Complex analysis notes

PADABHICOMPLEX 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 ofCauchy’s theorem and Cauchy’s integral formula, Morera’s theorem, analogy between entirefunction and polynomials, open mapping theorem.

Unit II

Counting zeroes, Meromorphic functions, the argument principle, Rouche’s theorem andits 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 removablesingularity, 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, Weierstrassfactorization 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 > 0such 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) = supν(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 ifboth Re γ and Im γ are of bounded variation.

Definition 3. A continuous function γ : [a, b]→ C is called a curve.1

PA

DA

BH

I

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 acontinuous function on a compact set [a, b], the set γ is a compact subset of C. So thatthe set G = C\γ is open in C. Also, G has only one unbounded component. Since γ isbounded, 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) =∫ ba|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 ε > 0 there is δ > 0 such that whenP : 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)]

∣∣∣∣∣ < ε

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 boundedvariation. Let α, β ∈ C. Then

(1)∫ ba(αf + βg)dγ = α

∫ bafdγ + β

∫ bagdγ.

(2)∫ bafd(αγ + βσ) = α

∫ bafdγ + β

∫ bafdσ.

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 makessense.

Definition 10. If γ : [a, b] → C is a rectifiable path and f is a function defined andcontinuous 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∫γf =

∫γf(z)dz.

Proposition 11. Let γ be a rectifiable curve and suppose that f is a function continuous onγ. Then

PADABHI

COMPLEX ANALYSIS LECTURE NOTES 3

(1)∫γf = −

∫−γ f ;

(2) |∫γf | ≤

∫γ|f ||dz| ≤ V (γ) sup[|f(z)| : z ∈ γ];

(3) If c ∈ C, then∫γf(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 → Cis continuous, then for every ε > 0 there is a polygonal path Γ in G such that Γ(a) = γ(a),Γ(b) = γ(b), and |

∫Γf −

∫γf | < ε.

Examples 13.

(1) Let γ : [0, 2π]→ C be γ(t) = eint, where n ∈ N. Then∫γ

1

zdz =

∫ 2π

0

1

eintineintdt = 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 − adz =

∫ 2π

0

1

Reintineintdt =

2nπi

R.

(3) Let n,m ∈ N. Let γ : [0, 2π]→ C be γ(t) = eint. Then∫γ

zmdz =

∫ 2π

0

ei(mn)tineintdt = 0.

(4) Let a, b ∈ C, n ∈ N, and let γ : [0, 1]→ C be γ(t) = tb+ (1− t)a. Then∫γ

zndz =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. Then1

2πi

∫γ

dzz−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

0Γ′(s)

Γ(s)−ads.

Then g(0) = 0, g(1) =∫ 1

0Γ′(s)

Γ(s)−ads =∫

Γdzz−a . We prove that 1

2πig(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. Thereforeg(1) = 2kπi for some k ∈ Z or 1

2πig(1) is an integer.

Now suppose that α = 12πi

∫γ

dzz−a is not an integer. Then there is ε > 0 such that B(α; ε)∩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

dzz−a −

12πi

∫γ

dzz−a | < ε. It means that | 1

2πi

∫Γ

dzz−a − α| < ε. This

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4 P. A. DABHI

shows that the integer 12πi

∫Γ

dzz−a ∈ B(α; ε). This is a contradiction. Hence 1

2πi

∫γ

dzz−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 − ais called the index of γ with respect to the point a. It is also sometimes called the windingnumber 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 curvesdefined 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 12< 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)− adγ(1− t)

=1

2πi

∫ 0

1

1

γ(s)− adγ(s) = − 1

2πi

∫ 1

0

1

γ(s)− adγ(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 abelonging to a component of G = C\γ. Also, n(γ; a) = 0 for a belonging to the unboundedcomponent of G.

Proof. Define f : G → C by f(a) = n(γ; a). We show that f is continuous. If this isdone, then it follows that f(D) is connected for each component D of G. But since f(G) iscontained in the set of integers, it follows that f(D) is a single point. That is, f is constanton D.We know that the component of G are open. Fix a ∈ G. Take any ε > 0. Let r = d(a, γ).Take δ = min r

2, πr2ε

2V (γ). Let b ∈ G with |a − b| < δ < r

2. Since d(a, γ) = r and

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|a− b| < δ ≤ r2, we have |z − b| ≥ r

2for 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

πr2ε

2V (γ)= ε.

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 ε > 0. Take a ∈ C such that |a| > R and |z − a| > V (γ)

2πεfor every z ∈ γ.

Then

|n(γ; a)| =

∣∣∣∣ 1

2πi

∫γ

dz

z − a

∣∣∣∣ ≤ 1

2π

∫γ

|dz||z − a|

≤ 1

2π

2πε

V (γ)V (γ) = ε.

Hence n(γ, a) = 0 as ε > 0 is arbitrary. Since f is constant on U , we have n(γ; a) = 0 forevery a ∈ U .

4. Cauchy’s Theorems

Theorem 19. Let γ be a rectifiable curve. Let ϕ be a function defined and continuous onγ. For each m ∈ N, let Fm(z) =

∫γϕ(w)(z−w)−mdw, z /∈ γ. Then each Fm is analytic

in C\γ and that F ′m(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 suchthat |ϕ(w)| ≤ M for every w ∈ γ. First we show that Fm is continuous. Take any ε > 0

and a ∈ C\γ. Let r = d(a, γ). Take δ = min r2, rm+1εmM2m+1V (γ)

. Let z ∈ G = C\γ with

|z − a| < δ. Since d(a, γ) = r and |z − a| < δ ≤ r2, we have |z −w| ≥ r

2for every w ∈ γ.

Now

|Fm(z)− Fm(a)| =

∣∣∣∣∫γ

ϕ(w)

(w − z)mdw −

∫γ

ϕ(w)

(w − a)mdw

∣∣∣∣=

∣∣∣∣∫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≤ ε.

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−kdw are continuous on C\γ. Applying limit z → a we get

limz→a

Fm(z)− Fm(a)

z − a= m

∫γ

ϕ(w)

(w − a)m+1dw = mFm+1(a).

Therefore Fm is differentiable at every a ∈ C\γ and F ′m(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) = 0for every w ∈ C\G, then for each a ∈ G\γ

n(γ; a)f(a) =1

2πi

∫γ

f(z)

z − adz.

Proof. Define ϕ : G × G → C by ϕ(z, w) = f(z)−f(w)z−w if z 6= w and ϕ(z, z) = f ′(z). Then

ϕ is continuous and for each w ∈ G the map z 7→ 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 inC. Also, H ∪G = C by hypothesis. Define g : C→ C by g(z) =

∫γϕ(z, w)dw if z ∈ G and

g(z) =∫γ(w − z)−1f(w)dw if z ∈ H. If z ∈ G ∩H, then

g(z) =

∫γ

f(w)− f(z)

w − zdw

=

∫γ

f(w)

w − zdw − f(z)

∫γ

dw

w − z

=

∫γ

f(w)

w − zdw − 2πif(z)n(γ; z)

=

∫γ

f(w)

w − zdw = 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 isanalytic on both G and H and hence on C, i.e., g is an entire function. The open set Hcontains 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 forgiven ε > 0 there is R1 > R > 0 such that |w−z|−1 < ε

sup[|f(θ)|:θ∈γ]V (γ) for every z ∈ C with

|z| > R1. Take any z ∈ C with |z| > R1. Then |g(z)| ≤ ε. It means that lim|z|→∞ g(z) = 0.The above shows that g is bounded on z ∈ C : |z| > R1. Since g is continuous andz ∈ C : |z| ≤ R1 is compact, g is bounded on z ∈ C : |z| ≤ R1. So, g is a bounded

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function. By Liouville’s theorem the function g is constant. Since lim|z|→∞ g(z) = 0, g is azero map. If a ∈ G\γ, then

0 =

∫γ

f(z)− f(a)

z − adz =

∫γ

f(z)

z − adz − f(a)

∫γ

dz

z − a

=

∫γ

f(z)

z − adz − f(a)2πin(γ; a).

Therefore n(γ; a)f(a) = 12πi

∫γf(z)z−adz 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 ∈ C\G. If a lies in the interior of γ, then f(a) = 12πi

∫γf(z)z−adz.

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 thatn(γ1;w) + n(γ2, w) + · · ·+ n(γm, w) = 0 for all w ∈ C\G. If a ∈ G\ ∪mi=1 γi, then

f(a)m∑k=1

n(γk, a) =m∑k=1

1

2πi

∫γk

f(z)

z − adz.

Proof. The proof is same as above. Define g(z) =∑n

k=1

∫γkϕ(z, w)dw if z ∈ G and g(z) =∑n

k=1

∫γk

(w − z)−1f(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, andlet f : G → C be analytic. Let γ1, γ2, . . . , γm be closed rectifiable curves in G such thatn(γ1;w) + n(γ2;w) + · · ·+ n(γk;w) = 0 for every w ∈ C\G. Then

m∑k=1

∫γk

f = 0.

Proof. Fix any a ∈ G\∪mk=1γk. 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 anopen subset of C, and let f : G → C be analytic. Let γ1, γ2, . . . , γm be closed rectifiablecurves in G such that n(γ1;w) + n(γ2;w) + · · · + n(γk;w) = 0 for every w ∈ C\G. Ifa ∈ G\ ∪mk=1 γk, then

f (k)(a)m∑j=1

n(γj, a) = k!m∑j=1

1

2πi

∫γj

f(z)

(z − a)k+1dz.

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 − adz, a ∈ G\ ∪mk=1 γk

Fix j ∈ a, 2, . . . ,m. Since the map f is continuous on γj, the map ϕj(a) =∫γj

f(z)z−adz is

analytic on G\γj and ϕ′j(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+1dz 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) = 12πi

m∑j=1

∫γj

f(z)z−adz, we obtain f (k)(a)

m∑j=1

n(γj, a) = k!m∑j=1

12πi

∫γj

f(z)(z−a)k+1dz for

every a ∈ G\ ∪mi=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 IntegralFormula for derivatives, we get 0 = f (k)(a)n(γ; a) = k!

2πi

∫γ

1(z−a)k+1dz for every k ∈ N. Hence∫

γ1

(z−a)ndz = 0 for every integer n ≥ 2.

Exercise 26. Let G be a simply connected region. If f : G → C is analytic, then f has aprimitive in G.

Theorem 27 (Analytic extension of a the exponential solution). Let G be a simply connectedregion and f : G→ C be analytic with f(z) 6= 0 for every z ∈ G. Then there is an analyticfunction g : G→ C such that eg = f .

Proof. Since f(z) 6= 0 for every z ∈ G, the function f ′

fis analytic on G. So f ′

fadmits an

antiderivative on G. Let g be the antiderivative of f ′

fon 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′ = egg′ = hg′. Nowddz

(fh) = hf ′−h′f

h2= hf ′−hg′f

h2= hf ′−hf ′

h2= 0. Therefore f

his a constant function. Therefore

there is c ∈ C such that f = ch. Since f(z) 6= 0 for every z ∈ G, c 6= 0. There is a ∈ Csuch that ea = c. Thus f = eaeg = eg+a. So the function g1 = g + a is the required analyticfunction.

Example 28. Let n ∈ N. Evaluate∫γ( zz−1

)ndz, 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 thatf (n−1)(1) = n!. By Cauchy’s Integral Formula

n! = f (n−1)(1) =(n− 1)!

2πi

∫γ

f(z)

(z − 1)ndz =

(n− 1)!

2πi

∫γ

(z

z − 1

)ndz.

Thus∫γ( zz−1

)ndz = 2πin.

Verify that∫σ( zz−1

)ndz = 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 acircle |z| = R. If 0 ≤ r < R, then

f(reiθ) =1

2π

∫ 2π

0

(R2 − r2)f(Reiϕ)

R2 − 2Rr cos(θ − ϕ) + r2dϕ.

The map P (r, θ) = (R2−r2)R2−2Rr cos(θ−ϕ)+r2

is known as Poisson kernel.

Proof. If r = 0, then 12π

∫ 2π

0(R2−r2)f(Reiϕ)

R2−2Rr cos(θ−ϕ)+r2dϕ = 1

2π

∫ 2π

0f(Reiϕ)dϕ = 1

2πi

∫|z|=R

f(z)zdz =

f(0). Let 0 6= a = reiθ ∈ B(0;R). Then 0 < r < R. By using Cauchy’s Integral Formula we

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have f(a) = 12πi

∫|z|=R

f(z)z−adz. 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 − adz − 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)iReiϕdϕ

=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) + r2dϕ.

Exercise 30. If the power series∑∞

n=0 an(z−a)n converges in B(a;R) for some R > 0, thenit converges absolutely for every z ∈ B(a;R).

Theorem 31 (Parseval’s Identity). Let f be analytic on |z| < R. If a = reiθ ∈ z ∈ C :|z| < R. Then

1

2π

∫ 2π

0

|f(reiθ)|2dθ =∞∑n=0

|an|2rn,

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 6= a = reiθ ∈ B(a;R). Then f(reiθ) =∑∞

n=0 anrneinθ and

so f(reiθ) =∑∞

n=0 anrne−inθ. Thus |f(reiθ)|2 =

(∑∞n=0 anr

neinθ) (∑∞

m=0 anrne−imθ

). Since

both the power series converges absolutely, the product is convergent. In fact, the productconverges uniformly as a function of θ. Therefore we may apply term by term integration.

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10 P. A. DABHI

Applying term by term integration,

∫ 2π

0

|f(reiθ)|2dθ =

∫ 2π

0

(∞∑n=0

anrneinθ

)(∞∑m=0

anrne−imθ

)dθ

=

∫ 2π

0

(∞∑n=0

∞∑m=0

anamrm+nei(n−m)θ

)dθ

=∞∑n=0

∞∑m=0

anamrm+n

(∫ 2π

0

ei(n−m)θdθ

)

= 2π∞∑n=0

|an|2r2n,

as∫ 2π

0ei(n−m)θdθ = 0 if n 6= m and

∫ 2π

0ei(n−m)θdθ = 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−iθdθ,

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 anrneinθ. So, f(a+ reiθ) =

∑∞n=0 anr

ne−inθ. Since the series f(a + reiθ) =∑∞n=0 anr

ne−inθ converges absolutely, it converges uniformly in the variable θ. So, we maytake term by term integration. Now

1

2πi

∫γ

f(z)

(z − a)2dz =

1

2πi

∫ 2π

0

∑∞n=0 anr

ne−inθ

r2e2iθreiθdθ

=1

2π

∫ 2π

0

(∞∑n=0

anrn−1e−i(n+1)θ

)dθ

=1

2π

∞∑n=0

anrn−1

(∫ 2π

0

e−i(n+1)θdθ

)= 0. [∵

∫ 2π

0

e−i(n+1)θdθ = 0 for all n ∈ N ∪ 0]

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Thus

f ′(a) = f ′(a) + 0

=1

2πi

∫γ

f(z)

(z − a)2dz +

1

2πi

∫γ

f(z)

(z − a)2dz

=1

2πi

∫γ

f(z) + f(z)

(z − a)2dz

=1

πi

∫γ

Ref(z)

(z − a)2dz

=1

πi

∫ 2π

0

Ref(a+ reiθ)

r2e2iθreiθidθ

=1

πi

∫ 2π

0

Ref(a+ reiθ)e−iθdθ

=1

πr

∫ 2π

0

Pa,r(θ)e−iθdθ.

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 functionsuch that

∫Tf(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 beanalytic 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 6= z0. Since∫Tf = 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 ε > 0. Since f is continuous at z0, there is δ > 0 such that |f(w) − f(z0)| < εwhenever |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|

≤ ε.

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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 off . It follows from the Cauchy’s Integral Formula that F ′ = f is also analytic on G. Hencethe 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 suchthat p(z) = (z − a)q1(z). If a is a zero of q1, then there is a polynomial q2 such thatq1(z) = (z − a)q2(z). Hence p(z) = (z − a)2q2(z). Continuing this way we get a polynomialq with q(a) 6= 0 and m ∈ N such that p(z) = (z − a)mq(z). Also, deg p = m + deg q. Thismotivates 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 ∈ Gwith f(a) = 0. Then a is a zero of multiplicity m ≥ 1 if there is an analytic functiong : G→ C such that g(a) 6= 0 and f(z) = (z − a)mg(z), z ∈ G.

Theorem 36. Let G be a region, and let f : G → C be analytic. Then the followingstatements 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 > 0be such that B(a;R) ⊂ G. As a is a limit point of Z(f) and f is continuous, we havef(a) = 0. Let n be a positive integer such that f(a) = f ′(a) = · · · = f (n−1)(a) = 0but f (n)(a) 6= 0. Since f is analytic on G and hence on B(a;R), it has a Taylor se-

ries expansion f(z) =∑∞

k=0f (k)(a)k!

(z − a)k, z ∈ B(a;R). Since f(a) = f ′(a) = · · · =

f (n−1)(a) = 0 and f (n)(a) 6= 0, we have f(z) =∑∞

k=nf (k)(a)k!

(z − a)k, z ∈ B(a;R), i.e.,

f(z) = (z − a)n∑∞

k=nf (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!6= 0. Since g is continuous and g(a) 6= 0, there is r > 0 (0 < r < R) such

that g(z) 6= 0 for every z ∈ B(a; r). As a is a limit point of Z(f), there is a point b suchthat 0 < |b − a| < r and f(b) = 0. Also, f(b) = (b − a)ng(a) 6= 0. This is a contradiction.Therefore our assumption that there is some n ∈ N such that f (n)(a) 6= 0 is false. Hencef (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 6= ∅.We prove that the set A is both open and closed in G. Then the connectedness of G willimply that A = G and then it will follow that f ≡ 0. Let (zk) be a sequence in A convergingto z ∈ G. Since each f (n) is continuous, it follows that f (n)(z) = limk→∞ f

(n)(zk) = 0. Hencez ∈ 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=0f (n)(a)n!

(z − a)n, z ∈

PA

DA

BH

I


B(a;R). Since each f (n)(a) = 0, f(z) = 0 for every z ∈ B(a;R). This shows that f (n)(z) = 0for every n ∈ N ∪ 0 and for every z ∈ B(a;R). Therefore B(a;R) ⊂ A. This proves thatA is open in G.

Remark 37.

(1) The condition that “G is connected” is necessary in above theorem. LetG = B(0; 1) ∪ B(3; 1). Define f(z) = 0 if z ∈ B(0; 1) and f(z) = 1 if B(3; 1). Thenf is analytic on G. The set z ∈ G : f(z) = 0 has a limit point in G but f is notidentically zero.

(2) The condition that “the set of zeros of f has a limit point in G” isnecessary. Define f : B(1; 1)→ C by f(z) = sin(1

z). Then f is analytic. Note that

sin(1z) = 0 if and only if z = 1

nπfor some n ∈ N. Hence Z(f) = z ∈ B(1; 1) : f(z) =

0 = 1nπ

: n ∈ N. The set Z(f) has a limit point 0 which is not in G. Also, f isnot identically zero.

Corollary 38. Let G be a region, and let f, g : G → C be analytic function. If the setz ∈ 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 satisfyingf( 1

n) = 1

1+nfor all n ≥ 2.

Let f : D→ C be an analytic function such that f( 1n) = 1

1+nfor all n ≥ 2. Define

g : D→ C by g(z) = zz+1

. Then g is analytic on D. Also, g( 1n) = 1

1+n. Then f − g is

analytic on a region D and 1n

: n ≥ 2 ⊂ z ∈ D : f(z) = g(z). We see that 0 ∈ Dis a limit point of z ∈ D : f(z) = g(z). Hence f(z) = z

1+zfor all z ∈ D.

(2) Is there an analytic function f : D→ C which vanish only on 1n

: 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 1n

: 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 6= 0. Then there is a ∈ G such that f(a) 6= 0. By continuity of f

there is R > 0 such that f(z) 6= 0 for every z ∈ B(a;R) ⊂ G. Since fg = 0 on G, wehave g(z) = 0 for every z ∈ B(a;R). Now B(a;R) ⊂ z ∈ G : g(z) = 0 has a limitpoint 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 showthat either f is constant or g = 0.

Suppose that g 6= 0. Then there is a ∈ G such that g(a) 6= 0. By continuity ofg, there is R > 0 such that g(z) 6= 0 for every z ∈ B(a;R) ⊂ G. Since g(z) 6= 0 forevery z ∈ B(a;R), then map 1

gis analytic on B(a;R). As fg is analytic on G, the

map f = fg 1g

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 andB(a;R) ⊂ z ∈ G : f(z)− c = 0. Therefore the zeros of f − c has a limit point a inG. This proves that f(z) = c for every z ∈ G. i.e., f is a constant map.

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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 notidentically zero. Then for each a ∈ G with f(a) = 0, there is n ∈ N and an analyticg : G→ C with g(a) 6= 0 such that f(z) = (z − a)ng(z), z ∈ G.

Proof. Let a ∈ G be such that f(a) = 0. Since f is not identically zero, by above theoremthere is m ∈ N such that f (m)(a) 6= 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)−nf(z) if

z 6= a and g(a) = f (n)(a)n!

. Clearly, g is analytic on G\a and g(a) 6= 0. To show g analyticon G, we not show that g is differentiable at a. Let R > 0 be such that B(a;R) ⊂ G. Let

f(z) =∑∞

k=0f (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=nf (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 6= a, then g(z) = (z − a)−nf(z) implies f(z) =(z − a)ng(z). If z = a, then f(a) = 0 and (z − a)ng(z) = 0 at a. Hence f(z) = (z − a)ng(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. Leta ∈ C with f(a) = 0. Then there is R > 0 such that B(a : R) ⊂ G and f(z) 6= 0 for everyz ∈ B(a;R)\a.

Proof. By above result there is an analytic function g : G→ C with g(a) 6= 0 and a naturalnumber n such that f(z) = (z − a)ng(z), z ∈ G. Since g is continuous and g(a) 6= 0,there is R > 0 such that g(z) 6= 0 for every z ∈ B(a;R). Let z ∈ B(a;R)\a. Thenf(z) = (z − a)ng(z) 6= 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 γbe a simple closed curve in G such that interior of γ lies in G. Then the number of zeros off inside γ are finite.

Proof. Suppose that the zeros of f inside γ are infinite. Then the set of zeros of f inside isa bounded infinite subset of C (as interior of γ is bounded). Let z0 be a limit point of thisset. Then either z is inside γ or it is on γ. In any case, a limit point z0 of zeros of f is in Gand 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 besuch that f(a) = 0. Show that a is a zero of f of multiplicity m if and only iff(a) = f ′(a) = · · · = f (m−1)(a) = 0 but f (m)(a) 6= 0.

PADABHI


(2) Let f, g : G → C be analytic, let a ∈ G be such that f(a) = g(a) = 0. If a is a zeroof f of multiplicity m and a is a zero of g of multiplicity n, then show that a is a zeroof fg of multiplicity m+ n .

Theorem 46 (Counting Zero Principle). Let G be a region, and let f be analytic on G withzeros a1, a2, . . . , am repeated according to multiplicity. Let γ be a closed rectifiable curve inG which does not pass through any zero of f and that n(γ;w) = 0 for all w ∈ C\G. 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 afactorization 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) 6= 0 for any z ∈ G. Observe that f ′(z)f(z)

=∑m

k=11

z−ak+ g′(z)

g(z).

Since g is analytic and nowhere vanishing g′

gis analytic. Since n(γ;w) = 0 for all w ∈ C\G,

it follows from the Cauchy’s theorem that∫γg′

g= 0. Hence 1

2πi

∫γf ′

f=∑m

k=11

2πi

∫γ

1z−ak

dz =∑mk=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 closedrectifiable curve in G which does not pass through any zero of f(z)−α and that n(γ;w) = 0

for all w ∈ C\G. If a1, a2, . . . , am are zeros of f(z)−α, then 12πi

∫γ

f ′(z)f(z)−αdz =

∑mk=1 n(γ; ak).

Proof. Take g(z) = f(z)− α and apply the above theorem.

Let a, b ∈ C, and let b 6= 0. Let z ∈ C. Then Im(z−ab

)> 0 iff Im

(x−a1+i(y−a2)

b21+b22(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 allthe 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 6= K ∈ C. Note that p′(z)p(z)

= 1z−α1

+ 1z−α2

+ · · ·+ 1z−αn .

Let Im(z−ab

)< 0 be a half plane in which α1, α2, . . . , αn are lying. Therefore Im

(αk−ab

)< 0

for 1 ≤ k ≤ n. Let z0 be a zero of p′(z) not lying in this half plane, i.e., Im(z0−ab

)≥ 0.

Now Im(z0−αkb

)= Im

(z0−ab

)− Im

(αk−ab

)> 0 for 1 ≤ k ≤ n. Since Imz = −(Imz−1)|z|2

for all 0 6= z ∈ C, we have Im(z0−αkb

)= −Im

(z0−αkb

)−1 | z0−αkb|2 > 0. Now Im

(bp′(z0)p(z0)

)=

Im(∑n

k=1b

z0−αk

)=∑n

k=1 Im(

bz0−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]→ Cis 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 6= w and ϕ(z, z) = f ′(z). Then we know that ϕ

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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 suchthat |ϕ(z, w)| ≤M for every z, w ∈ γ. In other words, |f(z)− f(w)| ≤M |z−w| for everyz, 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))− αdγ(t) =

1

2πi

∫γ

f ′(z)

f(z)− αdz

=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 ofmultiplicity m at a, there there exist ε > 0 and δ > 0 such that for |ξ − α| < δ, the equationf(z) = ξ has exactly m simple roots in B(a; ε).

Proof. Since the zeros of an analytic function are isolated, there is ε > 0 such that f(z) = αhas no solution in 0 < |z − a| < 2ε and f ′(z) 6= 0 in 0 < |z − a| < 2ε. Let γ(t) = a + εe2πit,0 ≤ t ≤ 1. Put σ = f γ. Then α /∈ σ. So, there is δ > 0 such that B(α; δ) ∩ σ = ∅.Therefore B(α; ε) lies in a component of C\σ, i.e., if |ξ − α| < δ, then n(σ; ξ) = n(σ;α).

Note that m = 12πi

∫γ

f ′(z)f(z)−αdz. Now,

m =1

2πi

∫γ

f ′(z)

f(z)− αdz =

1

2πi

∫ 1

0

f ′(γ(t))

f(γ(t))− αdγ(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 off(z) = ξ in B(a; ε). Since f ′(z) 6= 0 for 0 < |z − a| < ε, each of these roots is simple.

Theorem 51 (Open Mapping Theorem). Let G be a region, and let f : G → C be anonconstant analytic function. Then f is an open map.

PADABHI


Proof. Let U be an open subset of G, and let a ∈ U . Take α = f(a). Then by f(z) − αhas a zero in B(a;R) for any R > 0 with B(a;R) ⊂ U . Then by above theorem, there existε > 0 and δ > 0 such that if |ξ − f(a)| < δ, then ξ ∈ f(B(a; ε)), i.e., f(a) ∈ B(f(a); δ) ⊂f(B(a; ε)) ⊂ 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) 6= 0 forany z ∈ G.

Suppose that f ′(a) = 0 for some a. Then a is a zero of f(z)− f(a) of multiplicityat least 2. By above result there is δ > 0 and ε > 0 such that if |ξ − f(a)| < δ, thenf(z) = ξ has at least two solutions in B(a; ε). 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 therange of f , then f−1 : Ω→ C is analytic.

By the Open Mapping Theorem f is an open map. Therefore Ω = f(G) is anopen subset of C. The map f : G → Ω is one one and onto and has an inversef−1 : Ω → G ⊂ C. Since f is an open map, the map f−1 is continuous. We needto check the differentiability of f−1 at every point of Ω. Let w = f(z) ∈ Ω. Byabove exercise, f ′(w) 6= 0. Let (wn) be a sequence in Ω\w converging to w. Thenwn = f(zn) for some zn ∈ G and also zn 6= 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

limwn→w

f−1(wn)− f−1(w)

wn − w= lim

f(zn)→f(z)

zn − zf(zn)− f(z)

= limzn→z

zn − zf(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 isδ > 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 thatg(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 off 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 aand 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 , fis analytic in B(a;R)\a for some R > 0. Define g : B(a;R) → C by g(z) = (z − a)f(z)

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18 P. A. DABHI

if z 6= 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 withT . If a /∈ ∆, then

∫Tg = 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 ε > 0 there is δ > 0 such that|g(z) − g(a)| < ε

`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 apolygon and n(P,w) = 0 for every w ∈ C\B(a;R). By Cauchy theorem,

∫Pg = 0. Now

|∫Tg| = |

∫T1g +

∫Pg| ≤ |

∫T1g| ≤ ε

`` = ε. Thus

∫Tg = 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]. Byabove

∫Tig = 0 for i = 1, 2, 3. Now

∫Tg =

∫T1g +

∫T2g +

∫T3g = 0. Hence it follows from

Morera’s theorem that g is analytic on B(a;R). Since g(a) = 0, there is an analytic functionh : 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 removablesingularity of f .

Proposition 56. Let a be an isolated singularity of f . If a is a pole of f , then a is aremovable singularity of 1

f.

Proof. Since a is a pole of f , limz→a |f(z)| = ∞. This implies that limz→a1

f(z)= 0 and

hence limz→a(z − a) 1f(z)

= 0. Since a is pole of f , f is analytic in a neighbourhood of a. As

limz→a |f(z)| =∞, f(z) 6= 0 for any z in some deleted neighbourhood of a, say, B(a; δ)\a.Thus 1

fis analytic in B(a; δ)\a and limz→a(z − a) 1

f(z)= 0. Therefore by above a is a

removable singularity of 1f.

Proposition 57. Let G be a region, and let a ∈ G. Let f be analytic on G\a, and let a besingularity of f . Then a is a pole of f if and only if there is an analytic function g : G→ Cwith g(a) 6= 0 and m ∈ N such that f(z) = g(z)

(z−a)mfor every z ∈ G\a.

Proof. Suppose that there is an analytic function g : G→ C with g(a) 6= 0 and m ∈ N such

that f(z) = g(z)(z−a)m

for every z ∈ G\a. Since g(a) 6= 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 aremovable 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→a1

f(z)= limz→a h(z),

there is an analytic k : B(a;R)→ C with k(a) 6= 0 and m ∈ N such that h(z) = (z−a)mk(z)for every z ∈ B(a;R). Define g : G → C by g(z) = k(z) if z = a and g(z) = (z − a)−mf(z)if z 6= 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) 6= 0 and m ∈ N such that f(z) = g(z)(z−a)m

for z 6= 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 onG\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) 6= 0, g(a2) 6= 0 such that f(z) = g(z)(z−a1)m1 (z−a2)m2

for every z ∈ G\a1, a2.

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Since a1 is a pole of f of order m1, there is an analytic function h1 : G\a2 → Cwith h1(a1) 6= 0 such that f(z) = h1(z)

(z−a1)m1for every z ∈ G\a1, a2. Then f(z) =

(z−a2)m2h1(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)m2h1(z)

is analytic on G\a1 and g1(a1) 6= 0. Similarly, since a2 is a pole of f of order m2, there

is an analytic function g2 : G\a2 → C with g2(a2) 6= 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) 6= 0,

g(a2) 6= 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 removablesingularity of f .

Proof. As f is bounded in 0 < |z − a| < δ, there is M > 0 such that |f(z)| ≤ M for everyz ∈ C with 0 < |z − a| < δ. Since f is analytic in 0 < |z − a| < δ, it has a Laurent seriesexpansion

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+1dz for every m ∈ N. Let m ∈ N. Then

|bm| =

∣∣∣∣∣ 1

2πi

∫Cρ

f(z)

(z − a)−m+1dz

∣∣∣∣∣ ≤ 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 actuallyvalid in |z − a| < δ and hence f is analytic at a. If f(a) 6= a0, then in order to make fanalytic 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 removablesingularity of f .

Exercise 62. If a is a removable singularity of f , then f is bounded and analytic in a deletedneighbourhood 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). Sincef = 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 onlyif f is analytic and bounded in some deleted neighbourhood of a.

Theorem 63 (Casorati - Weierstrass Theorem). Let a be an essential singularity of f . Thenfor any δ > 0 the set f(z) : 0 < |z − a| < δ is dense in C (OR Given c ∈ C, ε > 0 andδ > 0, there is z ∈ C with 0 < |z − a| < δ such that |f(z)− c| < ε).

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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| < δis dense in C.Suppose that there is δ > 0 with 0 < δ < R such that f(z) : 0 < |z − a| < δ is not densein C. Then there is c ∈ C and ε > 0 such that B(c; ε)

⋂f(z) : 0 < |z − a| < δ = ∅, i.e.,

|f(z)− c| ≥ ε for every z ∈ C with 0 < |z−a| < δ. Define g : B(a; δ)\ → C by g(z) = 1f(z)−c .

Then g is analytic in 0 < |z − a| < δ and |g(z)| = 1|f(z)−c| ≤

1ε. 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) 6= 0 or g(a) = 0. Letg(a) 6= 0. Since g(a) 6= 0 and g is continuous, g(z) 6= 0 for every z in a neighbourhood of a,say, B(a; η) and also 1

gis 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 bea 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 suchthat 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 with0 < |zn− a| < rn such that |f(zn)−w| < rn. Then the sequence (zn) converges to a and thesequence (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 underf 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 anessential singularity of f is called Picard’s exceptional point.The function f(z) = e

1z has an essential singularity at 0. The number 0 is the Picard’s

exceptional point for f .Let 0 6= w = reiθ ∈ C. Then there is x ∈ R such that ex = r. But then ex+iθ = exeiθ =reiθ = w. Take z = x + iθ. For every n ∈ N take zn = 1

z+2nπi. Then zn → 0 as n→∞ and

e1zn = ez+2nπi = w. Hence for every 0 6= w ∈ C, the equation e

1z = 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 anisolated singularity of f .

(1) a is a removable singularity of f if and only if f is analytic and bounded in a deletedneighbourhood of a.

(2) a is a pole of f if and only of limz→a |f(z)| =∞.

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(3) If a is an essential singularity of f , then f wanders through the complex plane in aneighbourhood 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(1z).

(2) a pole of f if 0 is a pole of f(1z).

(3) an essential singularity of f if 0 is an essential singularity of f(1z).

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’sIntegral Formula

|f (n)(0)| =

∣∣∣∣ n!

2πi

∫|z|=r

f(z)

zn+1dz

∣∣∣∣ ≤ 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 atmost 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 suchthat M ≤ u(x, y) for all (x, y) ∈ R2. Define g : C→ C by g(z) = exp(−f(z)). Then gis 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 Theoremg 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 isconstant.

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 removablesingularity 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(1z)| ≤M for 0 < |z| < δ.

Let z ∈ C with |z| > 1δ. Then |1

z| < δ and so |f(z)| ≤M . Since f is continuous, it is

bounded on the compact set B(0; 1δ), say, it is bounded by K. Thus |f(z)| ≤M +K

for every z ∈ C. By Liouville’s Theorem, the function f is constant.

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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 apolynomial of degree m.

Let f be a polynomial of degree m, say, f(z) = a0 + a1z+ · · ·+ amzm and am 6= 0.

Then f(1z) = a0 + a1

z+ · · · + am

zm. If n ∈ N ∪ 0, then znf(1

z) has a singularity at 0

and zmf(1z) 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 zmf(1

z) exists, say, c. This implies that there is

δ > 0 such that |zmf(1z) − c| ≤ 1 for all z ∈ C with 0 < |z| < δ. Let z ∈ C with

|z| > 1δ. Then 1

|z|m |f(z)| ≤ (1 + |c|), i.e., |f(z)| ≤ (|c| + 1)|z|m for all z ∈ C with

|z| > 1δ. 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 zkf(1z) will have a removable singularity at 0,

and hence ∞ will become a pole of f order k < m. This is a contradiction. Hencethe degree of f is m.

Examples 67.

(1) (Fourier series Expansion is associated with the Laurent series expansion of analyticf in an annular region z ∈ C : R1 < |z| < R2, where 0 < R1 < 1 < R2). Let0 < 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=−∞ anz

n, R1 < |z| < R2,

where an = 12πi

∫|z|=1

f(z)zn+1dz for all n ∈ Z. Let n ∈ Z. Then

an =1

2πi

∫|z|=1

f(z)

zndz =

1

2πi

∫ 2π

0

f(eiθ)

ei(n+1)θieiθdθ

=1

2π

∫ 2π

0

f(eiθ)e−inθdθ.

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

0F (t)e−intdt 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 Gexcept for poles is called a meromorphic function.

Theorem 69 (Argument Principle). Let G be a region, and let f be meromorphic in Gwith zeros z1, z2, . . . , zn and poles p1, p2, . . . , pm counted according to their multiplicities. Ifγ is a closed rectifiable curve in G which does not pass through any zero or pole of f and ifn(γ;w) = 0 for every w ∈ C\G, 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) 6= 0 for any z ∈ G. Since g is analytic on

PA

DA

BH

I


G and nowhere vanishing, the function g′

gis analytic on G. As n(γ;w) = 0 for every

w ∈ C\G, it follows from the Cauchy’s Theorem that∫γg′

g= 0. Observe that f ′(z)

f(z)=∑n

k=11

z−zk−∑m

k=11

z−pk+ g′(z)

g(z)for every z ∈ γ (In fact, f ′

fis analytic in a neighbourhood

of γ). Therefore 12πi

∫γf ′

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 eacha ∈ γ there is an open disc B(a; δa) which contains no zero or pole of f . Then the branchof 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 ε > 0 such that for each a ∈ γ we candefine a branch of log f(z) in B(a; ε). Since γ is uniformly continuous, there is a partition0 = t0 < t1 < · · · < tk = 1 of [0, 1] such that γ(t) ∈ B(γ(tj−1); ε) for tj−1 ≤ t ≤ tj. Let`j be the branch of log f defined on B(γ(tj−1); ε) 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 imaginarywe 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, . . . , pmand z1, z2, . . . , zn counted according to their multiplicities. If g is analytic on G and if γ isa closed rectifiable curve in G which does not pass through any zero or pole of f such thatn(γ;w) = 0 for every w ∈ C\G, then

1

2πi

∫γ

gf ′

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 totheir multiplicities, there is an analytic function h : G → C with h(z) 6= 0 for every z ∈ Gsuch that f(z) = (z−z1)···(z−zn)

(z−p1)···(z−pm)h(z) for every z ∈ G\p1, p2, . . . , pm. Since g and h′

hare

analytic on G and n(γ;w) = 0 for every w ∈ C\G, it follows from the Cauchy’s Theorem

that∫γg h′

h= 0. Observe that g(z)f

′(z)f(z)

=∑n

k=1g(z)z−zk

−∑m

k=1g(z)z−pk

. Applying integration

over γ and using Cauchy’s Integral Formula, we obtain 12πi

∫γg f′

f=∑n

k=1 g(zk)n(γ; zk) −∑mk=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, countedaccording to their multiplicities, respectively. We also note that n(γ; z) = 1 for every zero

and pole. It follows form the argument principle, that∫γf ′

f= −4πi.

Exercise 72. Let G be open in C, and let γ be a rectifiable curve in G with initial point α andend point β. If f : G→ C is continuous with primitive F : G→ C, then

∫γf = F (β)−F (α).

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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. IfZf , Zg (Pf , Pg) are the number of zeros (poles) of f and g inside γ counted according to theirmultiplicities 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 meromorphicfunction f

gmaps a neighbourhood of γ to Ω = C\[0,∞). Let ` be the branch of logarithm

defined on Ω. Then `(fg) 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 wereassumed 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|→∞[ a0zn

+ a1zn−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| = 32

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 Theoremf(z) has 4 zeros inside |z| = 2 as g(z) = −z4 has 4 zeros inside |z| = 2.

Now take h(z) = −6z. If |z| = 32, then |f(z) + h(z)| = |z4 + 1| ≤ (3

2)4 + 1 < 9 =

|h(z)|. Since h(z) = −6z has one zero inside |z| = 32, 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 nsolutions 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 nzeros inside |z| = 1, it follows from Rouche’s theorem that f(z) = 0 has n solutionsinside |z| = 1.

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(3) Let f be analytic in a neighbourhood of B(0; 1). If |f(z)| < 1 on |z| = 1, then showthat 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)|. Sinceg(z) = z has exactly one zero inside |z| = 1, h(z) = f(z) − z has exactly one zeroinside |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 neighbourhoodof 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 havelocal maximum.

Theorem 77 (Maximum Modulus Principle (Second Version)). Let G be a bounded openset. 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 analyticwith |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 6= z ∈ D or |f ′(0)| = 1, then there is c ∈ C with |c| = 1such that f(z) = cz for all z ∈ D.

Proof. Define g : D → C by g(z) = f(z)z

if z 6= 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

rfor all z ∈ C

with |z| = r, it follows form the maximum modulus principle that |g(z)|leq 1r

for all z ∈ Cwith |z| ≤ r. Since 0 < r < 1 is arbitrary, letting r → 1 we obtain |g(z)| ≤ 1 for all z ∈ D.If z 6= 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 6= z0 ∈ D or |g(0)| = |f ′(0)| = 1. Then either z0 of0 will be a local maximum of |g|. Again it follows from the maximum modulus principlethat g is constant. Since |g(z0)| = 1 or |g(0)| = 1, we have |c| = 1. Hence f(z) = cz for allz ∈ 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−a1−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

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26 P. A. DABHI

(1) ϕa(0) = −a;(2) ϕa(a) = 0;

(3) ϕ−a(z) = z+a1+az

;(4) ϕ−a(0) = a;

(5) ϕ−a(−a) = 0;(6) ϕ′a(0) = 1− |a|2;(7) ϕ′a(a) = (1− |a|2)−1.

Proposition 81. Let a ∈ C with |a| < 1. Then ϕa is a one one map of D onto itself; theinverse 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−a1−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−a1−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. Thenz = 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 functionfrom 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|21−|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 inequalityholds trivially. Let |α| < 1. Define g(z) = ϕα f ϕ−a(z) for all z ∈ D. Then g is analyticand |g(z)| ≤ 1 for all z ∈ D. Also, g(0) = 0. By Schwarz’s lemma |g(z)| ≤ |z| for all z ∈ Dand |g′(0)| ≤ 1. Now

g′(0) = ϕ′α(f(ϕ−a(0)))f ′(ϕ−a(0))ϕ′−a(0) = ϕ′α(α)f ′(a)ϕ′−a(0) = f ′(a)1− |a|2

1− |α|2.

Therefore |g′(0)| ≤ 1 implies that |f ′(a)| ≤ 1−|α|21−|a|2 .

If |g(z)| = |z| for some 0 6= z ∈ D or |g′(0)| = 1, then there is c ∈ C with |c| = 1 such thatg(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 inverseof 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 MappingTheorem the inverse g of f is also analytic map of D onto itself. Therefore g f(z) = z forall 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)| ≥ 11−|a|2 gives |f ′(a)| = 1

1−|a|2 . By above theorem (with α = 0), there is c ∈ Cwith |c| = 1 such that f(z) = cϕa(z) for all z ∈ D.

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Corollary 84. A map f : D→ D is analytic, one one and onto if and only if there is c ∈ Cwith |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−z01−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. Defineg on D by g(z) = ϕα f(z). Since both ϕ and f are analytic, one one and onto, the mapg : D→ D is analytic, one one and onto. Since g(0) = 0, by above result there is c ∈ C suchthat g(z) = cz for all z ∈ D, i.e., ϕα f(z) = cz for all z ∈ D. Applying ϕ−α both the sideswe obtain

f(z) = ϕ−α(cz) =cz + α

1 + αcz= c

z + αc

1 + αcz.

Take z0 = −αc. 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) 6= 0 for any z ∈ G, then f is conformal. If|a| < 1, then ϕ′a(z) 6= 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 andz0 ∈ D.

Examples 85.

(1) Does there exist an analytic function f : D→ D such that f(12) = 3

4and f ′(1

2) = 2

3?

Suppose that f : D → D be analytic such that f(12) = 3

4and f ′(1

2) = 2

3. Then

23

= |f ′(12)| ≤ 1−| 3

4|2

1−| 12|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) = 12

and f ′(0) = 14?

Take any c ∈ C with |c| = 1. Define f : D → D by f(z) = ϕ− 12(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)−a1−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 anestimate 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, thereis 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(12) = 0, then |f(1

4)| ≤ |ϕ 1

2(1

4)|.

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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 abovethe 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

∑ni=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 realnumbers t1, t2, . . . , tn ≥ 0,

∑ni=1 tizi ∈ A.

Proof. (1) Suppose that f is convex. We prove the result by induction. Let n = 1. Letx1 ∈ [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. Supposethat the result is true for n. Let x1, x2, . . . , xn+1 ∈ [a, b], and let t1, t2, . . . , tn+1 ≥ 0 with

PADABHI

COMPLEX ANALYSIS LECTURE NOTES 29∑n+1i=1 ti = 1. Then

f

(n+1∑i=1

tixi

)= f

(n∑i=1

tixi + tn+1xn+1

)

= f

((1− tn)

n∑i=1

ti1− tn

xi + tn+1xn+1

)

≤ (1− tn)f

(n∑i=1

ti1− tn

xi

)+ tnf(xn+1) [∵ f is convex]

≤ (1− tn)n∑i=1

ti1− 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 hypothesisthat f(tx1 + (1− t)x2) = f(t1x1 + t2x2) ≤ t1f(x1) + t2f(x2) = tf(x1) + (1− t)f(x2). Hencef 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−twhenever a ≤ s < t < u ≤ b.

Proof. Assume that f is convex. Let s, t, u ∈ [a, b] with s < t < u. Then t = u−tu−ss + t−s

u−su.

Note that u−tu−s > 0, fract− su− s > 0 and u−t

u−s + t−su−s = 1. Since f is convex, f(t) ≤

u−tu−sf(s)+ t−s

u−sf(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. Letx, 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−ty−xf(x) + t−x

y−xf(y). Since y−ty−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 anincreasing 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.

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30 P. A. DABHI

Conversely, assume that f ′ is increasing. Let x, u, y ∈ [a, b] with x < u < y. By applyingthe 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) maxf(a), f(b) = maxf(a), f(b).Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar - 388120, Gu-jarat, IndiaE-mail address: [email protected]

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