This document provides lecture notes on complex analysis covering four units of content: 1) The index of a close curve, Cauchy's theorem, and entire functions. 2) Counting zeroes, meromorphic functions, and maximum principle. 3) Spaces of continuous and analytic functions, and behavior of functions. 4) Comparison of entire functions, analytic continuation, and harmonic functions. It also provides definitions and theorems regarding integrals over rectifiable curves, winding numbers, and Cauchy's theorem. Exercises and proofs are included.

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COMPLEX ANALYSIS LECTURE NOTES
PRAKASH A. DABHI
1. Content to be covered
Unit I
The index of a close curve, behavior of the index on the components, different versions of
Cauchy’s theorem and Cauchy’s integral formula, Morera’s theorem, analogy between entire
function and polynomials, open mapping theorem.
Unit II
Counting zeroes, Meromorphic functions, the argument principle, Rouche’s theorem and
its application, maximum principle, Schwarz lemma and its application, convex functions,
Hadmard’s theorem.
Unit III
Spaces of continuous functions C(G, Ω), topology of uniform convergence on compact sets,
space of analytic functions, Arzela-Ascoli theorem, Montel’s theorem, Hurwitz’s theorem,
Riemann mapping theorem. Behavior of the function and Riemann’s theorem on removable
singularity, Casorati-Weierstrass theorem.
Unit IV
Comparison of entire functions and polynomials with respect to singularity, analytic continu-
ation, Poisson’s integral formula on a circle, Luca’s theorem, Parseval’s identity, Weierstrass
factorization theorem, Genus and order of entire functions, Walli’s formula, Jensen’s inequal-
ity, Poisson-Jenson’s inequality, Runge’s theorem, harmonic functions.
2. Integral over rectifiable curves
Definition 1. A function γ : [a, b] → C is said to be of bounded variation if there is M > 0
such that for any partition P : a = t0 < t1 < · · · < tn = b of [a, b],
ν(γ; P) =
n
k=1
|γ(tk) − γ(tk−1)| ≤ M.
If γ is of bounded variation, then the total variation V (γ) of γ is defined by
V (γ) = sup{ν(γ; P) : P is a partition of [a, b]}.
Exercise 2. Let γ : [a, b] → C be a map. Show that γ is of bounded variation if and only if
both Re γ and Im γ are of bounded variation.
Definition 3. A continuous function γ : [a, b] → C is called a curve.
1

2. P
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2 P. A. DABHI
If γ : [a, b] → C is a curve, then γ(t) = x(t) + iy(t) for some continuous functions x, y :
[a, b] → R. We denote the trace of γ by {γ}, i.e., {γ} = {γ(t) : t ∈ [a, b]}. Since γ is a
continuous function on a compact set [a, b], the set {γ} is a compact subset of C. So that
the set G = C{γ} is open in C. Also, G has only one unbounded component. Since {γ} is
bounded, there is R > 0 such that {γ} ⊂ {z ∈ C : |z| ≤ R} and hence G ⊃ {z ∈ C : |z| > R}.
Definition 4. A curve γ : [a, b] → C is said to be rectifiable if γ is of bounded variation.
Definition 5. A curve γ : [a, b] → C is called piecewise smooth if there are points a = t0 <
t1 < · · · < tm = b such that γ is differentiable on each of (tk−1, tk).
Proposition 6. If γ : [a, b] → C is piecewise smooth, then γ is of bounded variation and
V (γ) =
b
a
|γ (t)|dt.
Theorem 7. Let γ : [a, b] → C be of bounded variation, and let f : [a, b] → C be continuous.
Then there is a complex number I such that for every > 0 there is δ > 0 such that when
P : a = t0 < t1 < · · · < tn = b is a partition of [a, b] with P = max{|tk − tk−1| : 1 ≤ k ≤
n} < δ, then
I −
n
k=1
f(τk)[γ(tk) − γ(tk−1)] <
for whatever choice of points τk, tk−1 ≤ τk ≤ tk.
The number I is called the integral of f with respect to γ over [a, b] and is designated by
I =
b
a
fdγ =
b
a
f(t)dγ(t).
Proposition 8. Let f, g : [a, b] → C be continuous, and let γ, σ : [a, b] → C be of bounded
variation. Let α, β ∈ C. Then
(1)
b
a
(αf + βg)dγ = α
b
a
fdγ + β
b
a
gdγ.
(2)
b
a
fd(αγ + βσ) = α
b
a
fdγ + β
b
a
fdσ.
Proposition 9. Let γ : [a, b] → C be piecewise smooth, and let f : [a, b] → C be continuous.
Then
b
a
fdγ =
b
a
f(t)γ (t)dt.
If γ : [a, b] → C is a rectifiable path with γ ⊂ E ⊂ C and f : E → C is a continuous function,
then f ◦γ is a continuous function on [a, b]. With this in mind the following definition makes
sense.
Definition 10. If γ : [a, b] → C is a rectifiable path and f is a function defined and
continuous on the trace of γ, then the (line) integral of f along γ is
b
a
f(γ(t))dγ(t).
This line integral is also denoted by γ
f = γ
f(z)dz.
Proposition 11. Let γ be a rectifiable curve and suppose that f is a function continuous on
{γ}. Then

3. P
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COMPLEX ANALYSIS LECTURE NOTES 3
(1) γ
f = − −γ
f;
(2) | γ
f| ≤ γ
|f||dz| ≤ V (γ) sup[|f(z)| : z ∈ {γ}];
(3) If c ∈ C, then γ
f(z)dz = γ+c
f(z − c)dz.
Lemma 12. If G is an open set in C, γ : [a, b] → G is a rectifiable path, and f : G → C
is continuous, then for every > 0 there is a polygonal path Γ in G such that Γ(a) = γ(a),
Γ(b) = γ(b), and | Γ
f − γ
f| < .
Examples 13.
(1) Let γ : [0, 2π] → C be γ(t) = eint
, where n ∈ N. Then
γ
1
z
dz =
2π
0
1
eint
ineint
dt = 2nπi.
(2) Let a ∈ C, R > 0, and let n ∈ N. Let γ : [0, 2π] → C be γ(t) = a + Reint
. Then
γ
1
z − a
dz =
2π
0
1
Reint
ineint
dt =
2nπi
R
.
(3) Let n, m ∈ N. Let γ : [0, 2π] → C be γ(t) = eint
. Then
γ
zm
dz =
2π
0
ei(mn)t
ineint
dt = 0.
(4) Let a, b ∈ C, n ∈ N, and let γ : [0, 1] → C be γ(t) = tb + (1 − t)a. Then
γ
zn
dz =
bn+1
− an+1
n + 1
.
3. Winding number
Proposition 14. Let γ : [0, 1] → C be a closed rectifiable curve, and let a /∈ {γ}. Then
1
2πi γ
dz
z−a
is an integer.
Proof. Let Γ : [0, 1] → C be any closed polygonal path in the open set C{a}. Then Γ is
piecewise smooth and hence it is rectifiable. Define g : [0, 1] → C by g(t) =
t
0
Γ (s)
Γ(s)−a
ds.
Then g(0) = 0, g(1) =
1
0
Γ (s)
Γ(s)−a
ds = Γ
dz
z−a
. We prove that 1
2πi
g(1) is an integer. Observe
that g (t) = Γ (s)
Γ(s)−a
. Now
d
dt
[(Γ(t) − a)e−g(t)
] = Γ (t)e−g(t)
+ (Γ(t) − a)(−e−g(t)
g (t))
= Γ (t)e−g(t)
− Γ (t)e−g(t)
= 0.
Therefore Γ(t)−a)e−g(t)
is a constant map on [0, 1]. It gives (Γ(0)−a)e−g(0)
= (Γ(1)−a)e−g(1)
.
Since Γ is a closed curve, Γ(0) = Γ(1). Also since g(0) = 0, we have eg(1)
= 1. Therefore
g(1) = 2kπi for some k ∈ Z or 1
2πi
g(1) is an integer.
Now suppose that α = 1
2πi γ
dz
z−a
is not an integer. Then there is > 0 such that B(α; )∩Z =
∅. By Lemma 12 there is a polygonal path Γ : [0, 1] → C{a} such that Γ(0) = γ(0),
Γ(1) = γ(1) and | 1
2πi Γ
dz
z−a
− 1
2πi γ
dz
z−a
| < . It means that | 1
2πi Γ
dz
z−a
− α| < . This

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4 P. A. DABHI
shows that the integer 1
2πi Γ
dz
z−a
∈ B(α; ). This is a contradiction. Hence 1
2πi γ
dz
z−a
is an
integer.
Definition 15. If γ is a closed rectifiable curve in C, then for a /∈ {γ} the integer
n(γ; a) =
1
2πi γ
dz
z − a
is called the index of γ with respect to the point a. It is also sometimes called the winding
number of γ around a.
Recall that if γ : [0, 1] → C is a curve, −γ or γ−1
is the curve defined by (−γ)(t) = γ(1 − t)
(this is actually a reparametrization of the original definition). Also if γ and σ are curves
defined on [0, 1] with γ(1) = σ(0), then γ + σ is the curve (γ + σ)(t) = γ(2t) if 0 < t ≤ 1
2
and (γ + σ)(t) = σ(2t − l) if 1
2
< t < 1.
Lemma 16. If γ and σ are closed rectifiable curves having the same initial points, then
(1) n(γ; a) = −n(−γ; a) for every a /∈ {γ};
(2) n(γ + σ; a) = n(γ; a) + n(σ; a) for every a /∈ {γ} ∪ {σ} .
Proof. Since {γ} = {−γ}, a /∈ {−γ}. now
n(−γ, a) =
1
2πi −γ
dz
z − a
=
1
2πi
1
0
1
γ(1 − t) − a
dγ(1 − t)
=
1
2πi
0
1
1
γ(s) − a
dγ(s) = −
1
2πi
1
0
1
γ(s) − a
dγ(s)
= −
1
2πi γ
dz
z − a
= −n(γ, a).
Exercise 17.
(1) Let (X, d) be a metric space. Then
(a) Each point of X is contained in a component of X;
(b) Two components of X are either disjoint or identical.
(2) Let G be an open subset of C. Then
(a) Components of G are open;
(b) G has countable components.
Theorem 18. Let γ be a closed rectifiable curve in C. Then n(γ; a) is constant for a
belonging to a component of G = C{γ}. Also, n(γ; a) = 0 for a belonging to the unbounded
component of G.
Proof. Define f : G → C by f(a) = n(γ; a). We show that f is continuous. If this is
done, then it follows that f(D) is connected for each component D of G. But since f(G) is
contained in the set of integers, it follows that f(D) is a single point. That is, f is constant
on D.
We know that the component of G are open. Fix a ∈ G. Take any > 0. Let r = d(a, {γ}).
Take δ = min{r
2
, πr2
2V (γ)
}. Let b ∈ G with |a − b| < δ < r
2
. Since d(a, {γ}) = r and

5. P
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COMPLEX ANALYSIS LECTURE NOTES 5
|a − b| < δ ≤ r
2
, we have |z − b| ≥ r
2
for every z ∈ {γ}. Now
|f(b) − f(a)| =
1
2π γ
((z − a)−1
− (z − b)−1
)dz
=
1
2π γ
a − b
(z − a)(z − b)
dz
≤
|a − b|
2π γ
|dz|
|z − a||z − b|
≤
2|a − b|V (γ)
πr2
[ |z − a|, |z − b| ≥ r >
r
2
]
<
2δV (γ)
πr2
≤
2V (γ)
πr2
πr2
2V (γ)
= .
Hence f is continuous.
Let U be an unbounded component of G. Then there is R > 0 such that U ⊃ {z ∈ C : |z| >
R}. Take any > 0. Take a ∈ C such that |a| > R and |z − a| > V (γ)
2π
for every z ∈ {γ}.
Then
|n(γ; a)| =
1
2πi γ
dz
z − a
≤
1
2π γ
|dz|
|z − a|
≤
1
2π
2π
V (γ)
V (γ) = .
Hence n(γ, a) = 0 as > 0 is arbitrary. Since f is constant on U, we have n(γ; a) = 0 for
every a ∈ U.
4. Cauchy’s Theorems
Theorem 19. Let γ be a rectifiable curve. Let ϕ be a function defined and continuous on
{γ}. For each m ∈ N, let Fm(z) = γ
ϕ(w)(z − w)−m
dw, z /∈ {γ}. Then each Fm is analytic
in C{γ} and that Fm(z) = mFm+1(z) for every z ∈ C{γ}.
Proof. Since ϕ is continuous on a compact set {γ}, it is bounded. So, there is M > 0 such
that |ϕ(w)| ≤ M for every w ∈ {γ}. First we show that Fm is continuous. Take any > 0
and a ∈ C{γ}. Let r = d(a, {γ}). Take δ = min{r
2
, rm+1
mM2m+1V (γ)
}. Let z ∈ G = C{γ} with
|z − a| < δ. Since d(a, {γ}) = r and |z − a| < δ ≤ r
2
, we have |z − w| ≥ r
2
for every w ∈ {γ}.
Now
|Fm(z) − Fm(a)| =
γ
ϕ(w)
(w − z)m
dw −
γ
ϕ(w)
(w − a)m
dw
=
γ
ϕ(w)[
1
(w − z)m
−
1
(w − a)m
]dw
=
γ
ϕ(w)(z − a)
m
k=1
1
(w − a)k(w − z)m+1−k
dw
< δV (γ)Mm
2m+1
rm+1
≤ .
Therefore Fm is continuous.

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6 P. A. DABHI
Take any z ∈ C{γ}. Then
Fm(z) − Fm(a)
z − a
=
m
k=1 γ
ϕ(w)
(w − a)k(w − z)m+1−k
dw
=
m
k=1 γ
ϕ(w)
(w−a)k
(w − z)m+1−k
dw.
Since a /∈ {γ}, the functions ϕ(w)
(w−a)k are continuous on {γ} for every 1 ≤ k ≤ m. Hence the
functions γ
ϕ(w)
(w−a)k
(w−z)m+1−k dw are continuous on C{γ}. Applying limit z → a we get
lim
z→a
Fm(z) − Fm(a)
z − a
= m
γ
ϕ(w)
(w − a)m+1
dw = mFm+1(a).
Therefore Fm is differentiable at every a ∈ C{γ} and Fm(a) = mFm+1(a).
Theorem 20 (Cauchy’s Integral Formula (First Version)). Let G be an open subset of C,
and let f : G → C be analytic. If γ is a closed rectifiable curve in G such that n(γ; w) = 0
for every w ∈ CG, then for each a ∈ G{γ}
n(γ; a)f(a) =
1
2πi γ
f(z)
z − a
dz.
Proof. Define ϕ : G × G → C by ϕ(z, w) = f(z)−f(w)
z−w
if z = w and ϕ(z, z) = f (z). Then
ϕ is continuous and for each w ∈ G the map z → ϕ(z, w) is analytic. Let H = {w ∈ C :
n(γ; w) = 0}. Since n(γ; w) is an integer valued continuous function, the set H is open in
C. Also, H ∪ G = C by hypothesis. Define g : C → C by g(z) = γ
ϕ(z, w)dw if z ∈ G and
g(z) = γ
(w − z)−1
f(w)dw if z ∈ H. If z ∈ G ∩ H, then
g(z) =
γ
f(w) − f(z)
w − z
dw
=
γ
f(w)
w − z
dw − f(z)
γ
dw
w − z
=
γ
f(w)
w − z
dw − 2πif(z)n(γ; z)
=
γ
f(w)
w − z
dw = h(z). [ n(γ; z) = 0, z ∈ H]
Therefore the function g is well defined. It follows from last lemma with m = 1 that g is
analytic on both G and H and hence on C, i.e., g is an entire function. The open set H
contains the unbounded component of C{γ} and so there is R > 0 such that {z ∈ C : |z| >
R} ⊂ H. Since {γ} is compact, sup{|w − z|−1
: w ∈ {γ}} → 0 as |z| → ∞. Therefore for
given > 0 there is R1 > R > 0 such that |w−z|−1
< sup[|f(θ)|:θ∈{γ}]
V (γ) for every z ∈ C with
|z| > R1. Take any z ∈ C with |z| > R1. Then |g(z)| ≤ . It means that lim|z|→∞ g(z) = 0.
The above shows that g is bounded on {z ∈ C : |z| > R1}. Since g is continuous and
{z ∈ C : |z| ≤ R1} is compact, g is bounded on {z ∈ C : |z| ≤ R1}. So, g is a bounded

7. P
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COMPLEX ANALYSIS LECTURE NOTES 7
function. By Liouville’s theorem the function g is constant. Since lim|z|→∞ g(z) = 0, g is a
zero map. If a ∈ G{γ}, then
0 =
γ
f(z) − f(a)
z − a
dz =
γ
f(z)
z − a
dz − f(a)
γ
dz
z − a
=
γ
f(z)
z − a
dz − f(a)2πin(γ; a).
Therefore n(γ; a)f(a) = 1
2πi γ
f(z)
z−a
dz for every a ∈ G{γ}.
Corollary 21. Let G be an open subset of C, and let γ be a simple closed curve in G with
n(γ; w) = 0 for all w ∈ CG. If a lies in the interior of γ, then f(a) = 1
2πi γ
f(z)
z−a
dz.
Theorem 22 (Cauchy Integral Formula (Second Version)). Let G be an open subset of C,
and let f : G → C be analytic. Let γ1, γ2, . . . γm be closed rectifiable curves in G such that
n(γ1; w) + n(γ2, w) + · · · + n(γm, w) = 0 for all w ∈ CG. If a ∈ G ∪m
i=1 {γi}, then
f(a)
m
k=1
n(γk, a) =
m
k=1
1
2πi γk
f(z)
z − a
dz.
Proof. The proof is same as above. Define g(z) = n
k=1 γk
ϕ(z, w)dw if z ∈ G and g(z) =
n
k=1 γk
(w − z)−1
f(w)dw if z ∈ H = {z ∈ C : n(γ1; z) + n(γ2; z) + · · · + n(γk; z) = 0}.
Theorem 23 (Cauchy’s Theorem (First Version)). Let G be an open subset of C, and
let f : G → C be analytic. Let γ1, γ2, . . . , γm be closed rectifiable curves in G such that
n(γ1; w) + n(γ2; w) + · · · + n(γk; w) = 0 for every w ∈ CG. Then
m
k=1 γk
f = 0.
Proof. Fix any a ∈ G∪m
k=1 {γk}. Define g : G → C by g(z) = (z−a)f(z). Then g is analytic.
The result now follows from the Second version of the Cauchy’s Integral Formula.
Theorem 24 (Cauchy’s Integral Formula for Derivatives (Second Version)). Let G be an
open subset of C, and let f : G → C be analytic. Let γ1, γ2, . . . , γm be closed rectifiable
curves in G such that n(γ1; w) + n(γ2; w) + · · · + n(γk; w) = 0 for every w ∈ CG. If
a ∈ G ∪m
k=1 {γk}, then
f(k)
(a)
m
j=1
n(γj, a) = k!
m
j=1
1
2πi γj
f(z)
(z − a)k+1
dz.
Proof. Using Cauchy’s Integral Formula (Second Version), we get
f(a)
m
j=1
n(γj; a) =
1
2πi
m
j=1 γj
f(z)
z − a
dz, a ∈ G ∪m
k=1 {γk}
Fix j ∈ {a, 2, . . . , m}. Since the map f is continuous on {γj}, the map ϕj(a) = γj
f(z)
z−a
dz is
analytic on G{γj} and ϕj(a) = γj
f(z)
(z−a)2 dz for every a ∈ G{γj}. Applying induction we
obtain ϕ
(k)
j = k! γj
f(z)
(z−a)k+1 dz for a ∈ G{γj}. Hence differentiating k times the equation

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8 P. A. DABHI
f(a)
m
j=1
n(γj; a) = 1
2πi
m
j=1γj
f(z)
z−a
dz, we obtain f(k)
(a)
m
j=1
n(γj, a) = k!
m
j=1
1
2πi
γj
f(z)
(z−a)k+1 dz for
every a ∈ G ∪m
i=1 {γi}.
Corollary 25. Let γ be a closed rectifiable curve in C, and let a /∈ {γ}. Then γ
1
(z−a)n dz = 0
for every integer n ≥ 2.
Proof. Take f(z) = 1. Then f is an entire function. It follows from the Cauchy’s Integral
Formula for derivatives, we get 0 = f(k)
(a)n(γ; a) = k!
2πi γ
1
(z−a)k+1 dz for every k ∈ N. Hence
γ
1
(z−a)n dz = 0 for every integer n ≥ 2.
Exercise 26. Let G be a simply connected region. If f : G → C is analytic, then f has a
primitive in G.
Theorem 27 (Analytic extension of a the exponential solution). Let G be a simply connected
region and f : G → C be analytic with f(z) = 0 for every z ∈ G. Then there is an analytic
function g : G → C such that eg
= f.
Proof. Since f(z) = 0 for every z ∈ G, the function f
f
is analytic on G. So f
f
admits an
antiderivative on G. Let g be the antiderivative of f
f
on G. Then g = f
f
. Define h : G → C
by h(z) = eg(z)
. Then h is analytic and nonzero on G. Note that h = eg
g = hg . Now
d
dz
(f
h
) = hf −h f
h2 = hf −hg f
h2 = hf −hf
h2 = 0. Therefore f
h
is a constant function. Therefore
there is c ∈ C such that f = ch. Since f(z) = 0 for every z ∈ G, c = 0. There is a ∈ C
such that ea
= c. Thus f = ea
eg
= eg+a
. So the function g1 = g + a is the required analytic
function.
Example 28. Let n ∈ N. Evaluate γ
( z
z−1
)n
dz, where γ(t) = 1 + eit
, 0 ≤ t ≤ 2π.
Here n(γ; 1) = 1. Take f(z) = zn
. Then f is analytic within and on |z − 1| = 1. Note that
f(n−1)
(1) = n!. By Cauchy’s Integral Formula
n! = f(n−1)
(1) =
(n − 1)!
2πi γ
f(z)
(z − 1)n
dz =
(n − 1)!
2πi γ
z
z − 1
n
dz.
Thus γ
( z
z−1
)n
dz = 2πin.
Verify that σ
( z
z−1
)n
dz = 4πin, where σ(t) = 1 + e2it
, 0 ≤ t ≤ 2π.
5. Some results are of independent interest
Theorem 29 (Poisson Integral Formula for a circle). Let f be analytic within and on a
circle |z| = R. If 0 ≤ r < R, then
f(reiθ
) =
1
2π
2π
0
(R2
− r2
)f(Reiϕ
)
R2 − 2Rr cos(θ − ϕ) + r2
dϕ.
The map P(r, θ) = (R2−r2)
R2−2Rr cos(θ−ϕ)+r2 is known as Poisson kernel.
Proof. If r = 0, then 1
2π
2π
0
(R2−r2)f(Reiϕ)
R2−2Rr cos(θ−ϕ)+r2 dϕ = 1
2π
2π
0
f(Reiϕ
)dϕ = 1
2πi |z|=R
f(z)
z
dz =
f(0). Let 0 = a = reiθ
∈ B(0; R). Then 0 < r < R. By using Cauchy’s Integral Formula we

9. P
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COMPLEX ANALYSIS LECTURE NOTES 9
have f(a) = 1
2πi |z|=R
f(z)
z−a
dz. Since |R2
a
| > R, the function f(z)
z−R2
a
is analytic within and on
|z| = R. By Cauchy’s Theorem |z|=R
f(z)
z−R2
a
dz = 0. So,
f(reiθ
) = f(a) − 0 =
1
2πi |z|=R
f(z)
z − a
dz −
1
2πi |z|=R
f(z)
z − R2
a
dz
=
1
2πi |z|=R
1
z − a
−
1
z − R2
a
f(z)dz
=
1
2πi |z|=R
(|a|2
− R2
)f(z)
(z − a)(az − R2)
dz
=
1
2πi
2π
0
(r2
− R2
)
(Reiϕ − reiθ)(re−iθReiϕ − R2)
f(Reiϕ
)iReiϕ
dϕ
=
1
2π
2π
0
(R2
− r2
)f(Reiϕ
)
(R − rei(θ−ϕ))(R − re−i(θ−ϕ))
dϕ
=
1
2π
2π
0
(R2
− r2
)f(Reiϕ
)
R2 − 2Rr cos(θ − ϕ) + r2
dϕ.
Exercise 30. If the power series ∞
n=0 an(z −a)n
converges in B(a; R) for some R > 0, then
it converges absolutely for every z ∈ B(a; R).
Theorem 31 (Parseval’s Identity). Let f be analytic on |z| < R. If a = rei
θ ∈ {z ∈ C :
|z| < R}. Then
1
2π
2π
0
|f(reiθ
)|2
dθ =
∞
n=0
|an|2
rn
,
where an = f(n)(0)
n!
for all n ∈ N ∪ {0}.
Proof. Since f is analytic in |z| < R, it has a Taylor series expansion f(z) = ∞
n=0 anzn
,
z ∈ B(a; R), where an = f(n)(0)
n!
for all n ∈ N ∪ {0}. If a = 0, i.e., r = 0, then both
the side will be |f(0)|. Let 0 = a = reiθ
∈ B(a; R). Then f(reiθ
) = ∞
n=0 anrn
einθ
and
so f(reiθ) = ∞
n=0 anrn
e−inθ
. Thus |f(reiθ
)|2
= ∞
n=0 anrn
einθ ∞
m=0 anrn
e−imθ
. Since
both the power series converges absolutely, the product is convergent. In fact, the product
converges uniformly as a function of θ. Therefore we may apply term by term integration.

10. P
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10 P. A. DABHI
Applying term by term integration,
2π
0
|f(reiθ
)|2
dθ =
2π
0
∞
n=0
anrn
einθ
∞
m=0
anrn
e−imθ
dθ
=
2π
0
∞
n=0
∞
m=0
anamrm+n
ei(n−m)θ
dθ
=
∞
n=0
∞
m=0
anamrm+n
2π
0
ei(n−m)θ
dθ
= 2π
∞
n=0
|an|2
r2n
,
as
2π
0
ei(n−m)θ
dθ = 0 if n = m and
2π
0
ei(n−m)θ
dθ = 2π if m = n.
Theorem 32. Let f be analytic in |z − a| < R. Then
f (a) =
1
πr
2π
0
Pa,r(θ)e−iθ
dθ,
where Pa,r(θ) = Re f(a + reiθ
), 0 < r < R.
Proof. Let γ(t) = a+reiθ
, 0 ≤ θ ≤ 2π and 0 < r < R. By Cauchy’s Integral Formula f (a) =
1
2πi γ
f(z)
(z−a)2 dz. Since f is analytic, f has a Taylor series expansion f(z) = ∞
n=0 an(z − a)n
,
z ∈ B(a; R), where an = f(n)(a)
n!
for all n ∈ N ∪ {0}. Since a + reiθ
∈ B(a; R), f(a +
reiθ
) = ∞
n=0 anrn
einθ
. So, f(a + reiθ) = ∞
n=0 anrn
e−inθ
. Since the series f(a + reiθ
) =
∞
n=0 anrn
e−inθ
converges absolutely, it converges uniformly in the variable θ. So, we may
take term by term integration. Now
1
2πi γ
f(z)
(z − a)2
dz =
1
2πi
2π
0
∞
n=0 anrn
e−inθ
r2e2iθ
reiθ
dθ
=
1
2π
2π
0
∞
n=0
anrn−1
e−i(n+1)θ
dθ
=
1
2π
∞
n=0
anrn−1
2π
0
e−i(n+1)θ
dθ
= 0. [
2π
0
e−i(n+1)θ
dθ = 0 for all n ∈ N ∪ {0}]

11. P
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COMPLEX ANALYSIS LECTURE NOTES 11
Thus
f (a) = f (a) + 0
=
1
2πi γ
f(z)
(z − a)2
dz +
1
2πi γ
f(z)
(z − a)2
dz
=
1
2πi γ
f(z) + f(z)
(z − a)2
dz
=
1
πi γ
Ref(z)
(z − a)2
dz
=
1
πi
2π
0
Ref(a + reiθ
)
r2e2iθ
reiθ
idθ
=
1
πi
2π
0
Ref(a + reiθ
)e−iθ
dθ
=
1
πr
2π
0
Pa,r(θ)e−iθ
dθ.
This completes the proof.
6. Morera’s Theorem
Morera’s Theorem is the converse of Cauchy’s Theorem.
Definition 33. A closed polygonal path with three sides is called a triangular path.
Theorem 34 (Morera’s Theorem). Let G be a region. Let f : G → C be continuous function
such that T
f(z)dz = 0 for every triangular path T in G. Then f is analytic in G.
Proof. If we can show that f is analytic in each open disc contained in G, then f will be
analytic in G. So, without loss of generality we may assume that G is an open disc, say,
B(a; R). Define F : G → C by F(z) = [a,z]
f(w)dw. We prove that F is a primitive of
f. Let z0, z ∈ G with z = z0. Since T
f = 0 for every triangular path in G, we have
[a,z0]
f(w)dw + [z0,z]
f(w)dw + [z,a]
f(w)dw = 0. Therefore [a,z]
f(w)dw = [a,z0]
f(w)dw +
[z0,z]
f(w)dw, i.e., F(z) − F(z0) = [z0,z]
f(w)dw. This gives F(z)−F(z0)
z−z0
= 1
z−z0 [z0,z]
f(w)dw.
Take any > 0. Since f is continuous at z0, there is δ > 0 such that |f(w) − f(z0)| <
whenever |w − z0| < δ. Let z ∈ G such that |z − z0| < δ. Then
F(z) − F(z0)
z − z0
− f(z0) =
1
z − z0 [z0,z]
f(w)dw − f(z0)
=
1
z − z0 [z0,z]
(f(w) − f(z0))dw
≤
1
|z − z0| [z0,z]
|f(w) − f(z0)||dw|
≤ .

12. P
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12 P. A. DABHI
Therefore F is differentiable at z0 and F (z0) = f(z0). This proves that F is a primitive of
f. It follows from the Cauchy’s Integral Formula that F = f is also analytic on G. Hence
the theorem.
7. Zeros of an analytic function
Let p be a complex polynomial. If a is a zero of p, then there is a polynomial q1 such
that p(z) = (z − a)q1(z). If a is a zero of q1, then there is a polynomial q2 such that
q1(z) = (z − a)q2(z). Hence p(z) = (z − a)2
q2(z). Continuing this way we get a polynomial
q with q(a) = 0 and m ∈ N such that p(z) = (z − a)m
q(z). Also, deg p = m + deg q. This
motivates us to define a zero of an analytic function.
Definition 35. Let G be an open subset of C, and let f : G → C be analytic. Let a ∈ G
with f(a) = 0. Then a is a zero of multiplicity m ≥ 1 if there is an analytic function
g : G → C such that g(a) = 0 and f(z) = (z − a)m
g(z), z ∈ G.
Theorem 36. Let G be a region, and let f : G → C be analytic. Then the following
statements are equivalent.
(1) f ≡ 0.
(2) There is a ∈ G such that f(n)
(a) = 0 for every n ∈ N ∪ {0}.
(3) The set {z ∈ G : f(z) = 0} has a limit point in G.
Proof. (1) ⇒ (3) If f ≡ 0, then {z ∈ G : f(z) = 0} = G and hence it has a limit point in G
(as any point of G is a limit point of G).
(3) ⇒ (2) Let a ∈ G be a limit point of Z(f) = {a ∈ G : f(z) = 0}, and let R > 0
be such that B(a; R) ⊂ G. As a is a limit point of Z(f) and f is continuous, we have
f(a) = 0. Let n be a positive integer such that f(a) = f (a) = · · · = f(n−1)
(a) = 0
but f(n)
(a) = 0. Since f is analytic on G and hence on B(a; R), it has a Taylor se-
ries expansion f(z) = ∞
k=0
f(k)(a)
k!
(z − a)k
, z ∈ B(a; R). Since f(a) = f (a) = · · · =
f(n−1)
(a) = 0 and f(n)
(a) = 0, we have f(z) = ∞
k=n
f(k)(a)
k!
(z − a)k
, z ∈ B(a; R), i.e.,
f(z) = (z − a)n ∞
k=n
f(k)(a)
k!
(z − a)k−n
, z ∈ B(a; R). Define g : B(a; R) → C by g(z) =
∞
k=n
f(k)(a)
k!
(z − a)k−n
. Since the above power series converges, g is analytic on B(a; R) and
g(a) = f(n)(a)
n!
= 0. Since g is continuous and g(a) = 0, there is r > 0 (0 < r < R) such
that g(z) = 0 for every z ∈ B(a; r). As a is a limit point of Z(f), there is a point b such
that 0 < |b − a| < r and f(b) = 0. Also, f(b) = (b − a)n
g(a) = 0. This is a contradiction.
Therefore our assumption that there is some n ∈ N such that f(n)
(a) = 0 is false. Hence
f(n)
(a) = 0 for every n ∈ N ∪ {0}.
(2) ⇒ (1) Let A = {z ∈ G : f(n)
(a) = 0 for every n ∈ N ∪ {0}}. Then by hypothesis A = ∅.
We prove that the set A is both open and closed in G. Then the connectedness of G will
imply that A = G and then it will follow that f ≡ 0. Let (zk) be a sequence in A converging
to z ∈ G. Since each f(n)
is continuous, it follows that f(n)
(z) = limk→∞ f(n)
(zk) = 0. Hence
z ∈ A. This proves that A is closed in G. Let a ∈ A. Let R > 0 be such that B(a; R) ⊂ G.
Then f has a Taylor series expansion in B(a; R) namely f(z) = ∞
n=0
f(n)(a)
n!
(z − a)n
, z ∈

13. P
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COMPLEX ANALYSIS LECTURE NOTES 13
B(a; R). Since each f(n)
(a) = 0, f(z) = 0 for every z ∈ B(a; R). This shows that f(n)
(z) = 0
for every n ∈ N ∪ {0} and for every z ∈ B(a; R). Therefore B(a; R) ⊂ A. This proves that
A is open in G.
Remark 37.
(1) The condition that “G is connected” is necessary in above theorem. Let
G = B(0; 1) ∪ B(3; 1). Define f(z) = 0 if z ∈ B(0; 1) and f(z) = 1 if B(3; 1). Then
f is analytic on G. The set {z ∈ G : f(z) = 0} has a limit point in G but f is not
identically zero.
(2) The condition that “the set of zeros of f has a limit point in G” is
necessary. Define f : B(1; 1) → C by f(z) = sin(1
z
). Then f is analytic. Note that
sin(1
z
) = 0 if and only if z = 1
nπ
for some n ∈ N. Hence Z(f) = {z ∈ B(1; 1) : f(z) =
0} = { 1
nπ
: n ∈ N}. The set Z(f) has a limit point 0 which is not in G. Also, f is
not identically zero.
Corollary 38. Let G be a region, and let f, g : G → C be analytic function. If the set
{z ∈ G : f(z) = g(z)} has a limit point in G, then f = g.
Examples 39.
(1) Let D = {z ∈ C : |z| < 1}. Find all analytic function f : D → C satisfying
f(1
n
) = 1
1+n
for all n ≥ 2.
Let f : D → C be an analytic function such that f(1
n
) = 1
1+n
for all n ≥ 2. Define
g : D → C by g(z) = z
z+1
. Then g is analytic on D. Also, g(1
n
) = 1
1+n
. Then f − g is
analytic on a region D and {1
n
: n ≥ 2} ⊂ {z ∈ D : f(z) = g(z)}. We see that 0 ∈ D
is a limit point of {z ∈ D : f(z) = g(z)}. Hence f(z) = z
1+z
for all z ∈ D.
(2) Is there an analytic function f : D → C which vanish only on {1
n
: n ≥ 2}.
If such a function f exists, then the zeros of f has a limit point 0 ∈ D. But then
f = 0. So, it cannot vanish only on {1
n
: n ≥ 2}. Hence no such function exists.
(3) Let G be a region, and let f, g : G → C be analytic. If fg = 0, then show that either
f = 0 or g = 0.
Suppose that f = 0. Then there is a ∈ G such that f(a) = 0. By continuity of f
there is R > 0 such that f(z) = 0 for every z ∈ B(a; R) ⊂ G. Since fg = 0 on G, we
have g(z) = 0 for every z ∈ B(a; R). Now B(a; R) ⊂ {z ∈ G : g(z) = 0} has a limit
point a in G and hence g = 0.
(4) Let G be a region. Let f, g : G → C be analytic. If fg is analytic on G, then show
that either f is constant or g = 0.
Suppose that g = 0. Then there is a ∈ G such that g(a) = 0. By continuity of
g, there is R > 0 such that g(z) = 0 for every z ∈ B(a; R) ⊂ G. Since g(z) = 0 for
every z ∈ B(a; R), then map 1
g
is analytic on B(a; R). As fg is analytic on G, the
map f = fg1
g
is analytic on B(a; R). Thus both f and f are analytic on B(a; R)
and hence f is constant on B(a; R), say, c. Note that f(z) − c is analytic on G and
B(a; R) ⊂ {z ∈ G : f(z) − c = 0}. Therefore the zeros of f − c has a limit point a in
G. This proves that f(z) = c for every z ∈ G. i.e., f is a constant map.

14. P
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14 P. A. DABHI
Corollary 40. Let G be a region. Suppose that f : G → C is analytic and that f is not
identically zero. Then for each a ∈ G with f(a) = 0, there is n ∈ N and an analytic
g : G → C with g(a) = 0 such that f(z) = (z − a)n
g(z), z ∈ G.
Proof. Let a ∈ G be such that f(a) = 0. Since f is not identically zero, by above theorem
there is m ∈ N such that f(m)
(a) = 0. Let n be the smallest positive integer such that f(a) =
f (a) = · · · = f(n−1)
(a) = 0 but f(n)
(a) = 0. Define g : G → C by g(z) = (z − a)−n
f(z) if
z = a and g(a) = f(n)(a)
n!
. Clearly, g is analytic on G{a} and g(a) = 0. To show g analytic
on G, we not show that g is differentiable at a. Let R > 0 be such that B(a; R) ⊂ G. Let
f(z) = ∞
k=0
f(k)(a)
k!
(z − a)k
be the Taylor series expansion of f in B(a; R). Since f(a) =
f (a) = · · · = f(n−1)
(a) = 0 and f(n)
(a) = 0, we have f(z) = (z − a)n ∞
k=n
f(k)(a)
k!
(z − a)k−n
,
z ∈ B(a; R). Let z ∈ B(a; R){a}. Then
g(z) − g(a)
z − a
=
1
z − a
∞
k=n
f(k)
(a)
k!
(z − a)k−n
−
f(n)
(a)
n!
−→
f(n+1)
(a)
(n + 1)!
as z → a.
Hence g is analytic at a. If z ∈ G and z = a, then g(z) = (z − a)−n
f(z) implies f(z) =
(z − a)n
g(z). If z = a, then f(a) = 0 and (z − a)n
g(z) = 0 at a. Hence f(z) = (z − a)n
g(z)
for every z ∈ G.
Remark 41. The above result says that a zero of an analytic function has a finite multi-
plicity.
Corollary 42. Let G be a region. suppose that f : G → C is analytic and nonconstant. Let
a ∈ C with f(a) = 0. Then there is R > 0 such that B(a : R) ⊂ G and f(z) = 0 for every
z ∈ B(a; R){a}.
Proof. By above result there is an analytic function g : G → C with g(a) = 0 and a natural
number n such that f(z) = (z − a)n
g(z), z ∈ G. Since g is continuous and g(a) = 0,
there is R > 0 such that g(z) = 0 for every z ∈ B(a; R). Let z ∈ B(a; R){a}. Then
f(z) = (z − a)n
g(z) = 0.
Remark 43. Above result says that the zeros of a nonconstant analytic functions are iso-
lated.
Lemma 44. Let G be a region, and let f : G → C be nonconstant analytic function. Let γ
be a simple closed curve in G such that interior of γ lies in G. Then the number of zeros of
f inside γ are finite.
Proof. Suppose that the zeros of f inside γ are infinite. Then the set of zeros of f inside is
a bounded infinite subset of C (as interior of γ is bounded). Let z0 be a limit point of this
set. Then either z is inside γ or it is on γ. In any case, a limit point z0 of zeros of f is in G
and hence f ≡ 0. This is a contradiction.
Exercises 45.
(1) Let G be an open subset of C, and let f : G → C be analytic. Let a ∈ G be
such that f(a) = 0. Show that a is a zero of f of multiplicity m if and only if
f(a) = f (a) = · · · = f(m−1)
(a) = 0 but f(m)
(a) = 0.

15. P
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COMPLEX ANALYSIS LECTURE NOTES 15
(2) Let f, g : G → C be analytic, let a ∈ G be such that f(a) = g(a) = 0. If a is a zero
of f of multiplicity m and a is a zero of g of multiplicity n, then show that a is a zero
of fg of multiplicity m + n .
Theorem 46 (Counting Zero Principle). Let G be a region, and let f be analytic on G with
zeros a1, a2, . . . , am repeated according to multiplicity. Let γ be a closed rectifiable curve in
G which does not pass through any zero of f and that n(γ; w) = 0 for all w ∈ CG. Then
1
2πi γ
f (z)
f(z)
dz =
m
k=1
n(γ; ak).
Proof. Since a1, a2, . . . , am are zeros of f (counted according to multiplicity), f admits a
factorization of the form f(z) = (z − a1)(z − a2) · · · (z − am)g(z), z ∈ G, for some analytic
function g : G → C such that g(z) = 0 for any z ∈ G. Observe that f (z)
f(z)
= m
k=1
1
z−ak
+ g (z)
g(z)
.
Since g is analytic and nowhere vanishing g
g
is analytic. Since n(γ; w) = 0 for all w ∈ CG,
it follows from the Cauchy’s theorem that γ
g
g
= 0. Hence 1
2πi γ
f
f
= m
k=1
1
2πi γ
1
z−ak
dz =
m
k=1 n(γ; ak).
Corollary 47. Let G be a region, and let f : G → C be analytic. Let α ∈ C. Let γ be a closed
rectifiable curve in G which does not pass through any zero of f(z) − α and that n(γ; w) = 0
for all w ∈ CG. If a1, a2, . . . , am are zeros of f(z)−α, then 1
2πi γ
f (z)
f(z)−α
dz = m
k=1 n(γ; ak).
Proof. Take g(z) = f(z) − α and apply the above theorem.
Let a, b ∈ C, and let b = 0. Let z ∈ C. Then Im z−a
b
> 0 iff Im x−a1+i(y−a2)
b2
1+b2
2
(b1 − ib2) > 0
iff (y − a2)b1 − (x − a1)b2 > 0 iff −b2x + b1y > a2b1 − a1b2.
Theorem 48 (Luca’s Theorem). If a polynomial has all the zeros in a half plane, then all
the zeros of its derivative are also in the same half plane.
Proof. Let p(z) be a polynomial of degree n, and let α1, α2, . . . , αn be its roots. Then
p(z) = K(z −α1) · · · (z −αn) for some 0 = K ∈ C. Note that p (z)
p(z)
= 1
z−α1
+ 1
z−α2
+· · ·+ 1
z−αn
.
Let Im z−a
b
< 0 be a half plane in which α1, α2, . . . , αn are lying. Therefore Im αk−a
b
< 0
for 1 ≤ k ≤ n. Let z0 be a zero of p (z) not lying in this half plane, i.e., Im z0−a
b
≥ 0.
Now Im z0−αk
b
= Im z0−a
b
− Im αk−a
b
> 0 for 1 ≤ k ≤ n. Since Imz = −(Imz−1
)|z|2
for all 0 = z ∈ C, we have Im z0−αk
b
= −Im z0−αk
b
−1
|z0−αk
b
|2
> 0. Now Im bp (z0)
p(z0)
=
Im n
k=1
b
z0−αk
= n
k=1 Im b
z0−αk
< 0. This is a contradiction to the fact that p (z0)
p(z0)
= 0.
This proves the theorem.
8. Open Mapping Theorem
Example 49. Let G be an open subset of C, and let f : G → C be analytic, If γ : [0, 1] → C
is a closed rectifiable curve in G, then f ◦ γ is a closed rectifiable curve.
Since both γ and f are continuous and the range of γ is a subset of the domain of f,
f ◦ γ : [0, 1] → C is continuous, i.e., f ◦ γ is a curve. Since γ is closed,f ◦ γ is closed. Define
ϕ : G × G → C by ϕ(z, w) = f(z)−f(w)
z−w
if z = w and ϕ(z, z) = f (z). Then we know that ϕ

16. P
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16 P. A. DABHI
is a continuous map. Since {γ} × {γ} is a compact subset of G × G, there is M > 0 such
that |ϕ(z, w)| ≤ M for every z, w ∈ {γ}. In other words, |f(z) − f(w)| ≤ M|z − w| for every
z, w ∈ {γ}. Let 0 = t0 < t1 < · · · < tn = 1 be any partition of [0, 1]. Then
n
k=1
|f ◦ γ(tk) − f ◦ γ(tk−1)| =
n
k=1
|f(γ(tk)) − f(γ(tk−1))|
≤
n
k=1
M|γ(tk) − γ(tk−1)| ≤ MV (γ).
Therefore f ◦ γ is rectifiable.
Let γ : [0, 1] → C be a closed rectifiable curve in a region G, and let f : G → C be analytic.
Then σ = f ◦ γ is a closed rectifiable curve. Let α ∈ C such that α /∈ {σ} = f({γ}). Then
n(σ; α) =
1
2πi σ
dw
w − α
=
1
2πi
1
0
d(f ◦ γ(t))
f ◦ γ(t) − α
=
1
2πi
1
0
f (γ(t))
f(γ(t)) − α
dγ(t) =
1
2πi γ
f (z)
f(z) − α
dz
=
m
k=1
n(γ; ak),
where a1, a2, . . . , am ∈ G satisfy f(aj) = α.
Theorem 50. Let f be analytic on B(a; R), and let f(a) = α. If f(z) − α has a zero of
multiplicity m at a, there there exist > 0 and δ > 0 such that for |ξ − α| < δ, the equation
f(z) = ξ has exactly m simple roots in B(a; ).
Proof. Since the zeros of an analytic function are isolated, there is > 0 such that f(z) = α
has no solution in 0 < |z − a| < 2 and f (z) = 0 in 0 < |z − a| < 2 . Let γ(t) = a + e2πit
,
0 ≤ t ≤ 1. Put σ = f ◦ γ. Then α /∈ {σ}. So, there is δ > 0 such that B(α; δ) ∩ {σ} = ∅.
Therefore B(α; ) lies in a component of C{σ}, i.e., if |ξ − α| < δ, then n(σ; ξ) = n(σ; α).
Note that m = 1
2πi γ
f (z)
f(z)−α
dz. Now,
m =
1
2πi γ
f (z)
f(z) − α
dz =
1
2πi
1
0
f (γ(t))
f(γ(t)) − α
dγ(t)
= n(σ; α) = n(σ; ξ) =
1
2πi σ
dw
w − ξ
=
p
k=1
n(γ, zk(ξ)),
where zk(ξ) are zeros of f(z) − ξ. Since n(γ; z) = 0 or 1, there are exactly m solutions of
f(z) = ξ in B(a; ). Since f (z) = 0 for 0 < |z − a| < , each of these roots is simple.
Theorem 51 (Open Mapping Theorem). Let G be a region, and let f : G → C be a
nonconstant analytic function. Then f is an open map.

17. P
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COMPLEX ANALYSIS LECTURE NOTES 17
Proof. Let U be an open subset of G, and let a ∈ U. Take α = f(a). Then by f(z) − α
has a zero in B(a; R) for any R > 0 with B(a; R) ⊂ U. Then by above theorem, there exist
> 0 and δ > 0 such that if |ξ − f(a)| < δ, then ξ ∈ f(B(a; )), i.e., f(a) ∈ B(f(a); δ) ⊂
f(B(a; )) ⊂ f(U). Thus if f(a) ∈ U, then there is δ > 0 such that f(a) ∈ B(f(a); δ) ⊂ f(U).
This shows that f(U) is an open subset of C. Therefore f is an open map.
Exercise 52.
(1) Let G be a region, and let f : G → C be analytic. If f is one one, then f (z) = 0 for
any z ∈ G.
Suppose that f (a) = 0 for some a. Then a is a zero of f(z) − f(a) of multiplicity
at least 2. By above result there is δ > 0 and > 0 such that if |ξ − f(a)| < δ, then
f(z) = ξ has at least two solutions in B(a; ). This shows that f is not one one.
(2) Let G be a region, and let f : G → C be one one and analytic. If Ω = f(G) is the
range of f, then f−1
: Ω → C is analytic.
By the Open Mapping Theorem f is an open map. Therefore Ω = f(G) is an
open subset of C. The map f : G → Ω is one one and onto and has an inverse
f−1
: Ω → G ⊂ C. Since f is an open map, the map f−1
is continuous. We need
to check the differentiability of f−1
at every point of Ω. Let w = f(z) ∈ Ω. By
above exercise, f (w) = 0. Let (wn) be a sequence in Ω{w} converging to w. Then
wn = f(zn) for some zn ∈ G and also zn = z as f is one one. Since f−1
is continuous,
zn → z as n → ∞. In other words zn → z if and only if f(zn) → f(z). Now
lim
wn→w
f−1
(wn) − f−1
(w)
wn − w
= lim
f(zn)→f(z)
zn − z
f(zn) − f(z)
= lim
zn→z
zn − z
f(zn) − f(z)
=
1
f (z)
=
1
f (f−1(w))
.
Therefore f−1
is differentiable at each w ∈ Ω and hence analytic and (f−1
) (w) =
1
f (f−1(w))
.
9. Singularities of a complex function
Definition 53. A singularity a of a function f is called an isolated singularity if there is
δ > 0 such that f is analytic in a deleted neighbourhood 0 < |z − a| < δ of a but not at a.
Definition 54. An isolated singularity a of f is called
(1) a removable singularity if there is an analytic function g : B(a; R) → C such that
g(z) = f(z) for every z ∈ B(a : R){a}.
(2) a pole if limz→a |f(z)| = ∞.
(3) an essential singularity if it neither a removable singularity nor a pole.
Proposition 55. Let a be an isolated singularity of f. Then a is a removable singularity of
f if and only if limz→a(z − a)f(z) = 0.
Proof. Let a be a removable singularity of f. Then there is an analytic function g : B(a; R) →
C, for some R > 0, such that g(z) = f(z) for every z ∈ B(a; R){a}. Since g is analytic at a
and hence continuous at a, the limit limz→a(z − a)f(z) = limz→a(z − a)g(z) exists and is 0.
Conversely, assume that limz→a(z − a)f(z) = 0. Since a is an isolated singularity of f, f
is analytic in B(a; R){a} for some R > 0. Define g : B(a; R) → C by g(z) = (z − a)f(z)

18. P
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18 P. A. DABHI
if z = a and g(a) = 0. Then g is continuous on B(a; R) and analytic on B(a; R){a}.
Let T be any triangle in B(a; R) with length . Let ∆ be the interior of T together with
T. If a /∈ ∆, then T
g = 0 as g is analytic on B(a; R){a}. Let a be a vertex of T.
Then T = [a, b, c, a]. Since g is continuous at a, given > 0 there is δ > 0 such that
|g(z) − g(a)| < is |z − a| < δ. Choose x on the edge [a, b] and y on the edge [a, c] so
that |a − x| < δ and |a − y| < δ. Let T1 = [a, x, y, a] and P = [x, b, c, y, x]. Then P is a
polygon and n(P, w) = 0 for every w ∈ CB(a; R). By Cauchy theorem, P
g = 0. Now
| T
g| = | T1
g + P
g| ≤ | T1
g| ≤ = . Thus T
g = 0. Let a be in the interior of
∆, and let T = [x, y, z, x]. Let T1 = [x, y, a, x], T2 = [y, z, a, y] and T3 = [z, x, a, z]. By
above Ti
g = 0 for i = 1, 2, 3. Now T
g = T1
g + T2
g + T3
g = 0. Hence it follows from
Morera’s theorem that g is analytic on B(a; R). Since g(a) = 0, there is an analytic function
h : B(a; R) → C such that g(z) = (z − a)h(z) for every z ∈ B(a; R). If z ∈ B(a; R){a},
then f(z) = h(z) as g(z) = (z−a)f(z). Hence f = h on B(a; R){a} and so a is a removable
singularity of f.
Proposition 56. Let a be an isolated singularity of f. If a is a pole of f, then a is a
removable singularity of 1
f
.
Proof. Since a is a pole of f, limz→a |f(z)| = ∞. This implies that limz→a
1
f(z)
= 0 and
hence limz→a(z − a) 1
f(z)
= 0. Since a is pole of f, f is analytic in a neighbourhood of a. As
limz→a |f(z)| = ∞, f(z) = 0 for any z in some deleted neighbourhood of a, say, B(a; δ){a}.
Thus 1
f
is analytic in B(a; δ){a} and limz→a(z − a) 1
f(z)
= 0. Therefore by above a is a
removable singularity of 1
f
.
Proposition 57. Let G be a region, and let a ∈ G. Let f be analytic on G{a}, and let a be
singularity of f. Then a is a pole of f if and only if there is an analytic function g : G → C
with g(a) = 0 and m ∈ N such that f(z) = g(z)
(z−a)m for every z ∈ G{a}.
Proof. Suppose that there is an analytic function g : G → C with g(a) = 0 and m ∈ N such
that f(z) = g(z)
(z−a)m for every z ∈ G{a}. Since g(a) = 0, limz→a |f(z)| = limz→a | g(z)
(z−a)m | = ∞.
Therefore a is a pole of f.
Conversely, assume that a is a pole of f, i.e., limz→a |f(z)| = ∞. Then by above a is a
removable singularity of 1
f
. Let h : B(a; R) → C be analytic such that 1
f(z)
= h(z) for every
z ∈ B(a; R){a} for some R > 0 with B(a; R) ⊂ G. Since 0 = limz→a
1
f(z)
= limz→a h(z),
there is an analytic k : B(a; R) → C with k(a) = 0 and m ∈ N such that h(z) = (z −a)m
k(z)
for every z ∈ B(a; R). Define g : G → C by g(z) = k(z) if z = a and g(z) = (z − a)−m
f(z)
if z = a. Then g is the required function.
Definition 58. Let a be a pole of f. Then there is an analytic function g in a neighbourhood
of a with g(a) = 0 and m ∈ N such that f(z) = g(z)
(z−a)m for z = a. The natural number m is
called the order of the pole a of f.
Exercise 59. Let G be an open subset of C, and let a1, a2 ∈ G. Let f be analytic on
G{a1, a2}, and let a1 and a2 be poles of f of order m1 and m2 respectively. Then there is
an analytic function g : G → C with g(a1) = 0, g(a2) = 0 such that f(z) = g(z)
(z−a1)m1 (z−a2)m2
for every z ∈ G{a1, a2}.

19. P
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COMPLEX ANALYSIS LECTURE NOTES 19
Since a1 is a pole of f of order m1, there is an analytic function h1 : G{a2} → C
with h1(a1) = 0 such that f(z) = h1(z)
(z−a1)m1
for every z ∈ G{a1, a2}. Then f(z) =
(z−a2)m2 h1(z)
(z−a1)m1 (z−a2)m2
= g1(z)
(z−a1)m1 (z−a2)m2
for every z ∈ G{a1, a2}, where g1(z) = (z − a2)m2
h1(z)
is analytic on G{a1} and g1(a1) = 0. Similarly, since a2 is a pole of f of order m2, there
is an analytic function g2 : G{a2} → C with g2(a2) = 0 such that f(z) = g2(z)
(z−a1)m1 (z−a2)m2
for every z ∈ G{a2}. Define g : G → C by g(z) = g1(z) if z ∈ G{a2} and g(z) = g2(z)
if z ∈ G{a1}. Since g1(z) = g2(z) on G{a1, a2}, the function g is analytic, g(a1) = 0,
g(a2) = 0 and f(z) = g(z)
(z−a1)m1 (z−a2)m2
for every z ∈ G{a1, a2}.
Theorem 60 (Riemann’s Theorem on removable singularity). Let f be defined on B(a; δ).
If f is analytic and bounded in B(a; δ){a}, then either f is analytic at a or a is removable
singularity of f.
Proof. As f is bounded in 0 < |z − a| < δ, there is M > 0 such that |f(z)| ≤ M for every
z ∈ C with 0 < |z − a| < δ. Since f is analytic in 0 < |z − a| < δ, it has a Laurent series
expansion
f(z) =
∞
n=0
an(z − a)m
+
∞
m=1
bm(z − a)−m
, 0 < |z − a| < δ.
Let Cρ be the positively oriented circle |z − a| = ρ, where 0 < ρ < δ. Then bm =
1
2πi Cρ
f(z)
(z−a)−m+1 dz for every m ∈ N. Let m ∈ N. Then
|bm| =
1
2πi Cρ
f(z)
(z − a)−m+1
dz ≤
1
2π Cρ
|f(z)|
|z − a|−m+1
|dz|
=
1
2π
M2πρ
ρ−m+1
= Mρm
→ 0 as ρ → 0.
Thus f(z) = ∞
n=0 an(z −a)n
, 0 < |z −a| < δ. If f(a) = a0, then the power series is actually
valid in |z − a| < δ and hence f is analytic at a. If f(a) = a0, then in order to make f
analytic one should define f(a) = a0. Therefore a is a removable singularity of f.
Corollary 61. Let f be bounded and analytic in 0 < |z − a| < δ. Then a is a removable
singularity of f.
Exercise 62. If a is a removable singularity of f, then f is bounded and analytic in a deleted
neighbourhood of a.
Since a is a removable singularity of f, there is an analytic g : B(a; R) → C, for some R > 0,
such that g(z) = f(z) for every z ∈ B(a; R){a}. Let 0 < δ < R. Then g is analytic on
B(a; δ). Since g is continuous on B(a; R), it is bounded on the compact subset B(a; δ). Since
f = g on B(a; δ){a}, f is bounded and analytic on B(a; δ){a}.
It follows from above theorem and exercise that a is a removable singularity of f if and only
if f is analytic and bounded in some deleted neighbourhood of a.
Theorem 63 (Casorati - Weierstrass Theorem). Let a be an essential singularity of f. Then
for any δ > 0 the set {f(z) : 0 < |z − a| < δ} is dense in C (OR Given c ∈ C, > 0 and
δ > 0, there is z ∈ C with 0 < |z − a| < δ such that |f(z) − c| < ).

20. P
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20 P. A. DABHI
Proof. Since a is an isolated singularity, there is R > 0 such that f is analytic in B(a; R){a}.
It is sufficient to prove that for any δ > 0 with 0 < δ < R, the set {f(z) : 0 < |z − a| < δ}
is dense in C.
Suppose that there is δ > 0 with 0 < δ < R such that {f(z) : 0 < |z − a| < δ} is not dense
in C. Then there is c ∈ C and > 0 such that B(c; ) {f(z) : 0 < |z − a| < δ} = ∅, i.e.,
|f(z)−c| ≥ for every z ∈ C with 0 < |z −a| < δ. Define g : B(a; δ) → C by g(z) = 1
f(z)−c
.
Then g is analytic in 0 < |z − a| < δ and |g(z)| = 1
|f(z)−c|
≤ 1
. Therefore g is bounded and
analytic in a neighbourhood of a and hence a is removable singularity of g.
Let g(a) be defined so that g is analytic at a. Then we have g(a) = 0 or g(a) = 0. Let
g(a) = 0. Since g(a) = 0 and g is continuous, g(z) = 0 for every z in a neighbourhood of a,
say, B(a; η) and also 1
g
is bounded on this neighbourhood. But then f(z) = 1
g(z)
+ c will be
analytic and bounded on B(a; η){a}. This will imply that a is a removable singularity of f
(because of Riemann’s Theorem). This is a contradiction.
Assume that g(a) = 0. Then a will be a zero of g of finite multiplicity and hence a will be
a pole of 1
g
. But then a will be a pole of f = 1
g
+ c. This is again a contradiction. This
completes the proof.
The above theorem says that f takes values arbitrarily close to every complex num-
ber in every neighbourhood of essential singularity of f.
Corollary 64. Let a be an essential singularity of f. If w ∈ C, then there is a sequence
(zn) converging to a such that f(zn) → w.
Proof. Since a is an isolated singularity, f will be analytic in a deleted neighbourhood of a,
say, B(a; R){a}. Let (rn) be a strictly deceasing sequence of positive real numbers such
that rn < R for all n and rn → 0. Since a is an essential singularity of f, by Casorati -
Weierstrass Theorem {f(z) : 0 < |z − a| < rn} is dense in C. Therefore there is zn ∈ C with
0 < |zn − a| < rn such that |f(zn) − w| < rn. Then the sequence (zn) converges to a and the
sequence (f(zn)) converges to w as (rn) converges to 0.
In 1879, with the aid of elliptical modular functions, Picard proved a deep result - the so -
called Picard’s Great Theorem: “If f has an essential singularity at a, then the range under
f of any deleted neighbourhood of a is the whole complex plane with at most exception”.
The complex number which is unassumed by a function in a deleted neighbourhood of an
essential singularity of f is called Picard’s exceptional point.
The function f(z) = e
1
z has an essential singularity at 0. The number 0 is the Picard’s
exceptional point for f.
Let 0 = w = reiθ
∈ C. Then there is x ∈ R such that ex
= r. But then ex+iθ
= ex
eiθ
=
reiθ
= w. Take z = x + iθ. For every n ∈ N take zn = 1
z+2nπi
. Then zn → 0 as n → ∞ and
e
1
zn = ez+2nπi
= w. Hence for every 0 = w ∈ C, the equation e
1
z = w has infinitely many
solutions in a neighbourhood of 0.
We note the following important facts about the nature of singularity of f. Let a be an
isolated singularity of f.
(1) a is a removable singularity of f if and only if f is analytic and bounded in a deleted
neighbourhood of a.
(2) a is a pole of f if and only of limz→a |f(z)| = ∞.

21. P
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COMPLEX ANALYSIS LECTURE NOTES 21
(3) If a is an essential singularity of f, then f wanders through the complex plane in a
neighbourhood of a.
Definition 65. Let f analytic in {z ∈ C : |z| > R} for some R > 0. Then ∞ is called
(1) a removable singularity of f if 0 is a removable singularity of f(1
z
).
(2) a pole of f if 0 is a pole of f(1
z
).
(3) an essential singularity of f if 0 is an essential singularity of f(1
z
).
Examples 66.
(1) Let f be an entire function. Let M > 0. If |f(z)| ≤ M|z|m
for every z ∈ C with
|z| > R for some R > 0. Then f is a polynomial of degree at most m.
Since f is entire function, it has a Taylor series valid for every z ∈ C, i.e., f(z) =
∞
n=0
f(n)(0)
n!
zn
, z ∈ C. Let r > R. Take any n ∈ N with n > m. Then by Cauchy’s
Integral Formula
|f(n)
(0)| =
n!
2πi |z|=r
f(z)
zn+1
dz ≤
n!
2π |z|=r
|f(z)|
|z|n+1
|dz|
≤
n!
2π |z|=r
M|z|m
|z|n+1
|dz| =
Mn!
rn−m
→ 0 as r → ∞.
This proves that f(n)
(0) = 0 for every n > m. Hence f is a polynomial of degree at
most m.
(2) Let f be an entire function. If the real part of f is bounded above or bounded below,
then show that f is a constant map.
Let f = u+iv. Suppose that Ref = u is bounded below. Then there is M ∈ R such
that M ≤ u(x, y) for all (x, y) ∈ R2
. Define g : C → C by g(z) = exp(−f(z)). Then g
is an entire function. Let z = x + iy ∈ C. Then |g(z)| = | exp(−u(x, y) − iv(x, y))| =
exp(−u(x, y)) ≤ e−M
. Thus g is a bounded entire function. By Liouville’s Theorem
g is constant, say, c. Differentiating g we get exp(−f(z))f (z) = 0 for all z ∈ C.
Since exp(−f(z)) is never zero, f (z) = 0 for all z ∈ C. Since C is connected, f is
constant.
If u is bounded above, then consider the map exp(f(z)).
The same is true either Im f is bounded above or below. (take g(z) = exp(if(z))
or g(z) = exp(−if(z))).
(3) Let f be an entire function. Then ∞ is a removable singularity of f if and only if f
is a constant map.
Assume that f is constant, say, c. Then f(1
z
) = c for all z ∈ C{0}. Clearly, f(1
z
)
is bounded and analytic in a deleted neighbourhood of 0. Therefore 0 is removable
singularity of f(1
z
) and hence ∞ is a removable singularity of f.
Conversely, assume that ∞ is removable singularity of f, i.e., 0 is a removable sin-
gularity of f(1
z
). Then f(1
z
) is bounded and analytic in some deleted neighbourhood
of 0, say, B(0; δ){0}. Therefore there is M > 0 such that |f(1
z
)| ≤ M for 0 < |z| < δ.
Let z ∈ C with |z| > 1
δ
. Then |1
z
| < δ and so |f(z)| ≤ M. Since f is continuous, it is
bounded on the compact set B(0; 1
δ
), say, it is bounded by K. Thus |f(z)| ≤ M + K
for every z ∈ C. By Liouville’s Theorem, the function f is constant.

22. P
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22 P. A. DABHI
(4) Let f be an entire function. Then ∞ is a pole of f of order m if and only if f is a
polynomial of degree m.
Let f be a polynomial of degree m, say, f(z) = a0 + a1z + · · · + amzm
and am = 0.
Then f(1
z
) = a0 + a1
z
+ · · · + am
zm . If n ∈ N ∪ {0}, then zn
f(1
z
) has a singularity at 0
and zm
f(1
z
) has a removable singularity at 0. Thus 0 is a pole of order m of f(1
z
),
i.e., ∞ is a pole of f of order m.
Conversely, assume that ∞ is a pole of f of order m. Then 0 is a pole of f(1
z
) of
order m. Therefore the limit limz→0 zm
f(1
z
) exists, say, c. This implies that there is
δ > 0 such that |zm
f(1
z
) − c| ≤ 1 for all z ∈ C with 0 < |z| < δ. Let z ∈ C with
|z| > 1
δ
. Then 1
|z|m |f(z)| ≤ (1 + |c|), i.e., |f(z)| ≤ (|c| + 1)|z|m
for all z ∈ C with
|z| > 1
δ
. The exercise above implies that f is a polynomial of degree at most m. If
the degree of f is k with k < n, the zk
f(1
z
) will have a removable singularity at 0,
and hence ∞ will become a pole of f order k < m. This is a contradiction. Hence
the degree of f is m.
Examples 67.
(1) (Fourier series Expansion is associated with the Laurent series expansion of analytic
f in an annular region {z ∈ C : R1 < |z| < R2}, where 0 < R1 < 1 < R2). Let
0 < R1 < 1 < R2, and let f be analytic on an annular region {z ∈ C : R1 < |z| < R2}.
Then f admits the Laurent series expansion f(z) = ∞
n=−∞ anzn
, R1 < |z| < R2,
where an = 1
2πi |z|=1
f(z)
zn+1 dz for all n ∈ Z. Let n ∈ Z. Then
an =
1
2πi |z|=1
f(z)
zn
dz =
1
2πi
2π
0
f(eiθ
)
ei(n+1)θ
ieiθ
dθ
=
1
2π
2π
0
f(eiθ
)e−inθ
dθ.
Thus if z = eiθ
, then the Laurent series of f will give f(eiθ
) = ∞
n=−∞ aneinθ
. Define
F : R → C by F(t) = f(eit
). Then F is a 2π- periodic function. By above we have
F(t) = ∞
n=−∞ aneint
, where an = 1
2π
2π
0
F(t)e−int
dt for every n ∈ Z.
10. Argument Principle and its applications
Definition 68. Let G be an open subset of C. A function f defined and analytic in G
except for poles is called a meromorphic function.
Theorem 69 (Argument Principle). Let G be a region, and let f be meromorphic in G
with zeros z1, z2, . . . , zn and poles p1, p2, . . . , pm counted according to their multiplicities. If
γ is a closed rectifiable curve in G which does not pass through any zero or pole of f and if
n(γ; w) = 0 for every w ∈ CG, then
1
2πi γ
f (z)
f(z)
dz =
n
k=1
n(γ; zk) −
m
k=1
n(γ; pk).
Proof. Since p1, p2, . . . , pm are poles of f and z1, z2, . . . , zn are zeros of f counted according to
their multiplicities, there is an analytic function g : G → C such that f(z) = (z−z1)···(z−zn)
(z−p1)···(z−pm)
g(z)
for every z ∈ G{p1, p2, . . . , pm} and g(z) = 0 for any z ∈ G. Since g is analytic on

23. P
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COMPLEX ANALYSIS LECTURE NOTES 23
G and nowhere vanishing, the function g
g
is analytic on G. As n(γ; w) = 0 for every
w ∈ CG, it follows from the Cauchy’s Theorem that γ
g
g
= 0. Observe that f (z)
f(z)
=
n
k=1
1
z−zk
− m
k=1
1
z−pk
+ g (z)
g(z)
for every z ∈ {γ} (In fact, f
f
is analytic in a neighbourhood
of {γ}). Therefore 1
2πi γ
f
f
= n
k=1 n(γ; zk) − m
k=1 n(γ; pk).
Why is the Argument Principle called so? Since no zero or pole lies on γ, for each
a ∈ {γ} there is an open disc B(a; δa) which contains no zero or pole of f. Then the branch
of log f(z) can be defined on this neighbourhood. These balls form an open cover of {γ};
and so, by Lebesgue’s Covering Lemma, there is > 0 such that for each a ∈ {γ} we can
define a branch of log f(z) in B(a; ). Since γ is uniformly continuous, there is a partition
0 = t0 < t1 < · · · < tk = 1 of [0, 1] such that γ(t) ∈ B(γ(tj−1); ) for tj−1 ≤ t ≤ tj. Let
j be the branch of log f defined on B(γ(tj−1); ) for 1 ≤ j ≤ k. Also, since the j-th and
(j+1)- st sphere both contain γ(tj) we can choose 1, 2, . . . , k such that 1(γ(t1)) = 2(γ(t1));
(γ(t2)) = 3(γ(t2)); . . . ; k−1(γ(tk−1)) = k(γ(tk−1)). If γj is the path γ restricted to [tj−1, tj],
then γj
f
f
= j(γ(tj)) − j(γ(tj−1)) for 1 ≤ j ≤ k as j = f
f
. Summing both the sides of this
equation we get
γ
f
f
= k(a) − 1(a),
where a = γ(0) = γ(1). Therefore k(a) − 1(a) = 2πiK. Because 2πiK is purely imaginary
we get Im k(a) − Im 1(a) = 2πK. This makes precise our contention that as z traces out γ,
arg f(z) changes by 2πK.
Corollary 70. Let G be a region, and let f be meromorphic on G with poles p1, p2, . . . , pm
and z1, z2, . . . , zn counted according to their multiplicities. If g is analytic on G and if γ is
a closed rectifiable curve in G which does not pass through any zero or pole of f such that
n(γ; w) = 0 for every w ∈ CG, then
1
2πi γ
g
f
f
=
n
k=1
g(zk)n(γ; zk) −
m
k=1
g(pk)n(γ; pk).
Proof. Since z1, z2, . . . , zn are zeros and p1, p2, . . . , pm are poles of f, counted according to
their multiplicities, there is an analytic function h : G → C with h(z) = 0 for every z ∈ G
such that f(z) = (z−z1)···(z−zn)
(z−p1)···(z−pm)
h(z) for every z ∈ G{p1, p2, . . . , pm}. Since g and h
h
are
analytic on G and n(γ; w) = 0 for every w ∈ CG, it follows from the Cauchy’s Theorem
that γ
gh
h
= 0. Observe that g(z)f (z)
f(z)
= n
k=1
g(z)
z−zk
− m
k=1
g(z)
z−pk
. Applying integration
over γ and using Cauchy’s Integral Formula, we obtain 1
2πi γ
gf
f
= n
k=1 g(zk)n(γ; zk) −
m
k=1 g(pk)n(γ; pk).
Examples 71. Evaluate |z|=3
f (z)
f(z)
dz if f(z) = (z2+1)2
(z2+3z+2)3 .
We note that the zeros and poles of f are i, i, −i, −1 and −1, −1, −1, −2, −2, −2, counted
according to their multiplicities, respectively. We also note that n(γ; z) = 1 for every zero
and pole. It follows form the argument principle, that γ
f
f
= −4πi.
Exercise 72. Let G be open in C, and let γ be a rectifiable curve in G with initial point α and
end point β. If f : G → C is continuous with primitive F : G → C, then γ
f = F(β)−F(α).

24. P
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24 P. A. DABHI
Theorem 73 (Rouche’s Theorem). Suppose that f and g are meromorphic in a neigh-
bourhood of B(a; R) with no zeros or poles on the circle γ = {z ∈ C : |z − a| = R}. If
Zf , Zg (Pf , Pg) are the number of zeros (poles) of f and g inside γ counted according to their
multiplicities and if |f(z) + g(z)| < |f(z)| + |g(z)| on γ, then Zf − Pf = Zg − Pg.
Proof. Let z ∈ {γ}, and let λ(z) = f(z)
g(z)
. If λ(z) is nonnegative, then |f(z) + g(z)| < |f(z)| +
|g(z)| will imply that λ(z) + 1 < λ(z) + 1. This is not possible. Therefore the meromorphic
function f
g
maps a neighbourhood of γ to Ω = C[0, ∞). Let be the branch of logarithm
defined on Ω. Then (f
g
) is well defined primitive of (f/g) (f/g)−1
in a neighbourhood of γ.
Thus
0 = (
f
g
(γ(1))) − (
f
g
(γ(0))) =
1
2πi γ
(f/g) (f/g)−1
=
1
2πi γ
f
f
−
1
2πi γ
g
g
= Zf − Pf − (Zg − Pg). [ Argument Principle]
This completes the proof.
The statement of Rouche’s Theorem was discovered by Irving Glicksberg (Amer. Math.
Monthly, 83(1976), 186 – 187). In the more classical statement of the theorem, f and g were
assumed to satisfy the stronger inequality |f + g| < |g| on γ (and so it is a weaker version).
It has following application.
Corollary 74 (Deduction of the Fundamental Theorem of Algebra from Rouche’s Theorem).
Proof. Let p be a polynomial of degree n ≥ 1. We may assume that p is of the form p(z) =
a0+a1z+· · ·+an−1zn−1
+zn
. We see that lim|z|→∞
p(z)
zn = lim|z|→∞[a0
zn + a1
zn−1 +· · ·+an−1
z
+1] = 1.
Therefore there is r > 0 such that |p(z)
zn − 1| < 1 if |z| > r. If R > r, then |p(z)
zn − 1| < 1 for
|z| = R, i.e., |p(z)−zn
| < |zn
| if |z| = R. Note that neither p(z) nor zn
has a zero on |z| = R.
It follows from Rouche’s Theorem that p(z) and zn
have the same number of zeros inside
|z| = R. Since zn
has exactly n zeros inside |z| = R, p(z) has n zeros inside |z| = R.
Examples 75.
(1) Determine the number of zeros of z4
+ 6z + 1 inside |z| = 3
2
and |z| = 2.
Take f(z) = z4
+ 6z + 1 and g(z) = −z4
. If |z| = 2, then |f(z) + g(z)| = |6z + 1| ≤
13 < 16 = |g(z)|. Also, neither f nor g has a zero on |z| = 2. By Rouche’s Theorem
f(z) has 4 zeros inside |z| = 2 as g(z) = −z4
has 4 zeros inside |z| = 2.
Now take h(z) = −6z. If |z| = 3
2
, then |f(z) + h(z)| = |z4
+ 1| ≤ (3
2
)4
+ 1 < 9 =
|h(z)|. Since h(z) = −6z has one zero inside |z| = 3
2
, f(z) has one zero inside |z| = 3
2
.
(2) Let a ∈ C with |a| > e, and let n ∈ N. Then the equation ez
− azn
= 0 has n
solutions in |z| = 1.
Take f(z) = ez
−azn
and g(z) = −azn
. Let z = x+iy ∈ C with x2
+y2
= |z|2
= 1.
Then |f(z) + g(z)| = |ez
| = |ex+iy
| = ex
≤ e < |a| = |g(z)|. Since g(z) = azn
has n
zeros inside |z| = 1, it follows from Rouche’s theorem that f(z) = 0 has n solutions
inside |z| = 1.

25. P
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COMPLEX ANALYSIS LECTURE NOTES 25
(3) Let f be analytic in a neighbourhood of B(0; 1). If |f(z)| < 1 on |z| = 1, then show
that f has exactly one fixed point inside |z| = 1.
Let h(z) = f(z) − z and g(z) = z. Then both h and g are analytic in a neigh-
bourhood of B(0; 1). Let |z| = 1. Then |h(z) + g(z)| = |f(z)| < 1 = |g(z)|. Since
g(z) = z has exactly one zero inside |z| = 1, h(z) = f(z) − z has exactly one zero
inside |z| = 1, i.e., f has exactly one fixed point inside |z| = 1.
11. Schwarz’s Lemma
We recall the Maximum Modulus Principle.
Theorem 76 (Maximum Modulus Principle (First Version)). Let G be a region, and let f :
G → C be analytic. If there is some a in G such that |f(a)| ≥ |f(z)| in some neighbourhood
of a, then f is a constant function.
In other words, if f is a nonconstant analytic function on a region G, then f does not have
local maximum.
Theorem 77 (Maximum Modulus Principle (Second Version)). Let G be a bounded open
set. Let f be continuous on the closure of G, and let f be analytic on G. Then
max{|f(z)| : z ∈ G} = max{|f(z)| : z ∈ ∂G}.
Theorem 78 (Schwarz’s Lemma). Let D = {z ∈ C : |z| < 1}. Let f : D → C be analytic
with |f(z)| ≤ 1 for all z ∈ D and f(0) = 0. Then |f(z)| ≤ |z| for all z ∈ D and |f (0)| ≤ 1.
Moreover, if |f(z)| = |z| for some 0 = z ∈ D or |f (0)| = 1, then there is c ∈ C with |c| = 1
such that f(z) = cz for all z ∈ D.
Proof. Define g : D → C by g(z) = f(z)
z
if z = 0 and g(0) = f (0). Then g is analytic. Let
0 < r < 1. Since g is analytic within and on |z| = r and |g(z)| = |f(z)
z
| ≤ 1
r
for all z ∈ C
with |z| = r, it follows form the maximum modulus principle that |g(z)|leq1
r
for all z ∈ C
with |z| ≤ r. Since 0 < r < 1 is arbitrary, letting r → 1 we obtain |g(z)| ≤ 1 for all z ∈ D.
If z = 0, then |g(z)| ≤ 1 implies that |f(z)| ≤ |z|. As f(0) = 0, clearly |f(0)| ≤ |0|. Hence
|f(z)| ≤ |z| for all z ∈ D. Also, since g(0) = f (0), we have |f (0)| ≤ 1.
Suppose that |f(z0)| = |z0| for some 0 = z0 ∈ D or |g(0)| = |f (0)| = 1. Then either z0 of
0 will be a local maximum of |g|. Again it follows from the maximum modulus principle
that g is constant. Since |g(z0)| = 1 or |g(0)| = 1, we have |c| = 1. Hence f(z) = cz for all
z ∈ D.
Definition 79. A function f : G → C which preserves angle and has limz→a
|f(z)−f(z)|
|z−a|
existing is called a conformal map.
We will apply Schwarz’s Lemma to characterize the conformal maps from D to itself.
Let a ∈ C with |a| < 1. Define ϕa : B(0; 1
|a|
) → C by ϕa(z) = z−a
1−az
. Then ϕa is analytic. We
also note that D ⊂ B(0; 1
|a|
). Hence ϕa is analytic on D too.
Lemma 80. Let a ∈ C with |a| < 1. Then

26. P
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26 P. A. DABHI
(1) ϕa(0) = −a;
(2) ϕa(a) = 0;
(3) ϕ−a(z) = z+a
1+az
;
(4) ϕ−a(0) = a;
(5) ϕ−a(−a) = 0;
(6) ϕa(0) = 1 − |a|2
;
(7) ϕa(a) = (1 − |a|2
)−1
.
Proposition 81. Let a ∈ C with |a| < 1. Then ϕa is a one one map of D onto itself; the
inverse of ϕa is ϕ−a. Further, ϕa maps the boundary of D onto itself.
Proof. We first note that |z| < 1 if and only if |ϕa(z)| < 1, Indeed, |ϕa(z)| < 1 iff | z−a
1−az
| < 1
iff (z − a)(z − a) < (1 − az)(1 − az) iff |z|2
− za − za + |a|2
< 1 − az − az + |a|2
|z|2
iff
|z|2
(1 − |a|2
) < 1 − |z|2
iff |z| < 1. Thus ϕa maps D to itself. If z, w ∈ D and ϕa(z) = ϕa(w),
then we get z = w, i.e., ϕa is one one. Let w ∈ D. Take z = w+a
1+aw
. Then ϕa(z) = w. This
shows that ϕa is onto and that ϕ−a is the inverse of ϕa. Let z ∈ C. Then |ϕa(z)| = 1 iff
| z−a
1−az
| = 1 iff (z − a)(z − a) = (1 − az)(1 − az) iff |z|2
− za − za + |a|2
= 1 − az − az + |a|2
|z|2
iff |z|2
(1 − |a|2
) = 1 − |z|2
iff |z| = 1. This shows that ϕa(∂D) ⊂ ∂D. Let w ∈ ∂D. Then
z = w+a
1+aw
∈ ∂D and ϕa(z) = w. Hence ϕa(∂D) = ∂D.
The following gives an estimate of an upper bound for the derivative of an analytic function
from D to D.
Theorem 82. Suppose that f is analytic on D with |f(z)| ≤ 1 for all z ∈ D. Let a ∈ C with
f(a) = α. Then |f (a)| ≤ 1−|α|2
1−|a|2 . Further, if |f (a)| = 1−|α|2
1−|a|2 , then there is c ∈ C with |c| = 1
such that f(z) = ϕ−α(cϕa(z)) for all z ∈ D.
Proof. If |α| = 1, then by maximum modulus principle f is constant; and the inequality
holds trivially. Let |α| < 1. Define g(z) = ϕα ◦ f ◦ ϕ−a(z) for all z ∈ D. Then g is analytic
and |g(z)| ≤ 1 for all z ∈ D. Also, g(0) = 0. By Schwarz’s lemma |g(z)| ≤ |z| for all z ∈ D
and |g (0)| ≤ 1. Now
g (0) = ϕα(f(ϕ−a(0)))f (ϕ−a(0))ϕ−a(0) = ϕα(α)f (a)ϕ−a(0) = f (a)
1 − |a|2
1 − |α|2
.
Therefore |g (0)| ≤ 1 implies that |f (a)| ≤ 1−|α|2
1−|a|2 .
If |g(z)| = |z| for some 0 = z ∈ D or |g (0)| = 1, then there is c ∈ C with |c| = 1 such that
g(z) = cz for all z ∈ D. Thus if z ∈ D, then ϕα ◦ f ◦ ϕ−a(z) = cz. Since |z| < 1 if and only if
|ϕa(z)| < 1, replacing z by ϕa(z) we obtain ϕα ◦ f(z) = cϕa(z). Applying ϕ−α, the inverse
of ϕα, we get f(z) = ϕ−α(cϕa(z)) for all z ∈ D.
Corollary 83. Let f : D → D be one one analytic map from D onto itself and that f(a) = 0.
Then there is c ∈ C with |c| = 1 such that f(z) = cϕa(z) for all z ∈ D.
Proof. Since f is one one analytic map of D onto itself, by an application of Open Mapping
Theorem the inverse g of f is also analytic map of D onto itself. Therefore g ◦ f(z) = z for
all z ∈ D. Since f(a) = 0 and g(0) = a, it follows from the last theorem that |f (a)| ≤ 1
1−|a|2
and |g (0)| ≤ 1 − |a|2
. Since (g ◦ f) (a) = 1, we get
1 = |(g ◦ f) (a)| = |g (f(a))f (a)| = |g (0)||f (a)| ≤ |f (a)|(1 − |a|2
).
Thus |f (a)| ≥ 1
1−|a|2 gives |f (a)| = 1
1−|a|2 . By above theorem (with α = 0), there is c ∈ C
with |c| = 1 such that f(z) = cϕa(z) for all z ∈ D.

27. P
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COMPLEX ANALYSIS LECTURE NOTES 27
Corollary 84. A map f : D → D is analytic, one one and onto if and only if there is c ∈ C
with |c| = 1 and z0 ∈ D such that f(z) = c z−z0
1−z0z
for all z ∈ D.
Proof. Suppose that there is c ∈ C and z0 ∈ D such that f(z) = c z−z0
1−z0z
for all z ∈ D. Then
clearly f is analytic, one one and onto.
Conversely, assume that f is analytic, one one and onto. Let f(0) = α. Then |α| < 1. Define
g on D by g(z) = ϕα ◦ f(z). Since both ϕ and f are analytic, one one and onto, the map
g : D → D is analytic, one one and onto. Since g(0) = 0, by above result there is c ∈ C such
that g(z) = cz for all z ∈ D, i.e., ϕα ◦ f(z) = cz for all z ∈ D. Applying ϕ−α both the sides
we obtain
f(z) = ϕ−α(cz) =
cz + α
1 + αcz
= c
z + α
c
1 + α
c
z
.
Take z0 = −α
c
. Then z0 ∈ D and f(z) = c z−z0
1−z0z
for all z ∈ D.
We know that if f : G → C is analytic and f(z) = 0 for any z ∈ G, then f is conformal. If
|a| < 1, then ϕa(z) = 0 for any z ∈ D. Thus if f : D → D is analytic, one one and onto,
then f is conformal and so it is of the form f(z) = c z−z0
1−z0z
for all z ∈ D for some |c| = 1 and
z0 ∈ D.
Examples 85.
(1) Does there exist an analytic function f : D → D such that f(1
2
) = 3
4
and f (1
2
) = 2
3
?
Suppose that f : D → D be analytic such that f(1
2
) = 3
4
and f (1
2
) = 2
3
. Then
2
3
= |f (1
2
)| ≤
1−| 3
4
|2
1−| 1
2
|2 = 7
12
, which is a contradiction. So, no such function exist.
(2) Does there exist an analytic function f : D → D such that f(0) = 1
2
and f (0) = 1
4
?
Take any c ∈ C with |c| = 1. Define f : D → D by f(z) = ϕ−1
2
(cz). Then f is a
required function.
(3) Let f : D → D be analytic. By considering the function g : D → D defined by
g(z) = f(z)−f(0)
1−f(0)f(z)
, prove that
|f(0)| − |z|
1 + |f(0)||z|
≤ |f(z)| ≤
|f(0)| + |z|
1 − |f(0)||z|
for |z| < 1.
If |f(0)| = 1, then f is constant (by Maximum Modulus Principle). Let |a| =
|f(0)| < 1. Then the function g : D → D defined by g(z) = f(z)−a
1−af(z)
is analytic and
g(0) = 0. By Schwarz’s Lemma, |g(z)| ≤ |z| for every z ∈ D. Now f(z)−a
1−af(z)
≤ |z|
gives |f(z) − a| ≤ |z|(|1 − af(z)|) ≤ |z| + |z||a||f(z)|. Therefore ±(|f(z)| − |a|) ≤
|z| + |z||a||f(z)|. This will give the required inequality (check!!!).
(4) Let f : D → C be analytic, and let |f(z)| ≤ 1 for all z ∈ D. If f(a) = 0, then find an
estimate of upper bound of |f(b)| for b ∈ D.
Define g : D → C by g(z) = f ◦ϕ−a(z). Then g is analytic, |g(z)| ≤ 1 and g(0) = 0.
By Schwarz’s Lemma, |g(z)| ≤ |z| for all z ∈ D. Since the map ϕ−a is onto, there
is z0 ∈ D such that ϕ−a(z0) = b or ϕa(b) = z0. Now |f ◦ ϕ−a(z)| ≤ |z| implies that
|f ◦ ϕ−a(z0)| ≤ |z0|, i.e., |f(b)| ≤ |ϕa(b)|.
In particular, if f : D → D is analytic and f(1
2
) = 0, then |f(1
4
)| ≤ |ϕ1
2
(1
4
)|.

28. P
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28 P. A. DABHI
12. Convex functions
Definition 86. A map f : [a, b] → R is called convex if f(tx+(1−t)y) ≤ tf(x)+(1−t)f(t)
for every x, y ∈ [a, b] and t ∈ [0, 1].
Geometrically, if f : [a, b] → R is convex, then the curve between the points (x, f(x)) and
(y, f(y)) lies below the segment joining them.
The following gives a relation between a convex function and a convex set.
Lemma 87. A map f : [a, b] → R is convex if and only if the set A = {(x, y) : x ∈
[a, b], f(x) ≤ y} is convex.
Proof. Assume that f is convex. Let (x1, y1), (x2, y2) ∈ A. Then f(x1) ≤ y1 and f(x2) ≤ y2.
Let t ∈ [0, 1]. Since f is a convex map, f(tx1 +(1−t)x2) ≤ tf(x1)+(1−t)f(x2) ≤ ty1 +(1−
t)y2, i.e., f(tx1 + (1 − t)x2) ≤ ty1 + (1 − t)y2. Hence the point t(x1, y1) + (1 − t)(x2, y2) ∈ A.
This proves that A is convex.
Conversely, assume that A is convex. Let x1, x2 ∈ [a, b], and let t ∈ [0, 1]. Clearly,
(x1, f(x1)), (x2, f(x2)) ∈ A. Since A is convex, (tx1 + (1 − t)x2, tf(x1) + (1 − t)f(x2)) ∈ A.
Therefore f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2). This shows that f is convex.
Above theorem shows that a function f : [a, b] → R is convex if and only if the region above
the graph of f is convex set.
Proposition 88.
(1) A function f : [a, b] → R is convex if and only if for any points x1, x2, . . . , xn in [a, b]
and real numbers t1, t2, . . . , tn ≥ 0 with n
i=1 ti = 1,
f
n
i=1
tixi ≤
n
i=1
tif(xi).
(2) A subset A of C is convex if and only if for any points z1, z2, . . . , zn in A and real
numbers t1, t2, . . . , tn ≥ 0, n
i=1 tizi ∈ A.
Proof. (1) Suppose that f is convex. We prove the result by induction. Let n = 1. Let
x1 ∈ [a, b], and let t1 = 1. Then f(t1x1) = f(x1) = t1f(x2). Since f is convex, f(t1x1 +
t2x2) ≤ t1f(x1) + t2f(x2) whenever x1, x2 ∈ [a, b] and t1, t2 ≥ 0 with t1 + t2 = 1. Suppose
that the result is true for n. Let x1, x2, . . . , xn+1 ∈ [a, b], and let t1, t2, . . . , tn+1 ≥ 0 with

29. P
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COMPLEX ANALYSIS LECTURE NOTES 29
n+1
i=1 ti = 1. Then
f
n+1
i=1
tixi = f
n
i=1
tixi + tn+1xn+1
= f (1 − tn)
n
i=1
ti
1 − tn
xi + tn+1xn+1
≤ (1 − tn)f
n
i=1
ti
1 − tn
xi + tnf(xn+1) [ f is convex]
≤ (1 − tn)
n
i=1
ti
1 − tn
f(xi) + tnf(xn+1) [ Induction hypothesis]
=
n+1
i=1
tif(xi).
This proves the required inequality for any n ∈ N.
Conversely, assume that the inequality holds for any n ∈ N. Let x1, x2 ∈ [a, b], and t ∈ [0, 1].
Take t1 = t and t2 = 1 − t. Then t1, t2 ≥ 0 and t1 + t2 = 1. It follows from the hypothesis
that f(tx1 + (1 − t)x2) = f(t1x1 + t2x2) ≤ t1f(x1) + t2f(x2) = tf(x1) + (1 − t)f(x2). Hence
f is convex.
(2) The proof is left to the reader.
Theorem 89. Let f : [a, b] → R. Then f is convex if and only if f(t)−f(s)
t−s
≤ f(u)−f(t)
u−t
whenever a ≤ s < t < u ≤ b.
Proof. Assume that f is convex. Let s, t, u ∈ [a, b] with s < t < u. Then t = u−t
u−s
s + t−s
u−s
u.
Note that u−t
u−s
> 0, fract − su − s > 0 and u−t
u−s
+ t−s
u−s
= 1. Since f is convex, f(t) ≤
u−t
u−s
f(s)+ t−s
u−s
f(u). Therefore (u−s)f(t) ≤ (u−t)f(s)+(u−s)f(u). Thus f(t)−f(s)
t−s
≤ f(u)−f(t)
u−t
.
Conversely, assume that the above inequality is satisfied for a ≤ s < t < u ≤ b. Let
x, y ∈ [a, b], and let x < y. Let λ ∈ (0, 1). Set t = λx + (1 − λ)y. Then x < t < y. Then
by the hypothesis, f(t)−f(x)
t−x
≤ f(y)−f(t)
y−t
, i.e., (y − t)(f(t) − f(x)) ≤ (t − x)(f(y) − f(t)). This
implies that f(t) ≤ y−t
y−x
f(x) + t−x
y−x
f(y). Since y−t
y−x
= λ and t−x
y−x
= 1 − λ, it follows that
f(λx + (1 − λ)y) = f(t) ≤ λf(x) + (1 − λ)f(y). Thus f is a convex function.
Corollary 90. A differentiable function f : [a, b] → R is convex if and only if f is an
increasing function.
Proof. Assume that f is a convex function. Let x, y ∈ [a, b], and let x < y. Let 0 < λ < 1.
Since f is convex, f(λx + (1 − λ)y) − f(x) ≤ λf(x) + (1 − λ)f(y) − f(x), i.e., f(λx + (1 −
λ)y) − f(x) ≤ (1 − λ)(f(y) − f(x)). Thus f(λx+(1−λ)y)−f(x)
(1−λ)(y−x)
≤ f(y)−f(x)
y−x
. Taking λ → 1, we
get f (x) ≤ f(y)−f(x)
y−x
. Again using the convexity of f, we have f(λx + (1 − λ)y) − f(y) ≤
λf(x) + (1 − λ)f(y) − f(y) = λ(f(x) − f(y)). Therefore f(λx+(1−λ)y)−f(y)
λ(x−y)
≥ f(x)−f(y)
x−y
. Taking
λ → 0, we obtain f (y) ≥ f(y)−f(x)
y−x
. Thus f (x) ≤ f (y). Therefore f is increasing.

30. P
A
DABHI
30 P. A. DABHI
Conversely, assume that f is increasing. Let x, u, y ∈ [a, b] with x < u < y. By applying
the Mean Value Theorem on the intervals [x, u] and [u, y] to f, there exist r ∈ (x, u) and
s ∈ (u, y) such that f (r) = f(u)−f(x)
u−x
and f (s) = f(y)−f(u)
u−y
. Since f is an increasing function,
we have f(u)−f(x)
u−x
≤ f(y)−f(u)
u−y
. Therefore f is a convex function.
Let f : [a, b] → R be convex. Let x ∈ [a, b]. Then there is λ ∈ [0, 1] such that x = λa+(1−λ)b.
Since f is convex, f(x) ≤ λf(a)+(1−λ)f(b) ≤ (λ+1−λ) max{f(a), f(b)} = max{f(a), f(b)}.
Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar - 388120, Gu-
jarat, India
E-mail address: lightatinfinite@gmail.com