded near a. Thus Φ is surjective iff U = B.]
Exercise-58: If f is a non-constant complex polynomial, then the zeroes of f lie in the convex hull of
the zeroes of f ′. [Hint : Let f(z) = c(z − a1) · · · (z − an), where c = 0. Thenf ′(z)
f(z)=∑n
j=1
1
z − aj.
If w ∈ C is such that f ′(w) = 0 and f(w) = 0, then 0 =f ′(w)
f(w)=∑n
j=1
1
w − aj=∑n
j=1
w − aj|w − aj |2
.
Letting αj =1
|w − aj |2, we see w = (
∑nj=1 αjaj)/(
∑nj=1 αj).]
Exercise-59: Let U ⊂ C be open and connected, and F : U × U → C be a function which is
holomorphic in each variable. Let A ⊂ U be a subset having a limit point in U and suppose
F (z, w) = 0 for every z, w ∈ A. Then F ≡ 0 on U × U . [Hint : For each z0 ∈ A, w 7→ F (z0, w) is
identically zero in U by [129]. Now fix w0 ∈ U and consider f : U → C defined as f(z) = F (z, w0).
Then f(z) = 0 for each z ∈ U by the first sentence. Since w0 is arbitrary, we get F ≡ 0.]
Remark: Considering F : C × C → C given by F (z, w) = ez+w − ezew and taking A = R, we
may deduce using Exercise-59 that ez+w = ezew for every z, w ∈ C. Similarly, we may also deduce
various trigonometric identities such as cos(z + w) = cos z cosw − sin z sinw for z, w ∈ C.
8 Some more fundamental theorems about holomorphic functions
[131] [Maximum modulus principle] Let U ⊂ C be open and connected, and f ∈ H(U) be non-
constant. Then, (i) |f | has no maximum in U .
(ii) If K ⊂ U is compact and a ∈ K is with |f(a)| = max{|f(z)| : z ∈ K}, then a ∈ ∂K.
Proof. (i) Suppose there is a ∈ U with |f(a)| = sup{|f(z)| : z ∈ U}. Choose R > 0 with
B(a,R) ⊂ U and let 0 < r < R. By Cauchy’s integral formula, f(a) =1
2πi
∫|z−a|=r
f(z)
z − adz, and
hence 2πr|f(a)| ≤∫|z−a|=r |f(z)||dz|. On the other hand, 2πr|f(a)| =
∫|z−a|=r |f(a)||dz|. Thus
we get∫|z−a|=r(|f(a)| − |f(z)|)|dz| ≤ 0. Since the integrand |f(a)| − |f(z)| is ≥ 0, we must have
38
|f(a)| = |f(z)| for every z with |z − a| = r. Since r ∈ (0, R) is arbitrary, we get |f(a)| = |f(z)|
for every z ∈ B(a,R). Thus f(B(a,R)) is contained in the circle ∂B(0, |f(a)|). By Exercise-38,
f |B(a,R) must be a constant. Since B(a,R) is a set having a limit point in U , [129] implies that f
is a constant in U , a contradiction.
(ii) If int(K) = ∅, then ∂K = K, and there is nothing to prove. So assume that V := int(K) = ∅.
Any connected component W of V is a connected open set in C and hence by (i), |f | restricted to
W has no maximum in W , which implies a /∈W . Therefore, a ∈ K \ V = ∂K.
Remark: Let a ∈ C, r > 0, and f ∈ H(B(a, r)) be non-constant. Using [131], it may be shown that
ϕ : [0, r) → R defined as ϕ(t) = max{|f(z)| : |z − a| = t} is (continuous and) strictly increasing.
Exercise-60: Prove Fundamental theorem of Algebra using Maximum modulus principle. [Hint :
Let p(z) =∑n
j=0 ajzj , where n ∈ N and an = 0. By Exercise-53, there is r > 0 such that
|anzn|/2 ≤ |p(z)| whenever |z| ≥ r. Here, r may be chosen large enough so that |p(0)| < |an|rn/2.
If p does not vanish in C, then for f := 1/p, we get |f(z)| < |f(0)| whenever |z| = r, a contradiction
to Maximuum modulus principle.]
Exercise-61: If a1, . . . , an ∈ ∂D, then there is z ∈ ∂D with∏n
j=1 |z − aj | > 1. [Hint : Since
p(z) :=∏n
j=1(z − aj) is non-constant, ∃ z ∈ ∂D with |p(z)| > |p(0)| =∏n
j=1 |aj | = 1 by [131](ii).]
Exercise-62: Let U ⊂ C be open and connected.
(i) Let f ∈ H(U). If there is a ∈ U with Re(f(a)) ≥ Re(f(z)) for every z ∈ U , then f is constant.
(ii) Let f ∈ H(U). If there is a ∈ U with |Re(f(a))| ≥ |Re(f(z))| ∀ z ∈ U , then f is constant.
(iii) Let f, g ∈ H(U) be non-vanishing. Suppose there is a closed ball B(a, r) ⊂ U such that
|f(z)| = |g(z)| whenever |z − a| = r. Then there is c ∈ ∂D with f(z) = cg(z) for every z ∈ U .
[Hint : (i) Let L = {z ∈ C : Re(z) = Re(f(a))}, W = {z ∈ C : Re(z) < Re(f(a))}, and let T be a
Mobius map such that T (L∪{∞}) = ∂D and T (W ) = D. Then |(T ◦ f)(z)| ≤ |(T ◦ f)(a)| for every
z ∈ U , and hence by [131], T ◦ f is constant. Since T is injective, f is also constant. (ii) Replacing
f with −f if necessary, assume Re(f(a)) ≥ 0. Then use (i) or argue as follows. Letting g = ef , we
have |g(z)| = eRe(f(z)) ≤ eRe(f(a)) = |g(a)|, and hence g is constant by [131]. So the connected set
f(U) is discrete and hence must be a singleton. (iii) h1 := f/g and h2 := g/f are holomorphic in
U . Since max{|hj(z)| : |z − a| = r} = 1 for j = 1, 2, we get |f(z)| = |g(z)| for every z ∈ B(a, r) by
[131]. Then h1(B(a, r)) ⊂ ∂D, and hence by Exercise-38, there is c ∈ ∂D with h1 ≡ c on B(a, r).
Then by [130], f/g = h1 ≡ c on U .]
Exercise-63: (i) Let a ∈ C, R > 0, and f ∈ H(B(a,R)) be f(z) =∑∞
n=0 an(z − a)n. If 0 < r < R
39
and M = max{|f(z)| : |z − a| = r}, then∑∞
n=0 |an|2r2n ≤M2.
(ii) Use (i) to give a proof of the Maximum modulus principle.
[Hint : (i) Let J =∫|z−a|=r
|f(z)|2
z − adz. Since |f(z)|2 = f(z)f(z) = f(z)
∑∞n=0 an(z − a)
nand since
the uniform convergence of the series on the circle |z−a| = r allows us to interchange the integral and
summation, J =∑∞
n=0 an
(∫|z−a|=r
f(z)(z − a)n
z − adz
)=∑∞
n=0 anr2n
(∫|z−a|=r
f(z)
(z − a)n+1dz
)=
2πi∑∞
n=0 |an|2r2n. On the other hand, |J | ≤∫|z−a|=r
M2
r|dz| = 2πM2. (ii) Let U ⊂ C be open
and connected and f ∈ H(U). Suppose there is a ∈ U is with |f(a)| ≥ |f(z)| for every z ∈ U . Let
B(a,R) ⊂ U , r ∈ (0, R) and M = max{|f(z)| : |z − a| = r}. If f(z) =∑∞
n=0 an(z − a)n in B(a,R),
then M ≤ |f(a)| = |a0|, and hence an = 0 for every n ≥ 1 by (i), which means f ≡ a0 in B(a,R).
Then by [130], f ≡ a0 in U .]
[132] [Weierstrass’ uniform convergence theorem] Let U ⊂ C be open, f : U → C be a function,
and (fn) be a sequence in H(U) converging to f uniformly on compact subsets of U . Then,
(i) f ∈ H(U).
(ii) For each k ∈ N, (f (k)n ) → f (k) as n→ ∞ uniformly on compact subsets of U .
Proof. (i) The hypothesis implies that for every open ball B with B ⊂ U , (fn) → f uniformly
on B, and this implies f is continuous in U . If ∆ ⊂ U is a solid triangle, then∫∂∆ fn = 0 for
every n ∈ N by Goursat’s lemma [119](i), and then∫∂∆ f = limn→∞
∫∂∆ fn = 0 by Exercise-42(iii).
Hence f ∈ H(U) by Morera’s theorem [125].
(ii) It suffices to show (f ′n) → f ′ uniformly on compact subsets of U (for then we may argue
inductively). Consider a compact subset A of U , and let ε > 0 be given. We need to find n0 ∈ N
such that |f ′(a) − f ′n(a)| ≤ ε for every n ≥ n0 and every a ∈ A. Choose r > 0 such that
K := {z ∈ C : dist(z,A) ≤ r} ⊂ U . SinceK is compact, there is n0 ∈ N such that |f(z)−fn(z)| ≤ εr
for every n ≥ n0 and every z ∈ K. Then by Cauchy’s estimate, for n ≥ n0 and a ∈ A we get that
|f ′(z)− f ′n(z)| ≤ max{|f(z)− fn(z)|r
: z ∈ B(a, r)} ≤ max{|f(z)− fn(z)|r
: z ∈ K} ≤ εr
r= ε.
Remark: Result [132] is another illustration of the strength of complex differentiability since [132]
says essentially that the operation f 7→ f ′ of complex differentiation on H(U) is continuous when
U ⊂ C is open. In contrast, differentiation of real functions is not a continuous operation: we
can construct a sequence (gn) of continuously differentiable functions from [0, 1] to R such that
∥gn∥∞ → 0 as n→ ∞ but ∥g′n∥∞ ≥ 1 for every n ∈ N, where ∥ · ∥∞ is the supremum norm.
Exercise-64: Let U ⊂ C be open, (fn) be a sequence in H(U), and f : U → C be a function.
40
(i) Let (f1+ · · ·+ fn) → f uniformly on compact subsets of U . Then f ∈ H(U), f =∑∞
n=1 fn, and
f (k) =∑∞
n=1 f(k)n for every k ∈ N, where the series converges uniformly on compact subsets of U .
(ii) Suppose f is continuous and let (fn) → f uniformly on the image [α] for each piecewise smooth
path α in U . Then f ∈ H(U) and (fn) → f uniformly on compact subsets of U .
[Hint : (ii) For every solid triangle ∆ ⊂ U , (fn) → f uniformly on ∂∆ and hence∫∂∆ f =
limn→∞∫∂∆ fn = 0 by Goursat’s lemma. Then f ∈ H(U) by Morera’s theorem. For any B(a, r) ⊂
U , (fn) → f uniformly on the circle |z− a| = r, and hence by Maximum modulus principle applied
to f − fn, we get (fn) → f uniformly on B(a, r). If K ⊂ U is compact, we may find finitely many
open balls Bj such that K ⊂∪m
j=1Bj ⊂ U , and hence (fn) → f uniformly on K.]
Exercise-65: Let g : C → C be a function, f ∈ H(C), and suppose (f (n)) → g uniformly on compact
subsets of C. Then there is c ∈ C such that g(z) = cez for every z ∈ C. [Hint : By [132], g ∈ H(C)
and g′ = limn→∞ f (n+1) = g. Writing g(z) =∑∞
n=0 anzn and using g′ = g, we get an = a0/n! for
every n and hence g(z) = a0ez.]
[133] [Open mapping theorem] Let U ⊂ C be open and connected, and f ∈ H(U) be non-constant.
Then f(U) is open in C.
Proof. Fix a ∈ U . We need to find ε > 0 such that B(f(a), ε) ⊂ f(U). Let R > 0 be with
B(a,R) ⊂ U . If for each r ∈ (0, R), there is zr ∈ U with |zr − a| = r and f(zr) = f(a), then
it will follow by Zeroes theorem that f ≡ f(a) on U , which is a contradiction to the hypothesis.
Therefore, there exist r ∈ (0, R) and ε > 0 such that |f(z) − f(a)| ≥ 2ε for every z ∈ U with
|z − a| = r by the compactness of the circle |z − a| = r. We claim that B(f(a), ε) ⊂ f(U).
Suppose the claim is false. Then there is w ∈ B(f(a), ε) such that w = f(z) for every z ∈ U .
Then g : U → C defined as g(z) = 1/(f(z)−w) is holomorphic. If |z−a| = r, then |f(z)−f(a)| ≥ 2ε
and |w − f(a)| < ε so that |f(z) − w| > ε, which implies |g(z)| < 1/ε. This is a contradiction to
the Maximum modulus principle since |g(a)| = 1/|f(a)− w| > 1/ε as w ∈ B(f(a), ε).
Remark: In the proof of Open mapping theorem, we used the Maximum modulus principle. It is
also possible to deduce the Maximum modulus principle from the Open mapping theorem - it is
easy to see this geometrically (do it).
[134] Let U ⊂ C be a star region (or more generally any connected open set such that every
f ∈ H(U) is integrable), and f ∈ H(U) be non-vanishing. Then,
(i) There is g ∈ H(U) such that eg = f .
41
(ii) [Existence of nth root] For each n ∈ N, there is h ∈ H(U) such that hn = f .
Proof. (i) For g ∈ H(U) to satisfy eg = f , it is necessary that we should have g′ = f ′/f . This gives
us a hint. Since U is a star region and f ′/f ∈ H(U), f ′/f has a primitive g1 ∈ H(U) by [119](ii),
i.e., g′1 = f ′/f . Then fe−g1 ∈ H(U) and (fe−g1)′ = (−fg′1+f ′)e−g1 ≡ 0. Hence there is c ∈ C with
fe−g1 ≡ c. Note that c = 0 since f and e−g1 are non-vanishing. Let b ∈ C be such that eb = c.
Then f(z) = eg1(z)+b for every z ∈ U . Take g(z) = g1(z) + b for z ∈ U .
(ii) Let g be as in (i) and put h = eg/n.
Exercise-66: Let f, g ∈ H(C) be with f(z)2+g(z)2 = 1 for every z ∈ C. Then there is ϕ ∈ H(C) such
that f(z) = cosϕ(z) and g(z) = sinϕ(z) for every z ∈ C. [Hint : 1 = (f(z)+ig(z))(f(z)−ig(z)) and
hence f(z) + ig(z) = 0. By [134](i), there is ϕ ∈ H(C) with eϕ(z) = f(z) + ig(z) = 1/(f(z)− ig(z))
for every z ∈ C. Then f(z) =eiϕ(z) + e−iϕ(z)
2= cosϕ(z) and g(z) =
eiϕ(z) − e−iϕ(z)
2i= sinϕ(z).]
[135] Let U ⊂ C be open and f ∈ H(U).
(i) If f is injective, then (f(U) is open in C) and f : U → f(U) is a biholomorphism.
(ii) If f(U) is open in C and f : U → f(U) is a biholomorphism, then f ′(z) = 0 for every z ∈ U .
(iii) If f ′(z) = 0 for every z ∈ U , then f is locally injective, i.e., for each a ∈ U , there is an open
ball B ⊂ U centered at a such that f |B is injective (however, we cannot expect f to be injective
on U ; for example, consider f(z) = ez on C).
Proof. (i) We may assume U is connected. By Open mapping theorem, f(U) is open, and g :=
f−1 : f(U) → U is continuous. Since f is injective, f ′ is not identically zero, and hence the set
A := {z ∈ U : f ′(z) = 0} is discrete and closed in U by [130]. As f is an injective open map,
f(A) is discrete and closed in f(U). So by Riemann continuation theorem [127], to show g is
holomorphic in f(U), it suffices to show g is holomorphic in f(U) \ f(A). Fix b ∈ U \ A, let
c = f(b), and (wn) be an arbitrary sequence in f(U) with (wn) → c. Then zn := g(wn) → g(c) = b,
and limn→∞g(wn)− g(c)
wn − c= limn→∞
zn − b
f(zn)− f(b)= 1/f ′(b) since f ′(b) = 0. This shows that g is
holomorphic at c with g′(c) = 1/f ′(b).
(ii) Letting g = f−1 : f(U) → U , we see 1 = (g ◦ f)′(z) = g′(f(z))f ′(z) and hence f ′(z) = 0.
(iii) Let a ∈ U . Since |f ′(a)| > 0 and f ′ is continuous at a, there is an open ball B ⊂ U
centered at a such that |f ′(b) − f ′(a)| < |f ′(a)|/2 for every b ∈ B. Now for z1, z2 ∈ B, observe
that |f(z2) − f(z1) − f ′(a)(z2 − z1)| = |∫[z1,z2]
(f ′(z) − f ′(a))dz| ≤∫[z1,z2]
|f ′(z) − f ′(a)||dz| ≤
|f ′(a)||z2 − z1|/2, from which it follows that f(z1) = f(z2) if z1 = z2. Thus f |B is injective.
42
Exercise-67: Let U, V ⊂ C be open and connected, f : U → V be injective and continuous,
g ∈ H(V ) be non-constant, and suppose g ◦ f ∈ H(U). Then f ∈ H(U). [Hint : K := {w ∈ V :
g′(w) = 0} is discrete and closed in V by hypothesis, and hence A := f−1(K) is discrete and closed
in U by the injectivity and continuity of f . By Riemann continuation theorem [127], it suffices to
show f is holomorphic in U \A. Let z ∈ U \A. Then f(z) /∈ K and hence g′(f(z)) = 0. By [135],
there is an open ball B ⊂ V centered at f(z) such that g : B → g(B) is a biholomorphism. Hence
f = g−1 ◦ (g ◦ f) on the neighborhood f−1(B) of z, and thus f is holomorphic at z.]
[136] [Non-constant holomorphic functions look like z 7→ zn locally] Let U ⊂ C be open and
connected, and f ∈ H(U) be non-constant. Then for every a ∈ U , there exist an open ball B ⊂ U
centered at a and a biholomorphism h : B → h(B) such that f(z) = f(a)+(h(z))m for every z ∈ B,
where m ∈ N is the order of the zero a for f(z)− f(a).
Proof. By [130](iii), there exist an open ball B1 ⊂ U centered at a and g ∈ H(B1) with g(a) = 0
such that f(z)−f(a) = (z−a)mg(z) for every z ∈ B1. By taking B1 small enough, we may suppose
that g is non-vanishing in B1. Then by [134](ii), there is h1 ∈ H(B1) with hm1 = g. Let h ∈ H(B1)
be defined as h(z) = (z−a)h1(z). Since h′(a) = h1(a) = 0, we may choose by [135](iii) a concentric
open ball B ⊂ B1 such that h|B is injective. Then by [135](i), h : B → h(B) is a biholomorphism
and f(z) = f(a) + (z − a)mg(z) = f(a) + (h(z))m for every z ∈ B.
9 Winding number and Cauchy’s general theory
Cauchy’s general theory of integration addresses the following questions: (i) Given an open set
U ⊂ C and piecewise smooth closed paths α, β in U , when can we say that∫α f =
∫β f for every
f ∈ H(U)? In particular, when can we say that∫α f = 0 for every f ∈ H(U)? (ii) Is it possible
to modify Cauchy’s integral formula for discs by replacing ‘∫|z−a|=r’ by ‘
∫α’, where α is a piecewise
smooth closed path?
[137] (i) Let U ⊂ C be open, f ∈ H(U), and α : [a, b] → U be a piecewise smooth closed path in
U . If f is non-vanishing on [α], then1
2πi
∫α
f ′(z)
f(z)dz ∈ Z.
(ii) Let α be a piecewise smooth closed path in C and p ∈ C \ [α]. Then 1
2πi
∫α
1
z − pdz ∈ Z.
Proof. (i) Since f is non-vanishing on the compact set [α], we can find an open neighborhood
V ⊂ U of [α] such that f is non-vanishing in V . Then, again by the compactness of [α], we may
choose finitely many points a = t0 < t1 < · · · < tn = b and open balls B1, . . . , Bn ⊂ V such that
43
αj := α|[tj−1,tj ] is smooth and [αj ] ⊂ Bj for 1 ≤ j ≤ n. Let zj = α(tj) for 0 ≤ j ≤ n. Since f is non-
vanishing in the star region Bj , there is gj ∈ H(Bj) with egj = f |Bj for 1 ≤ j ≤ n by [134](i). Then∫
αj(f ′/f) =
∫αjg′j = gj(zj)− gj(zj−1), and hence exp(
∫α(f
′/f)) = exp(∑n
j=1[gj(zj)− gj(zj−1)]) =∏nj=1 exp(gj(zj)− gj(zj−1)) =
∏nj=1
f(zj)
f(zj−1)= 1 since zn = z0. So,
∫α(f
′/f) ∈ ker(e) = 2πiZ.
(ii) Apply (i) with U = C \ {p} and f(z) = z − p.
Exercise-68: Let U = C \ [−1, 1]. Is there f ∈ H(U) with f(z)2 = z for every z ∈ U? [Hint : No.
Else, 2f(z)f ′(z) = 1 for z ∈ U . Dividing by 2f(z)2 = 2z, we getf ′(z)
f(z)=
1
2z. Now
∫|z|=2
f ′
f∈ 2πiZ
by [137], but∫|z|=2
dz
2z= πi, a contradiction.] [Remark: see also Equation (5.7.2) and Exercises
5.7.14 and 5.8.1 in D. Alpay, A Complex Analysis Problem Book.]
Definition: Let α be a piecewise smooth closed path in C. For p ∈ C \ [α], the winding number of
α around p (or the index of α with respect to p is defined as) n(α; p) =1
2πi
∫α
1
z − pdz ∈ Z. The
interior and exterior of α are defined as Int(α) = {p ∈ C \ [α] : n(α; p) = 0} and Ext(α) = {p ∈
C \ [α] : n(α; p) = 0}.
Remark: We explain why the winding number as defined above coincides with the geometric notion
of the number of times α winds around p in the anticlockwise direction. After a translation assume
p = 0 so that n(α; 0) =1
2πi
∫α
1
zdz. Since 0 /∈ [α], we may write α =
∑nj=1 αj in such a way that
each αj is smooth and there is an open ball Bj ⊂ C \ {0} with [αj ] ⊂ Bj . Let zj−1 and zj be the
starting point and end point of αj , and note that zn = z0 since α is closed. Let gj ∈ H(Bj) be a
branch of logarithm (see [108](ii)). Then gj(z) = log |z| + iarg(z) + 2πikj for some kj ∈ Z. Since
g′j(z) = 1/z, we have∫αj
1
zdz = gj(zj) − gj(zj−1) = log |zj | − log |zj−1| + i(arg(zj) − arg(zj−1)),
and hence1
2πi
∫α
1
zdz =
1
2πi
∑nj=1
∫αj
1
zdz =
1
2π
∑nj=1(arg(zj)− arg(zj−1)), which is an integer by
[137] and is clearly equal to the number of times α winds around p in the anticlockwise direction.
Exercise-69: Let α be a piecewise smooth closed path in C.
(i) If β is a piecewise smooth closed path in C with the same starting/end point as that of α,
then for every p ∈ C \ [α + β], we have that n(α + β; p) = n(α; p) + n(β; p), and in particular
n(−α; p) = −n(α; p).
(ii) p 7→ n(α; p) from C \ [α] to Z is continuous.
(iii) n(α; ·) is constant on each connected component of C \ [α]. Moreover, Int(α) and Ext(α) are
open in C \ [α] and hence open in C.
(iv) Int(α) is bounded. In fact, if r := max{|z| : z ∈ [α]}, then {p ∈ C : |p| > r} ⊂ Ext(α).
44
[Hint : (ii) Given p ∈ C\[α] and ε > 0, let λ = dist(p, [α])/2 and δ ∈ (0, λ) be withδl(α)
4πλ2< ε, where
l(α) :=∫α 1|dz| is the length of α (assume it is positive). Then for every q ∈ C\ [α] with |p−q| < δ,
we have |n(α; p)− n(α; q)| ≤ 1
2π
∫α |
p− q
(z − p)(z − q)||dz| ≤ 1
2π
∫α
δ
2λ2|dz| = δl(α)
4πλ2< ε. (iii) Use (ii)
and note Z is discrete. (iv) If p ∈ C is with |p| > r, choose r < s < |p|. Since f(z) := 1/(z − p) is
holomorphic in the star region B(0, s) containing [α], we get n(α; p) =1
2πi
∫α f = 0 by [119](ii).]
Exercise-70: Let U ⊂ C be open, α be a piecewise smooth closed path in C and let g : [α]×U → C
be a continuous function such that g(z, ·) ∈ H(U) for each z ∈ [α]. Then h : U → C defined as
h(w) =∫α g(z, w)dz is holomorphic. [Hint : Clearly, h is continuous. If ∆ ⊂ U is a solid triangle,
then∫∂∆ g(z, w)dw = 0 for each z ∈ [α] by Goursat’s lemma. As g is continuous, an interchange
of integrals is permitted by Fubini’s theorem, and hence∫∂∆ h(w)dw =
∫∂∆(∫α g(z, w)dz)dw =∫
α(∫∂∆ g(z, w)dw)dz = 0. By Morera’s theorem, h is holomorphic.]
[138] [Cauchy’s general integral theorem] Let U ⊂ C be open and let α be a piecewise smooth
closed path in U . Then the following are equivalent:
(i)∫α f = 0 for every f ∈ H(U).
(ii) n(α; p) = 0 for every p ∈ C \ U .
(iii) n(α; p)f(p) =1
2πi
∫α
f(z)
z − pdz for every f ∈ H(U) and every p ∈ U \ [α].
Proof. (i) ⇒ (ii): Apply (i) to f : U → C defined as f(z) = 1/(z − p).
(ii) ⇒ (iii): Since n(α;w)f(w) =1
2πi
∫α
f(w)
z − wdz, it suffices to show
∫α
f(z)− f(w)
z − wdz = 0 for
every w ∈ U \ [α]. Let V = Ext(α) = {p ∈ C \ [α] : n(α; p) = 0}. Define g1 : [α] × U → C
and g2 : [α] × V → C as follows: g1(z, z) = f ′(z), g1(z, w) =f(z)− f(w)
z − wfor z = w, and
g2(z, w) =f(z)
z − w. By Exercise-70, h1 : U → C and h2 : V → C defined as hj(w) =
∫α gj(z, w)dz for
j = 1, 2 are holomorphic. Moreover,∫α
f(w)
z − wdz = 0 for w ∈ U∩V by the definition of V = Ext(α),
and hence h1(w) = h2(w) for every w ∈ U ∩ V . By (ii), U ∪ V = C. It follows that h : C → C
defined as h|U = h1 and h|V = h2 is holomorphic. We claim that limw→∞ h(w) = 0.
Let M = max{|f(z)| : z ∈ [α]}, r = max{|z| : z ∈ [α]}, and l(α) denote the length of α. If
w ∈ C is with |w| > r, then w ∈ V and hence |h(w)| = |h2(w)| ≤∫α
|f(z)||z − w|
|dz| ≤ Ml(α)
dist(w, [α])→ 0
as w → ∞. This proves the claim. By [128](ii) (a corollary of Liouville’s theorem), h ≡ 0. In
particular, for every w ∈ U \ [α] we have that 0 = h(w) =∫α
f(z)− f(w)
z − wdz.
(iii) ⇒ (i): Apply (iii) to g : U → C defined as g(z) = (z − p)f(z).
[139] [Corollary] If U ⊂ C is open and connected, then the following are equivalent:
45
(i) Every f ∈ H(U) is integrable in U .
(ii) Every f ∈ H(U) has a primitive g ∈ H(U).
(iii)∫α f = 0 for every f ∈ H(U) and every piecewise smooth closed path α in U .
(iv) n(α; p) = 0 for every piecewise smooth closed path α in U and every p ∈ C \ U .
(v) n(α; p)f(p) =1
2πi
∫α
f(z)
z − pdz for every f ∈ H(U), every piecewise smooth closed path α in U
and every p ∈ U \ [α].
Remark: (i) In [139], we can add some more equivalent statements; see [144] below. (ii) Every star
region satisfies the statements of [139]. On the other hand, by taking p = 0, it may be seen that
the following open sets do not satisfy statement (iv) of [139] (and hence do not satisfy (i)-(iii) and
(v) also): C \ {0}, D \ {0}, and {z ∈ C : 1 < |z| < 2}.
Exercise-71: Let U ⊂ C be a connected open set and f ∈ H(U) \ {0}. Let a1, . . . , ak ∈ U be such
that aj is a zero of order mj ∈ N for f for 1 ≤ j ≤ k (there could be other zeroes also for f in
U). Then there is g ∈ H(U) such that f(z) = (z − a1)m1 · · · (z − ak)
mkg(z) for every z ∈ U and
g(aj) = 0 for 1 ≤ j ≤ k. [Hint : Argue by induction on k by using [130](iii).]
[140] [Counting zeroes] Let U ⊂ C be open and connected, f ∈ H(U) \ {0}, α be a piecewise
smooth closed path in U with n(α; p) = 0 for every p ∈ C \ U (i.e., Int(α) ⊂ U), and suppose f is
non-vanishing on [α]. Then, (i) f has at most finitely many zeroes in Int(α).
(ii) If f has no zeroes in Int(α), then∫α
f ′
f= 0. If a1, . . . , ak are the zeroes of f in Int(α) counted
with multiplicity, then∫α
f ′
f= 2πi
∑kj=1 n(α; aj).
Proof. Note that the compact set K := Int(α)∪ [α] = C \Ext(α) is contained in U by hypothesis.
(i) Since f is not identically zero, f has at most finitely many zeroes in K by [130].
(ii) First suppose f does not vanish in Int(α). Choose r > 0 such that V := {z ∈ C : dist(z,K) <
r} ⊂ U and f does not vanish in V . Since f ′/f is holomorphic in V and n(α; p) = 0 for every
p ∈ C \ V , we get∫α(f
′/f) = 0 by [138].
Next suppose a1, . . . , ak are the zeroes of f in Int(α) counted with multiplicity. By Exercise-71,
there is g ∈ H(U) such that f(z) = (z − a1) · · · (z − ak)g(z) for every z ∈ U and g(aj) = 0 for
1 ≤ j ≤ k. Since f does not vanish in [α], it follows by the choice of g that g does not vanish in
K. Choose s > 0 with W := {z ∈ C : dist(z,K) < s} ⊂ U and such that g does not vanish in W .
Since g′/g is holomorphic in W and n(α; p) = 0 for every p ∈ C \W , we get∫α(g
′/g) = 0 by [138].
To finish the argument, observe thatf ′(z)
f(z)=
1
z − a1+ · · ·+ 1
z − ak+g′(z)
g(z)for every z ∈W .
46
[141] Let U ⊂ C be open and connected, f ∈ H(U) \ {0} and B(a, r) ⊂ U . If f does not vanish
on the circle |z − a| = r, then1
2πi
∫|z−a|=r(f
′/f) is equal to the number of zeroes of f in B(a, r),
counted with multiplicity.
Proof. This is an immediate Corollary of [140].
Exercise-72: Evaluate∫|z|=2
2z + 1
z2 + z + 1dz. [Hint : Let f(z) = z2 + z+1. The zeroes of f are e2πi/3,
e4πi/3, and they lie in B(0, 2). Hence∫|z|=2
2z + 1
z2 + z + 1dz =
∫|z|=2(f
′/f) = 2πi× 2 = 4πi by [141].]
Exercise-73: Let a ∈ C, r > 0, f ∈ H(B(a, r)), and b = f(a). If m is the order of the zero a for
f(z) − b, then there are ε > 0, δ > 0 such that for every c ∈ B(b, ε) \ {b}, f(z) − c has exactly m
simple zeroes in B(a, δ). [Hint : Left as a reading assignment; see page 98 of J.B. Conway, Functions
of One Complex Variable.]
Our next aim is to present the ‘homotopy version’ of Cauchy’s integral theorem. Homotopy can
be defined in a more general setting, but the following definition will suffice for us:
Definition: Let U ⊂ C be open, and α, β : [0, 1] → U be (continuous) closed paths in U (need not
be piecewise smooth). We say α is homotopic to β in U if α can be continuously deformed in U
to β in the following sense: there is a continuous function h : [0, 1]2 → U (called a homotopy) such
that
(i) h(0, t) = α(t) and h(1, t) = β(t) for every t ∈ [0, 1], and
(ii) h(s, 0) = h(s, 1) for every s ∈ [0, 1] (i.e., t 7→ h(s, t) is a closed path in U for each s ∈ [0, 1]).
Note that ‘being homotopic’ is an equivalence relation in the collection of all closed paths in U .
Example: (i) Let U ⊂ C be a star region and c ∈ U be a center of U . Let α be any closed path
in U and β be the constant path at c. Then α is homotopic to β in U via the ‘convex’ homotopy
h : [0, 1]2 → U given by h(s, t) = (1− s)α(t) + sc. Hence any two closed paths in U are homotopic
to each other. (ii) Let 0 < r1 < r2 < r3 < r4, U = {z ∈ C : r1 < |z| < r4}, and α, β : [0, 1] → U
be given by α(t) = r2e2πit and β(t) = r3e
2πit. Then α is homotopic to β in U via the homotopy
h : [0, 1]2 → U given by h(s, t) = ((1− s)r2 + sr3)e2πit.
Exercise-74: [Integrals over two nearby closed paths are equal] Let U be open, α : [0, 1] → U be
a piecewise smooth closed path, and ε > 0 be such that V := {z ∈ C : dist(z, [α]) < 2ε} ⊂ U .
Then for every piecewise smooth closed path β : [0, 1] → U with |α(t) − β(t)| < ε for every
t ∈ [0, 1], we have that∫α f =
∫β f for every f ∈ H(U). [Hint : Choose n ∈ N large enough so
that |α(t1) − α(t2)| < ε and |β(t1) − β(t2)| < ε for every t1, t2 ∈ [0, 1] with |t1 − t2| ≤ 1/n. Let
47
Jk = [k−1n , kn ], αk = α|Jk , and βk = β|Jk for 1 ≤ k ≤ n. Let Pk be the piecewise smooth closed path
defined as Pk = αk + [α( kn), β(kn)]− βk + [β(k−1
n ), α(k−1n )], and note that [Pk] ⊂ B(α(k−1
n ), 2ε) ⊂ U
for 1 ≤ k ≤ n. If f ∈ H(U), then Cauchy’s integral theorem for the star region B(α(k−1n ), 2ε)
yields∫Pkf = 0, and hence 0 =
∑nk=1
∫Pkf =
∑nk=1
∫αkf −
∑nk=1
∫βkf =
∫α f −
∫β f .]
[142] [Homotopy version of Cauchy’s integral theorem] Let U ⊂ C be open, and α, β be piecewise
smooth closed paths in U . If α is homotopic to β in U , then∫α f =
∫β f for every f ∈ H(U).
Proof. Fix f ∈ H(U). Let h : [0, 1]2 → U be a homotopy from α to β, i.e., h(0, t) = α(t),
h(1, t) = β(t) and h(s, 0) = h(s, 1) for every s, t ∈ [0, 1]. For each s ∈ [0, 1], let hs : [0, 1] → U be
hs(t) = h(s, t). Then hs is a closed path in U but need not be piecewise smooth. The idea of the
proof is to approximate hs with a polygonal closed path (which is piecewise smooth).
Let [h] = h([0, 1]2), which is compact. Choose ε > 0 with V := {z ∈ C : dist(z, [h]) < 3ε} ⊂ U .
Let n ∈ N be large enough such that |h(s1, t) − h(s2, t)| < ε/3 for every t ∈ [0, 1] and every
s1, s2 ∈ [0, 1] with |s1 − s2| ≤ 1/n. For each k ∈ {0, 1, . . . , n}, let Pk be a polygonal closed
path in U such that |h( kn , t) − Pk(t)| < ε/3 for every t ∈ [0, 1] (∵ let m ∈ N be large enough
and choose Pk as the polygonal path with turning points h( kn ,0m), h( kn ,
1m), . . . , h( kn ,
mm)). Then
{z ∈ C : dist(z, [Pk]) < 2ε} ⊂ {z ∈ C : dist(z, [h]) < 2ε + ε/3} ⊂ V ⊂ U , and |Pk−1(t) − Pk(t)| ≤
|Pk−1(t)−h(k−1n , t)|+ |h(k−1
n , t)−h( kn , t)|+ |h( kn , t)−Pk(t)| < ε/3+ε/3+ε/3 = ε for every t ∈ [0, 1].
Therefore∫Pk−1
f =∫Pkf for 1 ≤ k ≤ n by Exercise-74. Moreover,
∫P0f =
∫α f and
∫Pnf =
∫β f
by applying Exercise-74 to the pairs P0, α and Pn, β (recall that α(t) = h(0, t) and β(t) = h(1, t)).
Hence we deduce that∫α f =
∫β f .
[143] [Corollary] Let U ⊂ C be open, α, β be piecewise smooth closed paths in U , and suppose α
is homotopic to β in U . Then, the winding number n(α; p) = n(β; p) for every p ∈ C \U (however,
n(α; p) need not be equal to n(β; p) for p ∈ U \ ([α] ∪ [β])).
Proof. Fix p ∈ C \ U and apply [142] to f(z) := 1/(z − p). For the statement in the bracket, let
U = C, α(t) = e2πit, β(t) = 3e2πit, and note that n(α; 2) = 0 = 1 = n(β; 2).
Remark: The converse of [143] is not true. There is a piecewise smooth closed path α in an open
set U ⊂ C such that n(α; p) = 0 for every p ∈ C \U but α is not homotopic to a constant path; see
Exercise 4.6.8 of J.B. Conway, Functions of One Complex Variable.
Definition: A connected open set U ⊂ C is said to be simply connected if every (continuous) closed
path in U is homotopic to a constant path. Since the image of a path in an open set U can be
48
covered by finitely many open balls in U , it may be seen that a connected open set U ⊂ C is
simply connected iff every piecewise smooth closed path in U is homotopic to a constant path. For
example, every star region in C is simply connected: if U ⊂ C is a star region and c ∈ U is a center
of U , then every closed path α : [0, 1] → U in U is homotopic to the constant path at c via the
homotopy h : [0, 1]2 → U given by h(s, t) = (1− s)α(t) + sc.
[144] Let U ⊂ C be a connected open set. Then the following are equivalent:
(i) U is simply connected.
(ii) C∞ \ U is connected.
(iii) n(α; p) = 0 for every piecewise smooth closed path α in U and every p ∈ C \ U .
(iv) Every f ∈ H(U) has a primitive.
(v) Every f ∈ H(U) is integrable.
(vi)∫α f = 0 for every f ∈ H(U) and every piecewise smooth closed path α in U .
(vii) n(α; p)f(p) =1
2πi
∫α
f(z)
z − pdz for every f ∈ H(U), every piecewise smooth closed path α in U
and every p ∈ U \ [α].
Proof. (i) ⇒ (iii): The given path α is homotopic to a constant path β in U since U is simply
connected. Hence for any p ∈ U \ [α], we have that n(α; p) = n(β; p) = 0 by [143].
We know by [139] that statements (iii)-(vii) are equivalent. For the rest of the proof and
some more equivalent statements, see Theorem 8.2.2 of J.B. Conway, Functions of One Complex
Variable.
Remark: Read [144](ii) carefully. If U = C \ {0}, then C \ U is connected but U is not simply
connected.
10 Singularities
Definition: Let U ⊂ C be open, a ∈ U and f ∈ H(U \ {a}). Then a is called an isolated singularity
of f . Moreover, (i) a is a removable singularity of f if f has a holomorphic extension to U ;
(ii) a is a pole of f if a is not a removable singularity of f but there is m ∈ N such that a is a
removable singularity of (z − a)mf(z) (here, the smallest such m is called the order of the pole a);
(iii) a is an essential singularity of f if a is neither a removable singularity nor a pole of f .
[145] Let U ⊂ C be open, a ∈ U and f ∈ H(U \ {a}). Then,
49
(i) a is a removable singularity of f ⇔ f has a continuous extension to U ⇔ there is an open ball
B ⊂ U centered at a such that f is bounded on B \ {a} ⇔ limz→a(z − a)f(z) = 0.
(ii) If a is a pole of order m ∈ N of f , then there is g ∈ H(U) such that (z − a)mf(z) = g(z) for
every z ∈ U \ {a} and g(a) = 0.
(iii) a is a pole of f ⇔ limz→a |f(z)| = ∞.
(iv) a is a pole of order m of f ⇔ limz→a(z − a)m+1f(z) = 0 but limz→a(z − a)mf(z) = 0.
Proof. (i) Use Riemann continuation theorem.
(ii) If g(a) = 0, then there is h ∈ H(U) with g(z) = (z − a)h(z) for every z ∈ U , and then h is a
holomorphic extension to U of (z − a)m−1f(z), which contradicts the assumption that a is a pole
of order m.
(iii) If a is a pole of orderm for f , then by (ii), limz→a |f(z)| = limz→a|g(z)|
|z − a|m= ∞ since g(a) = 0.
Conversely, assume limz→a |f(z)| = ∞. Then clearly a is not a removable singularity of f . Choose
an open ball B ⊂ U centered at a with |f(z)| ≥ 1 for every z ∈ B \ {a}. Then 1/f is holomorphic
and bounded in B \ {a}. So by (i), a is a removable singularity of 1/f in B \ {a}. Let g ∈ H(B) be
with 1/f(z) = g(z) for every z ∈ B \ {a}. Then g(z) = 0 for every z ∈ B \ {a}, and g(a) = 0 since
limz→a |f(z)| = ∞. Therefore, there are m ∈ N and h ∈ H(B) such that g(z) = (z − a)mh(z) for
every z ∈ B and h(a) = 0. Replacing B by a smaller concentric open ball, we may now suppose
h(z) = 0 for every z ∈ B. For z ∈ B \ {a}, we have 1/h(z) = (z − a)m/g(z) = (z − a)mf(z), which
shows that a is a pole of f in B (and hence in U since 1/h can be extended holomorphically to U
by prescribing its value to be (z − a)mf(z) for z ∈ U \B).
(iv) This follows from (i) and the definition of a pole of order m.
Example: (i) Let f(z) =sin z
z. Since limz→0 zf(z) = 0, 0 is a removable singularity of f . We
have f(z) = 1z (
z1! −
z3
3! +z5
5! − · · · ) so that we get a holomorphic extension by putting f(0) = 1.
(ii) For each m ∈ N, 0 is a pole of order m for f(z) := 1/zm. (ii) Let f(z) =z2 + 1
z3. Since
limz→0 |f(z)| = ∞, 0 is a pole of f , and it is a simple pole since limz→0 zf(z) exists in C. (iv) Let
f(z) = e1/z. Since limn→∞ f(1/n) = ∞ and limn→∞ |f(1/in)| = 1 = ∞, 0 is neither a removable
singularity nor a pole of f . Hence 0 is an essential singularity of f .
[146] [Casorati-Weierstrass theorem] Let U ⊂ C be open, and a ∈ U be an essential singularity of
f ∈ H(U \ {a}). Then for every open ball B ⊂ U centered at a, we have that f(B \ {a}) is (open
and) dense in C.
50
Proof. Fix B as given. Since a is not a removable singularity, f is not a constant on B \{a}. Hence
by Open mapping theorem, f(B \ {a}) is open in C. If f(B \ {a}) is not dense in C, there exist
b ∈ C and ε > 0 with |f(z) − b| ≥ ε for every z ∈ B \ {a}. Then g : B \ {a} → C defined as
g(z) =f(z)− b
z − asatisfies limz→a |g(z)| = ∞, and hence a is a pole of g by [145]. If m ∈ N is the
order of the pole a of g, then 0 = limz→a(z− a)m+1g(z) = limz→a(z− a)m(f(z)− b), which implies
limz→a(z − a)mf(z) = 0. This means, a is either a removable singularity (if m = 1) or a pole of
order ≤ m− 1 of f , a contradiction.
Example: Let f(z) = e1/z. We know 0 is an essential singularity of f . Note that for each n ∈ N,
f({z ∈ C : 0 < |z| < 1/n}) = {ew : |w| > n} = C\{0} since the exponential function is 2πi-periodic.
Exercise-76: Let U ⊂ C be open, and a ∈ U be an essential singularity of f ∈ H(U \ {a}). Then,
(i) For each w ∈ C, there is a sequence (an) in U \ {a} such that (an) → a and (f(an)) → w.
(ii) There is a dense Gδ subsetW ⊂ C such that for each w ∈W , there is a sequence (an) in U \{a}
such that (an) → a and f(an) = w for every n ∈ N.
[Hint : Let (Bn) be a decreasing sequence of open balls in U centered at a with∩∞
n=1Bn = {a}. (i)
By [146], we may choose an ∈ Bn with |w − f(an)| < 1/n. (ii) f(Bn \ {a}) is open and dense in C
by [146]. Hence W :=∩∞
n=1 f(Bn \ {a}) is a dense Gδ in C by Baire category theorem.]
Definition: f ∈ H(C) is called a transcendental entire function if f is not a polynomial.
[147] f ∈ H(C) is a transcendental entire function iff 0 is an essential singularity of f(1/z).
Proof. Let g : C \ {0} → C be g(z) = f(1/z). If f is constant, then 0 is a removable singularity
of g. If f is a non-constant polynomial, then limz→0 |g(z)| = limw→∞ |f(w)| = ∞ by Exercise-
53(ii), and hence 0 is a pole of g by [145](iii). Conversely, if 0 is not an essential singularity of
g, then there is k ∈ N such that zmg(z) has an extension to an entire function for every m ≥ k.
Therefore,∫|z|=1 z
mg(z)dz = 0 for every m ≥ k by Cauchy’s integral theorem for star regions. On
the other hand, if we write f(z) =∑∞
n=0 anzn, then
∫|z|=1 z
mg(z)dz =∫|z|=1(
∑∞n=0 anz
m−n)dz =∑∞n=0 an
∫|z|=1 z
m−ndz = 2πiam+1 for every m ≥ k (where the integral and summation are inter-
changed using the uniform convergence of the series), which implies am+1 = 0 for every m ≥ k.
Hence f must be a polynomial of degree ≤ k.
Remark: It follows from [147] that 0 is an essential singularity of e1/z, sin(1/z), and cos(1/z).
Definition: A Laurent series is a two-sided series of the form∑∞
n=−∞ an(z − a)n, where a, an ∈ C,
defined on some (open) subset of C \ {a}. A Laurent series∑∞
n=−∞ an(z−a)n is said to be conver-
51
gent/absolutely convergent/uniformly convergent on a subset J of C \ {a} if both∑∞
n=0 an(z − a)n
and∑∞
n=1 a−n(z−a)−n are convergent/absolutely convergent/uniformly convergent on J ; moreover,
if convergence happens at z ∈ J , then∑∞
n=−∞ an(z−a)n :=∑∞
n=0 an(z−a)n+∑∞
n=1 a−n(z−a)−n.
Notation: For a ∈ C and 0 < r < s, let A(a, r, s) = {z ∈ C : r < |z − a| < s} (open annulus
centered at a) and A(a, r, s) = {z ∈ C : r ≤ |z − a| ≤ s} (closed annulus centered at a).
[148] [Laurent series expansion] Let a ∈ C, 0 < r1 < r2 and U = A(a, r1, r2). If f ∈ H(U), then
there is a Laurent series∑∞
n=−∞ an(z−a)n which converges absolutely and uniformly in A(a, s1, s2)
for every s1, s2 ∈ R with r1 < s1 < s2 < r2 such that f(z) =∑∞
n=−∞ an(z − a)n for every z ∈ U .
Moreover, for each n ∈ Z, an =1
2πi
∫|z−a|=t
f(z)
(z − a)n+1dz for any t ∈ (r1, r2).
Proof. For n ∈ Z and t ∈ (r1, r2), define an =1
2πi
∫|z−a|=t
f(z)
(z − a)n+1dz. Observe that if t1, t2 ∈
(r1, r2), then the natural parametrizations of the circles |z− a| = t1 and |z− a| = t2 are homotopic
in U (via a convex homotopy), and hence the value of an is independent of the particular choice of
t by the Homotopy version of Cauchy’s integral theorem, [142]. Now consider s1, s2 with r1 < s1 <
s2 < r2. Let t1 ∈ (r1, s1) and t2 ∈ (s2, r2).
Fix w ∈ A(a, t1, t2) and define g : U → C as g(w) = f ′(w) and g(z) =f(z)− f(w)
z − wfor
z = w. By Riemann continuation theorem [127], g ∈ H(U). Hence∫|z−a|=t1
g =∫|z−a|=t2
g by [142].
Multiplying both sides by1
2πi, and using the fact that the winding number n(|z − a| = t1;w) = 0
and n(|z − a| = t2;w) = 1, we get1
2πi
∫|z−a|=t1
f(z)
z − wdz − 0 =
1
2πi
∫|z−a|=t2
f(z)
z − wdz − f(w), and
hence f(w) =−1
2πi
∫|z−a|=t1
f(z)
z − wdz +
1
2πi
∫|z−a|=t2
f(z)
z − wdz.
Let W1 = {w ∈ C : |w − a| > t1} and W2 = B(a, t2). Define hj : Wj → C for j = 1, 2 as
h1(w) =−1
2πi
∫|z−a|=t1
f(z)
z − wdz and h2(w) =
1
2πi
∫|z−a|=t2
f(z)
z − wdz. Then hj ∈ H(Wj) for j = 1, 2
by Exercise-70. Moreover, if we put W = W1 ∩ W2 = {w ∈ C : t1 < |w − a| < t2}, then
f(w) = h1(w) + h2(w) for every w ∈ W by the previous paragraph. To complete the proof, it is
enough to establish the following two claims.
Claim-1 : h1(w) =∑∞
n=1 a−n(w − a)−n for every w ∈ W1, and the series converges absolutely and
uniformly in A(a, s1, s2).
Claim-2 : h2(w) =∑∞
n=0 an(w − a)n for every w ∈ W2, and the series converges absolutely and
uniformly in A(a, s1, s2).
Consider w ∈ W1 so that |w − a| > t1. If |z − a| = t1, then1
z − w=
−1
(w − a)− (z − a)=
52
−1
(w − a)[1− (z − a)/(w − a)]= −
∑∞n=0
(z − a)n
(w − a)n+1= −
∑∞n=1
(z − a)n−1
(w − a)n= −
∑∞n=1
(w − a)−n
(z − a)−n+1.
Therefore, by interchanging the summation and integration, we get that h1(w)
=1
2πi
∫|z−a|=t1
f(z)
[∑∞n=1
(w − a)−n
(z − a)−n+1
]dz
=∑∞
n=1
[1
2πi
∫|z−a|=t1
f(z)
(z − a)−n+1dz
](w − a)−n =
∑∞n=1 a−n(w − a)−n.
Let M1 = max{|f(z)| : |z − a| = t1}. If s1 ≤ |w − a| ≤ s2, then for any n ∈ N, we see from the
expression for a−n that |a−n(w − a)−n| ≤ 1
2π
∫|z−a|=t1
|f(z)(z − a)n−1(w − a)−n||dz| ≤ M1tn1s
−n1 .
Since 0 < t1 < s1, it follows by Weierstrass M-test that the series for h1 converges absolutely and
uniformly in A(a, s1, s2). This proves claim-1.
Next consider w ∈ W2 so that |w − a| < t2. If |z − a| = t2, then we have that1
z − w=
1
(z − a)− (w − a)=∑∞
n=0
(w − a)n
(z − a)n+1. Therefore, by interchanging the summation and integral,
we may deduce h2(w) =∑∞
n=0 an(w−a)n. LetM2 = max{|f(z)| : |z−a| = t2}. If s1 ≤ |w−a| ≤ s2,
then for any n ≥ 0, we see that |an(w−a)n| ≤ 1
2π
∫|z−a|=t2
|f(z)(z−a)−n−1(w−a)n||dz| ≤M2t−n2 sn2 .
Since 0 < s2 < t2, it follows by Weierstrass M-test that the series for h2 converges absolutely and
uniformly in A(a, s1, s2). This proves claim-2.
Exercise-77: Let a ∈ C, r > 0, and suppose f ∈ H(B(a, r) \ {a}) has Laurent series expansion
f(z) =∑∞
n=−∞ an(z − a)n. Then, g :∈ C \ {a} → C defined as g(z) =∑∞
n=1 a−n(z − a)−n is
holomorphic in C \ {a}. [Hint : For each t ∈ (0, r), g is holomorphic in the region |z− a| > t by the
holomorphic property of h1 in the proof of [148].]
The Laurent series expansion can be used to classify isolated singularities:
[149] Let U ⊂ C be an open ball with center a ∈ C, and suppose f ∈ H(U \ {a}) has Laurent
series expansion f(z) =∑∞
n=−∞ an(z − a)n for z ∈ U \ {a}. Then,
(i) a is a removable singularity of f ⇔ a−n = 0 for every n ∈ N.
(ii) a is a pole of order m of f ⇔ a−m = 0 and a−n = 0 for every n ≥ m+ 1.
(iii) a is an essential singularity of f ⇔ a−n = 0 for infinitely many n ∈ N.
Proof. (i) Suppose a is a removable singularity of f . Choose r > 0 with B(a, r) ⊂ U , and note
that M := sup{|f(z)| : 0 < |z − a| ≤ r} < ∞. For any n ∈ N and t ∈ (0, r), we have that
|a−n| ≤1
2π
∫|z−a|=tMtn−1|dz| = Mtn, and hence a−n = 0 since t ∈ (0, r) can be arbitrarily small.
Conversely, if a−n = 0 for every n ∈ N, then the power series∑∞
n=0 an(z−a)n (which is convergent
in U \{a} and hence in the open ball U) defines a holomorphic function in U extending f . In other
53
words, f extends to U holomorphically by taking the value f(a) = a0.
The proof of (ii) is similar; work with (z − a)mf(z) instead of f . And (iii) follows from (i) and
(ii), and the definition of an essential singularity.
Seminar topic: Exercise 5.1.1 (page 110) of J.B. Conway, Functions of One Complex Variable.
Exercise-78: Let U ⊂ C be open, and f be holomorphic in U except for isolated singularities. Then
the set P (f) of poles of f in U is discrete and closed in U . [Hint : Since singularities are isolated by
definition, P (f) is discrete in U and no singularity of f is a limit point of P (f). If f is holomorphic
at w ∈ C, then |f(w)| = ∞ and hence w cannot be a limit point of P (f) by [145](iii).]
Definition: Let U ⊂ C. We say f is meromorphic on U if there is a discrete and closed subset
A ⊂ U such that f ∈ H(U \A) and each a ∈ A is a pole of f (briefly speaking, if f is holomorphic
in U except for poles).
Remark: Let U ⊂ C be open and connected, and M(U) be the collection of all meromorphic
functions on U . We can define - excluding the poles - pointwise addition and multiplication for
f, g ∈ M(U) in a natural way, and then M(U) becomes a field. It is a non-trivial fact that M(U) is
the field of fractions of the integral domain H(U); see page 318 of R. Remmert, Theory of Complex
Functions.
Exercise-79: Let U ⊂ C be open and connected, f, g ∈ M(U), and P (f), P (g) be their set of poles.
If W = U \ (P (f) ∪ P (g)), then the following are equivalent: (i) f = g.
(ii) {w ∈W : f(w) = g(w)} has a limit point in W .
(iii) There is w ∈W such that f (n)(w) = g(n)(w) for every n ≥ 0.
[Hint : Since P (f)∪P (g) is discrete and closed in U , the set W is open and connected in C. Apply
Zeroes theorem.]
11 Residues
Let U ⊂ C be open, a ∈ U , f ∈ H(U \{a}), and α be a piecewise smooth closed path in U \{a} such
that n(α; p) = 0 for every p ∈ C \ U . We ask whether∫α f can be determined. Some observations
in this context are the following:
(i) If a is a removable singularity of f , then∫α f = 0 by [138].
(ii) If a is an essential singularity of f , then in general it is difficult to evaluate∫α f .
54
(iii) Suppose B(a, r) ⊂ U , f(z) =∑∞
n=−∞ an(z − a)n in B(a, r) \ {a} and α is the natural
parametrization of the circle |z − a| = t for some t ∈ (0, r). Then by interchanging integration
and summation, we may see that1
2πi
∫α f = a−1.
Definition: Let U ⊂ C be open, a ∈ U , f ∈ H(U \ {a}), and suppose f has the Laurent series
expansion f(z) =∑∞
n=−∞ an(z − a)n in B \ {a} for some open ball B ⊂ U centered at a. Then
the coefficient a−1 is called the residue of f at a, and we write Res(f ; a) = a−1 (this definition and
terminology is motivated by observation (iii) above).
Concerning the question that we asked in the beginning of this section (about determining∫α f),
the result [150] gives an answer; it is useful mainly when all singularities are poles (i.e., when f is
meromorphic).
[150] [Residue theorem] Let U ⊂ C be open, w1, . . . , wk be finitely many distinct points in U , and
α be a piecewise smooth closed path in U \ {w1, . . . , wk} such that n(α; p) = 0 for every p ∈ C \U .
Then for every holomorphic function f : U \ {w1, . . . , wk} → C, we have that1
2πi
∫α f(z)dz =∑k
j=1 n(α;wj)Res(f ;wj).
Proof. Let Bj ⊂ U be an open ball centered at wj , and let f(z) =∑∞
n=−∞ an,j(z − wj)n be the
Laurent series expansion of f in Bj \ {wj} for 1 ≤ j ≤ k. By Exercise-77, gj : C \ {wj} → C
defined as gj(z) =∑∞
n=1 a−n,j(z − wj)−n is holomorphic in C \ {wj} for each j ∈ {1, . . . , k}.
Since f − gj and gm for m = j are holomorphic at wj , it follows that f −∑k
j=1 gj is holomor-
phic in U except for the removable singularities w1, . . . , wk. By Cauchy’s general integral theorem
[138], we conclude that∫α(f −
∑kj=1 gj) = 0. Also, observe that
1
2πi
∫α gj =
1
2πi
∫α
a−1,j
z − wjdz +
1
2πi
∫α [∑∞
n=2 a−n,j(z − wj)−n] dz = n(α;wj)Res(f ;wj) + 0, where the second integral is zero be-
cause the integrand in the square bracket has a primitive (which can be obtained by termwise
integration). Hence1
2πi
∫α f(z)dz =
∑kj=1
1
2πi
∫α gj =
∑kj=1 n(α;wj)Res(f ;wj).
While applying the Residue theorem, the following observations will be useful:
[151] Let U ⊂ C, a ∈ U and f ∈ H(U \ {a}).
(i) If a is a simple pole of f , then Res(f ; a) = limz→a(z − a)f(z).
(ii) If a is a pole of order m ≥ 2 of f and if g is a holomorphic extension of (z−a)mf(z) to U , then
Res(f ; a) =g(m−1)(a)
(m− 1)!.
(iii) Let a be a simple pole of f . Suppose there is an open ball B ⊂ U centered at a and g, h ∈ H(B)
such that f = g/h in B \ {a}, g(a) = 0, h(a) = 0, and h′(a) = 0. Then Res(f ; a) = g(a)/h′(a).
55
Proof. We may deduce (i) and (ii) from the Laurent series of f around a. To prove (iii), first note
that h has the power series expansion h(z) =∑∞
n=1
h(n)(a)
n!(z−a)n in B since h(a) = 0. Combining
this with (i), we get Res(f ; a) = limz→a(z − a)f(z) = limz→a(z − a)g(z)
h(z)=
g(a)
h′(a).
Exercise-80: (i) Let k ∈ Z and α : [0, 1] → C be α(t) = e2πikt and f(z) = e1/z. Evaluate∫α f .
(ii) Let α, β be the parametrizations of the circles |z − 1| = 1 and |z + 1| = 1 respectively, and let
f(z) =z2
(z + 1)2(z − 1). Evaluate
∫α f and
∫β f .
[Hint : (i) 0 is the only singularity; n(α; 0) = k and Res(f ; 0) = 1 (from the expansion of e1/z). By
[150],∫α f = 2πik. (ii) n(α; 1) = 1 = n(β;−1), n(α;−1) = 0 = n(β; 1), f ∈ H(C \ {−1, 1}), 1 is
a simple pole and −1 is a pole of order 2 of f , and g(z) := z2/(z − 1) is a holomorphic extension
of (z + 1)2f(z) in a neighborhood of −1. So by [151], we see Res(f ; 1) = limz→1(z − 1)f(z) = 1/4
and Res(f ;−1) = g′(−1)/1! = 3/4. Hence by [150],∫α f = πi/2 and
∫β f = 3πi/2.]
Exercise-81: (i) Let p, q be polynomials such that deg(p)+2 ≤ deg(q) and q(x) = 0 for every x ∈ R.
If A = {a ∈ C : Im(a) > 0 and g(a) = 0}, then 1
2πi
∫∞−∞
p(x)
q(x)dx =
∑a∈ARes(p/q; a).
(ii) Evaluate∫∞−∞
x2
1 + x4dx.
[Hint : (i) Let f(z) = p(z)/q(z). For R > 0, let αR : [0, 1] → C be αR(t) = Reπit and βR =
αR + [−R,R]. Then there is R0 > 0 such that for every R ≥ R0 we have that A ⊂ Int(βR)
with n(βR; a) = 1 for every a ∈ A, and hence by [150],1
2πi
∫βRf =
∑a∈ARes(f ; a). Also,
since deg(p) + 2 ≤ deg(q), we have |∫αRf | ≤ max|z|=R |f(z)| × πR → 0 as R → ∞, and therefore∫∞
−∞ f(x)dx = limR→∞∫[−R,R] f+limR→∞
∫αRf = limR→∞
∫βRf . (ii) Let p(x) = x2, q(x) = 1+x4,
f = p/q, and A = {a1, a2}, where a1 = eπi/4 and a2 = e3πi/4. Since a1, a2 are simple poles of f , we
see by [151] that Res(f ; a1) = p(a1)/q′(a1) = 1/(4a1) = a1/4 =
1− i
4√2, and similarly, Res(f ; a2) =
a2/4 =−1− i
4√2
. So by (i),∫∞−∞ f(x)dx = 2πi(Res(f ; a1) +Res(f ; a2)) =
πi(1− i− 1− i)
2√2
=π√2.]
Seminar topic: Examples 5.2.7 (p.115), 5.2.9 (p.117), 5.2.10 (p.117), and Exercises 5.2.1, 5.2.2
(p.121) of J.B. Conway, Functions of One Complex Variable.
12 More on zeroes and poles
[152] [Factorization of a meromorphic function] Let U ⊂ C be open and connected, and f be
meromorphic in U with only finitely many zeroes and poles in U . Let z1, . . . , zk be the zeroes of f
in U and w1, . . . , wm be the poles of f in U , counted with multiplicity. Then,
56
(i) There is a non-vanishing g ∈ H(U) with f(z) =(z − z1) · · · (z − zk)
(z − w1) · · · (z − wm)g(z) for every z ∈ U .
(ii) Consequently,f ′(z)
f(z)=∑k
j=1
1
z − zj−∑m
l=1
1
z − wl+g′(z)
g(z)for every z ∈ U\{z1, . . . , zk, w1, . . . , wm}.
Proof. Let U1 = U \ {w1, . . . , wm}. Since f ∈ H(U1), there is a non-vanishing f1 ∈ H(U1) by
factorization (Exercise-71) such that f(z) = (z − z1) · · · (z − zk)f1(z) for every z ∈ U1. Then
1/f1 ∈ H(U1) and limz→wl
1
|f1(z)|= limz→wl
|(z − z1) · · · (z − zk)||f(z)|
= 0 for 1 ≤ l ≤ m by [145](iii)
since wl is a pole of f . Hence there is f2 ∈ H(U) such that f2(z) = 1/f1(z) for z ∈ U1 and
f2(wl) = 0 for 1 ≤ l ≤ m. Factorizing f2, we may find a non-vanishing f3 ∈ H(U) such that
f2(z) = (z − w1) · · · (z − wm)f3(z) for every z ∈ U . Taking g = 1/f3 and substituting in the
expression for f1 via f2 we get (i), and then (ii) follows.
[153] [Argument principle: counting formula for zeroes and poles of a meromorphic function] Let
U ⊂ C be open and connected, and f be meromorphic in U with only finitely many zeroes and
poles in U . Let z1, . . . , zk be the zeroes of f in U and w1, . . . , wm be the poles of f in U , counted
with multiplicity. Let α be a piecewise smooth closed path in U \{z1, . . . , zk, w1, . . . , wm} such that
n(α; p) = 0 for every p ∈ C \ U . Then, (i)1
2πi
∫α
f ′(z)
f(z)dz =
∑kj=1 n(α; zj)−
∑ml=1 n(α;wl).
(ii) More generally,1
2πi
∫α
h(z)f ′(z)
f(z)dz =
∑kj=1 n(α; zj)h(zj)−
∑ml=1 n(α;wl)h(wl) ∀h ∈ H(U).
Proof. We may see that (i) is a special case of (ii) by taking h ≡ 1. To prove (ii), first note by [152]
thath(z)f ′(z)
f(z)=∑k
j=1
h(z)
z − zj−∑m
l=1
h(z)
z − wl+h(z)g′(z)
g(z)for every z ∈ U \{z1, . . . , zk, w1, . . . , wm}.
Apply ‘1
2πi
∫α’ and observe that
1
2πi
∫α
h(z)
z − zjdz = n(α; zj)h(zj),
1
2πi
∫α
h(z)
z − wldz = n(α;wl)h(wl),
and1
2πi
∫α
h(z)g′(z)
g(z)dz = 0 by Cauchy’s general integral theorem [138] for h, hg′/g ∈ H(U).
Exercise-82: (i) Evaluate∫|z|=2
eπz
z +1
z
dz.
(ii) [Formula for f−1] Let B(a, r) ⊂ C, f ∈ H(B(a, r)) be injective, and W = f(B(a, r)). Note that
W is open and f−1 ∈ H(W ) by [135](i). We have that f−1(w) =1
2πi
∫|z−a|=t
zf ′(z)
f(z)− wdz for every
t ∈ (0, r) and every w ∈W with f−1(w) ∈ B(a, t).
[Hint : (i) Let f(z) = z2 + 1, which has two simple zeroes ±i in B(0, 2), and let h(z) = eπz/2.
Then by [153],∫|z|=2
eπz
z +1
z
dz =∫|z|=2
h(z)f ′(z)
f(z)dz = 2πi(h(i) + h(−i)) = −2πi. (ii) Fix t ∈
(0, r) and w ∈ W with z0 := f−1(w) ∈ B(a, t). Then z0 is the only zero of f0 ∈ H(B(a, r))
defined as f0(z) = f(z) − w, and z0 is a simple zero of f0 because f ′0(z0) = f ′(z0) = 0 as f is a
57
biholomorphism. Let h(z) = z and α be the natural parametrization of |z − a| = t. By [153], we
see1
2πi
∫|z−a|=t
zf ′(z)
f(z)− wdz =
1
2πi
∫α
h(z)f ′0(z)
f0(z)dz = n(α; z0)h(z0) = 1 · z0 = f−1(w).]
[154] [Rouche’s theorem - comparing the difference of the number of zeroes and poles] Let U ⊂ C
be open, f, g be meromorphic in U and B(a, r) ⊂ U . Suppose that
(i) f and g have no zeroes or poles on the circle |z − a| = r, and
(ii) |f(z)− g(z)| < |f(z)|+ |g(z)| whenever |z − a| = r.
Then Zf − Pf = Zg − Pg, where Zf , Zg are the number of zeroes and Pf , Pg are the number of
poles of f, g respectively in B(a, r), counted with multiplicity.
Proof. We may assume U is connected since B(a, r) is included in some connected component of
U . Observe that as B(a, r) ⊂ U is compact, the numbers Zf , Zg, Pf , Pg are finite by (i), Zeroes
theorem, and Exercise-78. We need to show (Zf −Pf )− (Zg −Pg) = 0. By [153], this is equivalent
to showing that∫|z−a|=r(f
′/f−g′/g) = 0. Note that f ′/f−g′/g =f ′g − fg′
fg=f ′g − fg′
(f/g)g2=
(f/g)′
(f/g),
and hence it suffices to show∫|z−a|=r
(f/g)′
(f/g)= 0. And for this, we will show that there is an open
neighborhood V of the circle ∂B(a, r) such that(f/g)′
(f/g)has a primitive in V .
From (ii), we have that |f(z)g(z)
− 1| < |f(z)g(z)
| + 1 for every z ∈ ∂B(a, r), which implies that
f/g(∂B(a, r)) ⊂ C \ (−∞, 0]. By (i) and the compactness of ∂B(a, r), we may choose an ε > 0
such that the open annulus V := A(a, r − ε, r + ε) = {z ∈ C : r − ε < |z − a| < r + ε} satisfies
the following two properties: f and g have no zeroes or poles in V and f/g(V ) ⊂ C \ (−∞, 0]. Let
h : C \ (−∞, 0] → C be a branch of logarithm. Then (h ◦ (f/g))′ = (h′ ◦ (f/g)) · (f/g)′ = (f/g)′
(f/g)by
chain rule and the fact that h′(z) = 1/z. Thus h ◦ (f/g) ∈ H(V ) is a primitive of(f/g)′
(f/g)in V .
Remark: There is a more general version of Rouche’s theorem where the integral over the circle
|z− a| = r is replaced by the integral over a piecewise smooth closed path α; of course, ‘Zf −Pf =
Zg − Pg’ has to be modified to an expression in which the winding numbers also appear.
[155] [Hurwitz theorem - stability of the number of zeroes] Let U ⊂ C be open, and (fn) be a
sequence in H(U) converging uniformly on compact subsets of U to a function f : U → C. If
B(a, r) ⊂ U and f does not vanish on the circle |z − a| = r, then there is n0 ∈ N such that f and
fn have the same number of zeroes in B(a, r), counted with multiplicity.
Proof. By Weierstrass’ convergence theorem, f ∈ H(U). Let δ = min{|f(z)| : |z − a| = r}. Then
δ > 0 by hypothesis and the compactness of the circle |z − a| = r. Using the uniform convergence
58
of (fn) to f on the circle |z− a| = r, we may choose n0 ∈ N such that sup{|f(z)− fn(z)| : |z− a| =
r} < δ/2 for every n ≥ n0. Then for each n ≥ n0, fn does not vanish on the circle |z − a| = r and
|f(z)−fn(z)| < δ/2 ≤ |f(z)| ≤ |f(z)|+|fn(z)| whenever |z−a| = r. Hence by [154], the holomorphic
functions f and fn have the same number of zeroes in B(a, r), counted with multiplicity.
[156] Let U ⊂ C be a connected open set, and (fn) be a sequence in H(U) converging to a function
f : U → C uniformly on compact subsets of U . Then,
(i) If fn’s are non-vanishing in U , then either f is non-vanishing in U or f ≡ 0.
(ii) If fn’s are injective, then either f is injective or f is constant.
Proof. By Weierstrass convergence theorem, f ∈ H(U).
(i) Suppose f is not identically zero but f(a) = 0 for some a ∈ U . Since f−1(0) is discrete and
closed in U by [130], there is r > 0 such that B(a, r) ⊂ U and B(a, r) ∩ f−1(0) = {a}. Then by
Hurwitz theorem [155], there should exist n0 ∈ N such that fn has at least one zero in B(a, r) for
every n ≥ n0, a contradiction.
(ii) Fix a ∈ U , let g(z) = f(z) − f(a) and gn(z) = fn(z) − fn(a). Applying part (i) to gn, g on
U \ {a}, we deduce that either f(z) = f(a) for every z ∈ U \ {a} or f ≡ f(a).
Exercise-83: (i) Prove Fundamental theorem of Algebra using Rouche’s theorem.
(ii) Let n ∈ N and p(z) = zn +∑n−1
j=0 ajzj . Then there is z ∈ C with |z| = 1 and |p(z)| > 1.
[Hint : (i) Let p(z) =∑n
j=0 ajzj , where n ∈ N and an = 0. If we put q(z) = anz
n, then
limz→∞p(z)
q(z)= 1. Choose r > 0 large enough so that |p(z)
q(z)− 1| < 1 whenever |z| = r. Then
for z ∈ C with |z| = r, |p(z) − q(z)| < |q(z)| ≤ |p(z)| + |q(z)|. Hence by Rouche’s theorem,
the number of zeroes of p in B(0, r) is equal to that of q (counted with multiplicity), namely
n. (ii) Suppose |p(z)| ≤ 1 for every z ∈ ∂D. Let f(z) = zn and g(z) = zn − p(z). Then
|f(z) − g(z)| = |p(z)| ≤ 1 = |f(z)| ≤ |f(z)| + |g(z)| for every z ∈ ∂D. By Rouche’s theorem, the
number of zeroes of g in D (counted with multiplicity) should be equal to that of f in D, namely
n, which leads to a contradiction because deg(g) ≤ n− 1.]
Exercise-84: (i) Let a ∈ (1,∞) and f(z) = e−z + z − a. Find the number of zeroes (counted with
multiplicity) of f in U := {z ∈ C : Re(z) ≥ 0}.
(ii) Find the number of zeroes (counted with multiplicity) of f(z) := z4−6z+3 in U := A(0, 1, 2) =
{z ∈ Z : 1 < |z| < 2}.
[Hint : (i) If Re(z) = 0, then |e−z| = 1 and |z− a| > 1 so that f(z) = 0. If Re(z) > 0 and f(z) = 0,
then |e−z| < 1 and hence |z−a| < 1. That is, {z ∈ U : f(z) = 0} ⊂ B(a, 1). Let g(z) = z−a. Then
59
f, g ∈ H(U) are non-vanishing on |z − a| = r and |f(z) − g(z)| = |e−z| < 1 = |z − a| = |g(z)| ≤
|f(z)|+ |g(z)| whenever |z−a| = 1. By [154], f has exactly one zero in B(a, 1) and hence in U . (ii)
By considering the modulus of the three terms, note that f does not vanish on the circles |z| = 1
and |z| = 2. Let g(z) = z4 and h(z) = −6z. Then |f(z) − g(z)| = | − 6z + 3| ≤ 15 < 16 = |g(z)|
when |z| = 2 and |f(z)−h(z)| = |z4+3| ≤ 4 < 6 = |h(z)| when |z| = 1. By [154], f has four zeroes
in B(0, 2) and one zero in B(0, 1). Hence f has 4− 1 = 3 zeroes in U , counted with multiplicity.]
13 The automorphism group
Definition: For an open set U ⊂ C, let Aut(U) = {f : U → U : f is biholomorphic}.
We are interested in determining Aut(U) for some familiar open sets U . We need:
[157] [Injectivity and singularity] Let U ⊂ C be open, A ⊂ U be discrete and closed in U , and let
f ∈ H(U \A) be injective. Then,
(i) No point of A is an essential singularity of f .
(ii) If a ∈ A is a pole of f , then it is a simple pole.
(iii) If each a ∈ A is a removable singularity of f , then the holomorphic extension g ∈ H(U) of f
is also injective.
Proof. (i) Let a ∈ A and choose an open ball B ⊂ U centered at a such that A ∩ B = {a} and
the open set V := U \ (A ∪B) is nonempty. By Open mapping theorem and injectivity, f(V ) is a
nonempty open set in C disjoint with f(B \ {a}), and therefore f(B \ {a}) cannot be dense in C.
By Casorati-Weierstrass theorem, a cannot be an essential singularity of f .
(ii) We may assume U is connected, and then U \A is also open and connected. Since f is injective,
f−1(0) is discrete and closed in U \A by [130]. If a ∈ A is a pole of f , choose an open ball B ⊂ U
centered at a with A ∩ B = {a} and f(z) = 0 for every z ∈ B \ {a}. Then g : B → C defined as
g(a) = 0 and g(z) = 1/f(z) for z = a is holomorphic in B by [145](iii) and Riemann continuation
theorem. Moreover, g is injective in B, and hence g′ is non-vanishing in B by [135]. In particular,
g′(a) = 0, and thus a is a simple zero of g. Therefore, a is a simple pole of f .
(iii) Let z1, z2 ∈ U be distinct and suppose g(z1) = g(z2) =: w. Choose disjoint open balls
B1, B2 ⊂ U centered at z1, z2 respectively. By Open mapping theorem, W := g(B1) ∩ g(B2) is
an open neighborhood of w, and in particular, W is uncountable. Since A must be countable by
hypothesis, it follows that there are b1 ∈ B1 \ A and b2 ∈ B2 \ A such that g(b1) = g(b2), i.e.,
60
f(b1) = f(b2). This contradicts the injectivity of f as B1 ∩B2 = ∅.
Exercise-85: If f ∈ H(C) is injective, then f is a polynomial of degree one. [Hint : f(1/z) is injective
in C \ {0}, and hence 0 cannot be an essential singularity of f(1/z) by [157](i). Then f must be a
polynomial by [147], and clearly f is non-constant. To see deg(f) = 1, either note by [135] that f ′
is non-vanishing and hence a constant, or note by [157] that 0 must be a simple pole of f(1/z).]
[158] (i) Aut(C) = {z 7→ az + b : a ∈ C \ {0} and b ∈ C}.
(ii) Aut(C \ {0}) = {z 7→ az : a ∈ C \ {0}} ∪ {z 7→ a/z : a ∈ C \ {0}}.
Proof. (i) This follows by Exercise-85 and the observation that every polynomial of degree one
belongs to Aut(C).
(ii) Enough to show ⊂ since the other inclusion is clear. Consider f ∈ Aut(C \ {0}). By [157], 0
is either a removable singularity or a simple pole of f . In the first case, applying Exercise-85 to
the holomorphic extension of f to C and using the fact that f(C \ {0}) = C \ {0}, we deduce that
f(z) = az for some a ∈ C \ {0}. If 0 is a simple pole of f , then 0 is a removable singularity of 1/f ,
and hence by applying the previous case to 1/f , we may deduce that 1/f(z) = cz, or equivalently
f(z) = a/z for some a = 1/c ∈ C \ {0}.
Exercise-86: Aut(C \ {0, 1}) consists of the following six functions: z 7→ z, z 7→ 1/z, z 7→ 1 − z,
z 7→ 1/(1 − z), z 7→ z/(1 − z), z 7→ (z − 1)/z. [Hint : If f ∈ Aut(C \ {0, 1}), then each of 0,1 is
either a removable singularity or a simple pole of f .]
To determine Aut(D) for D = {z ∈ C : |z| < 1}, we will go through some preparation, and we
will also make a few related observations.
Definition: For a ∈ D, let ϕa be the Mobius map given by ϕa(z) =z − a
1− az.
Exercise-87: Fix a ∈ D, and let ϕa be as defined above. Then,
(i) ϕa is holomorphic in C \ {1/a}, and in particular in D.
(ii) For z ∈ ∂D, |ϕa(z)| =|z − a|
|z − a||z|= 1 and ϕa(a) = 0. So, ϕa(∂D) = ∂D by [114] and ϕa(D) ⊂ D.
(iii) ϕa(0) = −a and ϕa ∈ Aut(D) with ϕ−1a = ϕ−a.
(iv) ϕ′a(0) = 1− |a|2 and ϕ′a(a) = 1/(1− |a|2).
[159] (i) If b1, b2 ∈ D, then there is a Mobius map T ∈ Aut(D) with T (b1) = b2.
(ii) Let zj ∈ B(aj , Rj) ⊂ C for j = 1, 2. Then there is a biholomorphism f : B(a1, R1) → B(a2, R2)
with f(z1) = z2.
61
Proof. (i) Take T = ϕ−1b2
◦ ϕb1 and use Exercise-87.
(ii) Let gj : B(aj , Rj) → D be a biholomorphism and let bj = gj(zj) for j = 1, 2. By (i), there is
T ∈ Aut(D) with T (b1) = b2. Take f = g−12 ◦ T ◦ g1.
[160] [Schwarz lemma] Let f ∈ H(D) be with f(0) = 0 and |f(z)| ≤ 1 for every z ∈ D. Then,
(i) |f(z)| ≤ |z| for every z ∈ D and |f ′(0)| ≤ 1.
(ii) If either |f(z)| = |z| for some z ∈ D \ {0} or if |f ′(0)| = 1, then f is a rotation around the
origin, i.e., there is c ∈ ∂D such that f(z) = cz for every z ∈ D.
Proof. Note that limz→0f(z)
z=f(z)− f(0)
z − 0= f ′(0) since f(0) = 0. Hence g : D → C defined
as g(0) = f ′(0) and g(z) = f(z)/z for z = 0 is holomorphic in D by [127]. Fix r ∈ (0, 1). By
Maximum modulus principle, sup{|g(z)| : |z| ≤ r} = sup{|g(z)| : |z| = r} ≤ 1/r. Letting r → 1,
we get sup{|g(z)| : z ∈ D} ≤ 1, which implies (i). If either |f(z)| = |z| for some z ∈ D \ {0} or
if |f ′(0)| = 1, then it means |g| attains its maximum value 1 in D. Then by Maximum modulus
principle, g ≡ c for some c ∈ ∂D. Then f(z) = cz for every z ∈ D, as asserted in (ii).
Exercise-88: Let f ∈ H(D) be with f(0) = 0 and |f(z)| ≤ 1 for every z ∈ D. Suppose |f ′(a)| < 1
for some a ∈ D. Then for each r ∈ (0, 1), there is c ∈ [0, 1) such that |f(z)| ≤ c|z| whenever
|z| ≤ r. Consequently, for any z ∈ D, the sequence (zn) defined as z0 = z and zn = f(zn−1)
converges to 0. [Hint : Since |f ′(a)| < 1, f is not a rotation, and hence by [160](ii) we must have
|f(z)| < |z| for every z ∈ D \ {0} and |f ′(0)| < 1. Therefore, if g is as in the proof of [160], and
c := max{|g(z)| : |z| ≤ r}, then c < 1 and |f(z)| ≤ c|z| whenever |z| ≤ r.]
Notation: For c ∈ ∂D, let ψc : D → D be ψc(z) = cz, and Aut0(D) = {f ∈ Aut(D) : f(0) = 0}.
[161] (i) Aut0(D) = {ψc : c ∈ ∂D} = {all rotations around the origin} ∼= (∂D, ·).
(ii) Aut(D) = {ψc ◦ ϕa : c ∈ ∂D and a ∈ D}.
(iii) If f ∈ Aut(D) has at least two fixed points in D, then f = I.
(iv) Let Γ be a subgroup of Aut(D) containing Aut0(D). If Γ acts transitively on D (i.e., if for any
two points a, b ∈ D, there is g ∈ Γ with g(a) = b), then Γ = Aut(D).
Proof. (i) It suffices to prove ‘⊂’. Consider f ∈ Aut0(D). Applying [160](i) to both f and f−1, we
get |f(z)| = |z| for every z ∈ D. Then f must be a rotation by part (ii) of [160].
(ii) It suffices to prove ‘⊂’. Consider f ∈ Aut(D) and let a = f−1(0). Then f ◦ ϕ−1a (0) = 0 and
hence f ◦ ϕ−1a ∈ Aut0(D). By (i), there is c ∈ ∂D such that f ◦ ϕ−1
a = ψc and hence f = ψc ◦ ϕa.
62
(iii) Let a, b ∈ D be two distinct fixed points of f . Let g = ϕa ◦ f ◦ ϕ−1a and w = ϕa(b). Then 0, w
are two distinct fixed points of g in D. Since g ∈ Aut0(D), g is a rotation by part (i). Since the
rotation g has a non-zero fixed point, we must have g = I, which implies f = I.
(iv) Consider f ∈ Aut(D) and choose g ∈ Γ with g(f(0)) = 0. Then g ◦ f ∈ Aut0(D) ⊂ Γ and hence
f = g−1 ◦ (g ◦ f) ∈ Γ since Γ is a subgroup.
[162] [Pick’s lemma] Let f : D → D be holomorphic. Then,
(i) |f ′(z)| ≤ 1− |f(z)|2
1− |z|2for every z ∈ D.
(ii) f ∈ Aut(D) ⇔ equality holds in (i) for every z ∈ D ⇔ equality holds in (i) for some z ∈ D.
Proof. (i) Fix a ∈ D and let b = f(a). Let h = ϕb ◦ f ◦ ϕ−1a = ϕb ◦ f ◦ ϕ−a. Then h(D) ⊂ D
and h(0) = 0. By Schwarz lemma, 1 ≥ |h′(0)| = |ϕ′b(b)f ′(a)ϕ′−a(0)|. By Exercise-87(iv), ϕ′b(b) =
1/(1− |b|2) and ϕ′−a(0) = 1− |a|2. Hence 1− |b|2
1− |a|2≥ |f ′(a)|.
(ii) If f ∈ Aut(D), then for any a ∈ D and h defined above, we have h ∈ Aut0(D), and hence h is
a rotation by [161]. Hence 1 = |h′(0)| = |ϕ′b(b)f ′(a)ϕ′−a(0)|, from which it follows that1− |b|2
1− |a|2=
|f ′(a)|. Conversely, if |f ′(a)| = 1− |b|2
1− |a|2for some a ∈ D and b := f(a), then h(0) = 0 and |h′(0)| = 1,
which implies by [160](ii) that h is a rotation and hence f = ϕ−1b ◦ h ◦ ϕa ∈ Aut(D).
Exercise-89: (i) Let a ∈ C, R > 0 and f ∈ H(B(a,R)). Assume a is a zero of order m of f , and
M := sup{|f(z)| : z ∈ B(a,R)} <∞. Then |f(z)| ≤ M
Rm|z − a|m for every z ∈ B(a,R).
(ii) Let f : D → D be holomorphic and 0 be a zero of order m of f . Then |f(z)| ≤ |z|m for every
z ∈ D. Consequently, if m ≥ 2, then 0 is the only fixed point of f (as well as |f |).
(iii) If f : D → D is holomorphic and a ∈ D is a zero of order m of f , then |f(0)| ≤ |a|m.
[Hint : (i) Write f(z) = (z − a)mg(z). If 0 < r < R, then by Maximum modulus principle,
sup{|g(z)| : |z − a| ≤ r} = sup{|g(z)| : |z − a| = r} ≤ M/rm. Letting r → R, we get |g| ≤ M/Rm.
(iii) As 0 is a zero of order m for f ◦ ϕ−a, we have |f(0)| = |f ◦ ϕ−a(−a)| ≤ | − a|m = |a|m by (ii).]
Exercise-90: For each property given below, decide whether there a holomorphic function f : D → D
satisfying that property: (i) |f(0)| ≥ 3/4 and |f ′(0)| ≥ 2/3.
(ii) 0 is a zero of order three of f and |f(2/3)| ≥ 1/3.
(iii) 1/2 is a zero of order two of f and |f(0)| ≥ 1/3.
[Hint : The answer is ‘NO’ for all. For (i), note that the inequality in Pick’s lemma fails at z = 0.
For (ii) and (iii), see Exercise-89(ii) and Exercise-89(iii).]
63
Exercise-91: Let U ⊂ C be open, A ⊂ U be finite, and AutA(U) = {f ∈ Aut(U) : f(A) ⊂ A}.
Define η : AutA(U) → Aut(U \ A) as η(f) = f |U\A. Then η is an injective group homomorphism.
In general, η may not be surjective. [Hint : For injectivity, use Zeroes theorem or just continuity.
To see η may not be surjective, consider U = C, A = {0} and f(z) = 1/z.]
[163] (i) If A ⊂ D is finite, then AutA(D) is group-isomorphic to Aut(D \A) via f 7→ f |D\A.
(ii) Aut(D \ {0}) ∼= Aut0(D) = {all rotations around the origin} ∼= (∂D, ·).
(iii) If A ⊂ D is finite with n := |A| ≥ 2, then Aut(D \ A) is isomorphic to a subgroup of the
permutation group Sn, and in particular |Aut(D \A)| ≤ n! <∞.
Proof. (i) By Exercise-91, η : AutA(D) → Aut(D \A) defined as η(f) = f |D\A is an injective group
homomorphism. To show η is surjective, consider f ∈ Aut(D \ A). Since f is bounded, each point
of A is a removable singularity of f , and hence f has a holomorphic extension f : D → C. Since A
is a finite set, f(D) ⊂ D by continuity, and then f(D) ⊂ D by Open mapping theorem. Similarly,
if g : D → C is the holomorphic extension of f−1, then g(D) ⊂ D. Since f ◦ g(z) = z = g ◦ f(z) for
every z ∈ D \ A, it follows by Zeroes theorem that f ◦ g(z) = z = g ◦ f(z) for every z ∈ D. Hence
f ∈ AutA(D), and clearly η(f) = f .
(ii) This follows from part (i) and [161](i).
(iii) By (i), AutA(D) ∼= Aut(D \ A), and so we may work with AutA(D). Write A = {a1, . . . , an},
and define λ : AutA(D) → Sn as λ(f) = σ, where σ ∈ Sn is the unique permutation such that
f(aj) = aσ(j) for 1 ≤ j ≤ n. Clearly, λ is a group homomorphism. If f ∈ ker(λ), then every a ∈ A
is a fixed point of f , and therefore f = I by [161](iii) since n = |A| ≥ 2. Hence λ is injective.
Exercise-92: (i) If A ⊂ D is with |A| = 2, then Aut(D \A) ∼= {0, 1}.
(ii) If A ⊂ D is with |A| = 4, then |Aut(D \A)| ≤ 12.
(iii) Find a finite set A ⊂ D such that Aut(D \A) = {I}.
[Hint : Use the fact AutA(D) ∼= Aut(D \A) for all. (i) Since Aut(D) acts transitively on D, we may
suppose A = {0, b} for some b ∈ D \ {0} after a conjugation. Now, |AutA(D)| ≤ 2 by [163](iii)
and −ϕb ∈ AutA(D) \ {I}. (ii) AutA(D) is isomorphic to a subgroup of S4, and |S4| = 24. By
[161](iii), we cannot have AutA(D) ∼= S4. (iii) Let p, q ∈ D \ {0} be distinct and A = {0, p, q}.
If f ∈ AutA(D) \ {I}, then f is a non-identity permutation of A. In view of [161](ii), one of
the following four relations must hold: p = −q; 2q = p + pq2; 2p = q + qp2; |p| = |q| and
p2 + q2 = pq(1 + |q|2). Therefore, if p, q are chosen such that none of these is true (for example, let
p = 1/2, q = 3/4), then we will get AutA(D) = {I}.]
64
Exercise-93: [Another description for Aut(D)] Aut(D) = {z 7→ pz + q
qz + p: p, q ∈ C and |p| > |q|} =
{z 7→ pz + q
qz + p: p, q ∈ C and |p|2 − |q|2 = 1}. [Hint : See [161](ii). If f = ψc ◦ ϕa ∈ Aut(D), where
c ∈ ∂D and a ∈ D, then by choosing p ∈ ∂D with p2 = c and q = −pa, we see |p| > |q| and f(z) =cz − ca
1− az=
p2z + pq
1 + (q/p)z=pz + q
qz + psince pp = 1. Conversely, if f(z) =
pz + q
qz + p= (p/p)
z + q/p
1 + (q/p)z,
where |p| > |q|, then by taking c = p/p and a = −q/p, we see c ∈ ∂D, a ∈ D and f = ψc ◦ ϕa. ]
Remark: Let H = {z ∈ C : Im(z) > 0}. Note that h : H → D given by h(z) =z − i
z + iis a
biholomorphism (we choose h as a Mobius map with h(0) = −1, h(1) = −i, h(∞) = 1, and h(i) = 0)
with h−1(z) =iz + i
−z + 1. Therefore, Aut(H) ∼= Aut(D) ; f ∈ Aut(H) ⇔ h ◦ f ◦ h−1 ∈ Aut(D). The
matrix of the biholomorphism h : H → D is A :=
1 −i
1 i
(up to a non-zero scalar multiple),
and then A−1 =
1/2 1/2
i/2 −i/2
.
Exercise-94: Recall that SL(2,C) is the collection of all 2× 2 complex matrices with determinant
one (similarly SL(2,R)). Let Γ = {
p q
q p
: pq ∈ C and |p|2 − |q|2 = 1} ⊂ SL(2,C), and
A be the matrix above. Define F : SL(2,R) → SL(2,C) as F (B) = ABA−1. Then F is a
group homomorphism and F (SL(2,R)) = Γ. [Hint : If B =
a b
c d
∈ SL(2,R), then letting
p = (1/2)[(a+ d)+ i(b− c)] and q = (1/2)[(a− d)− i(b+ c)], we may see that ABA−1 =
p q
q p
and |p|2−|q|2 = 1. Conversely, if C :=
p q
q p
∈ Γ, then by taking a = Re(p+q), b = Im(p−q),
c = −Im(p+ q) and d = Re(p− q), we may see that B :=
a b
c d
∈ SL(2,R) and A−1CA = B,
or equivalently F (B) = C.]
[164] (i) Aut(H) = {z 7→ az + b
cz + d:
a b
c d
∈ SL(2,R)} and Aut(H) ∼= SL(2,R)/± I.
(ii) Aut(D) ∼= SL(2,R)/± I.
(iii) Any f ∈ Aut(H) with at least two fixed points in H must be the identity map.
(iv) Each of the following groups acts transitively on H: Aut(H), SL(2,R), SL(2,R)/± I.
Proof. (i) The first assertion follows by [161](ii), Exercise-94, and the Remark above Exercise-94.
For the second assertion note that two matrices B1, B2 ∈ SL(2,R) represent the same Mobius map
65
in Aut(H) iff B1 = −B2.
(ii) This is a corollary of (i) since Aut(D) ∼= Aut(H).
(iii) This follows from [161](iii) since Aut(D) ∼= Aut(H).
(iv) This follows from part (i) above and [161](iv) since Aut(D) ∼= Aut(H).
Remark: It is interesting to note that each of Aut(C), Aut(C \ {0}), Aut(D), Aut(D \ {0}), and
Aut(H) is a subgroup of the group of all Mobius maps.
Exercise-95: [An old question] Let U = {z ∈ C : 1 < |z| < 2}. Then no two of the following are
holomorphically equivalent: C, C \ {0}, D, D \ {0}, and U , [Hint : C and D are simply connected,
and the others are not. Liouville’s theorem and Exercise-52(iii) can be used to rule out three more
pairs: (C,D), (C \ {0},D \ {0}) and (C \ {0}, U). It remains to show there is no biholomorphism
f : D \ {0} → U . If f exists, then by the boundedness of U , 0 must be a removable singularity of
f ; and then the holomorphic extension f : D → C of f must satisfy f(0) /∈ U by [157](iii), which
implies f(D) cannot be open in C, a contradiction to Open mapping theorem.]
14 Harmonic functions
Definition: Let U ⊂ C ∼= R2 be open. A twice continuously differentiable function u : U → R is
said to be harmonic if the Laplace equation uxx + uyy = 0 is satisfied, where uxx =∂2u
∂x2.
[165] Let U ⊂ C be open and f = u+ iv ∈ H(U) (i.e, u = Re(f) and v = Im(f)). Then,
(i) u and v are harmonic functions.
(ii) Assume that either U is an open ball with center a = p + iq, or U = C and p+ iq ∈ C. Then∫ yq ux(x, t)dt−
∫ xp uy(s, q)ds = v(x, y)− v(p, q).
Proof. (i) Since f is infinitely often differentiable, so are u and v. Since ux = vy and uy = −vx by
Cauchy-Riemann equations and vxy = vyx (a property of C2-maps), we get uxx+uyy = vyx−vxy = 0.
Similarly, vxx + vyy = −uyx + uxy = 0.
(ii) Since ux = vy and −uy = vx, we get∫ yq ux(x, t)dt−
∫ xp uy(s, q)ds =
∫ yq vt(x, t)dt+
∫ xp vs(s, q)ds =
v(x, y)− v(x, q) + v(x, q)− v(p, q) = v(x, y)− v(p, q).
Definition: If U ⊂ C is open and f = u + iv ∈ H(U), then u and v are said to be harmonic
conjugates of each other. Note that if u and v are harmonic conjugates of each other and U is
connected, then a function v : U → R is a harmonic conjugate of u iff v − v is a constant on U .
66
[166] Let U ⊂ C be open and u : U → R be harmonic. Then,
(i) Assume that either U is an open ball or U = C. Then there is a harmonic function v : U → R
such that f := u+ iv ∈ H(U).
(ii) [Mean value property] If B(a, r) ⊂ U , then u(a) =1
2π
∫ 2π0 u(a+ reiθ)dθ.
(iii) [Maximum principle (no modulus here!)] If U is connected and b ∈ U is such that u(b) =
sup{u(z) : z ∈ U}. Then u ≡ u(b) in U .
Proof. (i) Let p + iq be the/a center of U . Using [165](ii) as a hint, define v : U → R as
v(x, y) =∫ yq ux(x, t)dt −
∫ xp uy(s, q)ds (we may ignore a constant). Then v is a C2-map since
u is. Differentiate v with respect to x and y, where we may differentiate under the integral sign.
Using the Fundamental theorem of calculus and the hypothesis that uxx = −uyy, we get vx(x, y) =∫ yq uxx(x, t)dt−uy(x, q) = −
∫ yq utt(x, t)dt−uy(x, q) = uy(x, q)−uy(x, y)−uy(x, q) = −uy(x, y) and
vy(x, y) = ux(x, y)−∫ xp
∂∂y (uy(s, b))ds = ux(x, y)−
∫ xp
∂ 0∂y ds = ux(x, y)− 0. Thus Cauchy-Riemann
equations hold, and therefore f := u+ iv ∈ H(U). Then v = Im(f) is harmonic by [165](i).
(ii) Let s > r be such that B(a, s) ⊂ U . Applying part (i) to the star region B(a, s), we may
find f ∈ H(B(a, s)) with Re(f) = u. By Cauchy’s integral formula, f(a) =1
2πi
∫|z−a|=r
f(z)
z − adz.
Writing z = a + reiθ, we havedz
z − a=
rieiθdθ
reiθ= idθ and hence f(a) =
1
2π
∫ 2π0 f(a + reiθ)dθ.
Equating the real parts of both sides we obtain u(a) =1
2π
∫ 2π0 u(a+ reiθ)dθ.
(iii) Let A = {z ∈ U : u(z) = u(b)}. Then A is a nonempty closed subset of U . It is enough
to show A is also open in U since U is connected. Consider a ∈ A, and choose s > 0 with
B(a, s) ⊂ U . Then for any r ∈ (0, s), we see by part (ii) that 0 = u(a) − 1
2π
∫ 2π0 u(a + reiθ)dθ =
1
2π
∫ 2π0 [u(a) − u(a + reiθ)]dθ. Since the integrand in the square bracket is continuous and ≥ 0 by
hypothesis, it must be identically zero, i.e., we must have u(z) = u(a) = u(b) whenever |z− a| = r.
Since r ∈ (0, s) is arbitrary, it follows that B(a, s) ⊂ A. This shows that A is open in U .
Example: Let u : R2 ∼= C → R be u(x, y) = 3x2y − y3. Then ux(x, y) = 6xy, uy(x, y) = 3x2 − 3y2,
and uxx(x, y) = 6y = −uyy(x, y) so that u is harmonic. Since the origin is a center of the star
region C, define v : R2 → R as in the proof of [166](i) as v(x, y) =∫ y0 ux(x, t)dt −
∫ x0 uy(s, 0)ds =∫ y
0 6xtdt−∫ xp 3s2ds = 3xy2 − x3, which is a harmonic conjugate of u in C and f := u+ iv ∈ H(C).
Since z3 = (x+ iy)3 = (x3 − 3xy2) + i(3x2y − y3), we may see that f(z) = −iz3.
Exercise-96: Let U ⊂ C be open, and f = u + iv ∈ H(U). Let ∇u = (ux, uy) = ux + iuy be the
gradient vector of u in R2 ∼= C). Then, (i) |∇u| = |f ′| = |∇v|.
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(ii) ∇v = i∇u so that ⟨∇u,∇v⟩ = 0.
[Hint : Use Cauchy-Riemann equations. (i) |∇u|2 = u2x+u2y = v2y+(−vx)2 = v2y+v
2x = det(Jf (z)) =
|f ′|2. (ii) i∇u = i(ux + iuy) = iux − uy = ivy + vx = ∇v.]
Exercise-97: (i) The Laplace equation u2xx+u2yy = 0 becomes urr+(1/r)ur+(1/r2)uθθ = 0 in polar
coordinates.
(ii) If U = C \ {0}, then u : U → C defined as u(z) = log |z| is harmonic, but there does not exist
any harmonic v : U → R with u+ iv ∈ H(U) (contrast this with [166](i)).
[Hint : (i) Write x = r cos θ and y = r sin θ. Then ur = uxxr + uyyr = ux cos θ + uy sin θ, etc.,
and hence we may show urr + (1/r)ur + (1/r2)uθθ = uxx + uyy. Or start with the right hand side
and apply a change of variable (this may require more work). (ii) Since u(r, θ) = log r, we have
ur = 1/r, urr = −1/r2 and uθθ = 0. Hence we may see u is harmonic by part (i) by checking that
urr + (1/r)ur + (1/r2)uθθ = 0. If there is a harmonic v : U → R with f = u+ iv ∈ H(U), derive a
contradiction as follows. Let g : C \ (−∞, 0] → C be g(z) = log |z|+ i arg(z) = u(z) + i arg(z), the
principal branch of logarithm. Since the purely imaginary function f − g ∈ H(C \ (−∞, 0]), there
is c ∈ R such that v(z) = arg(z) + c for every z ∈ C \ (−∞, 0]. But then v cannot be continuous at
any point of (−∞, 0), a contradiction.]
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