Analytic continuation

ANALYTIC CONTINUATION

A THESIS submitted by

Kishore Chandra Dalai

for

the partial fulfilment for the award of the degree of

Master of Science in Mathematics

under the supervision of

Dr. BAPPADITYA BHOWMIK

DEPARTMENT OF MATHEMATICS NIT ROURKELA

ROURKELA– 769 008

MAY 2013

DECLARATION

I declare that the topic “ANALYTIC CONTINUATION ” for completion for my master degree has not been submitted in any other institution or university for the award of any other degree or diploma.

Date: 13-05-2013

Place: Rourkela (Kishore Chandra Dalai)

Roll no: 411MA2072 Department of Mathematics

NIT Rourkela

THESIS CERTIFICATE

This is to certify that the project report entitled Analytic Continuationsubmitted by Kishore Chandra Dalai to the National Institute of Technology Rourkela, Orissa for the partial fulfilment of requirements for the degree of master of science in Mathematics is a bonafide record of review work carried out by her under my supervision and guidance.

The contents of this project, in full or in parts, have not been submitted to any other institute or university for the award of any degree or diploma.

( Bappaditya Bhowmik)

ACKNOWLEDGEMENTS

I desire to express my honor to my thesis advisor Dr. Bappaditya Bhowmik for the tenacious support, encouragement and strong leadership of my thesis.Through his carefull and patient guidance, Dr. Bappaditya Bhowmik inspired confidence in my research abil- ities and my mathematical artistry,giving me the courage to ask difficult and stimulating questions. I am also genuinely thankful for the time and effort spent by Dr.R.S. Tungla and Dr.S. Pradhan, members of my masters thesis board,in reviewing this paper.

Kishore Chandra Dalai

ABSTRACT

In this thesis, we study the following topics in complex analysis:- (1) Montel’s theorem.

(2) Riemann Mapping theorem. (3) Analytic continuation. We also study the celebrated Schwarz Reflection Principle and the Monodromy theorem.

Contents

ACKNOWLEDGEMENTS i

ABSTRACT ii

NOTATION iv

Chapter 1. INTRODUCTION 1

Chapter 2. SPACES OF ANALYTIC FUNCTIONS 2 Chapter 3. SPACES OF MEROMORPHIC FUNCTIONS AND THE

RIEMANN MAPPING THEOREM 10

1. Spaces of meromorphic functions 10

2. The Riemann Mapping Theorem 11

Chapter 4. ANALYTIC CONTINUATION AND MONODROMY

THEOREM 16

1. Analytic Continuation 16

2. Schwarz Reflection Principle 17

3. Analytic Continuation Along a Path 19

Bibliography 22

NOTATION

English Symbols

C the complex plane.

D the unit disk {z ∈C:|z|<1}.

B(a, R) the closed ball center at a and radius R.

H(G) set of analytic functions in G.

A⊂B A is a proper subset of B.

S class of normalized analytic univalent functions

CHAPTER 1

INTRODUCTION

Complex variable is a branch of mathematics which has something for all mathe- matician. In addition, to having application to other parts of analysis, it can exactly postulation to be an antecedent of many areas of mathematics. In fact, in this thesis, our purpose is to focuss on some topics in complex analysis such as spaces of analytic func- tions, normal family, spaces of meromorphic functions and celebrated Riemann mapping theorem. We finally review some theory of analytic continuation including the Schwarz reflection principle and the Monodromy theorem.

CHAPTER 2

SPACES OF ANALYTIC FUNCTIONS

In this chapter, we put a metric on the family of all analytic functions on a fixed domainG, and discuss compactness and convergence in this metric space. We also focus on spaces of analytic functions and give proof of the celebratedHurwitz’s theorem,normal families, Montel’s theorem.

Definition 2.1 (Spaces of analytic functions). Let G is an open set in C and (Ω, d) is a complete metric space then designate by C(G,Ω) the set of all continuous functions fromG to Ω .

Proposition 2.2. Let G is open in C then there is a sequence {K_n} of compact subsets of G such thatG= ∪^∞

n=1

K_n]. Moreover, the sets {K_n} can be chosen to satisfy the following conditions:

(a) K_n⊂int(K_n+1).

(b) K ⊂G and K compact implies K ⊂K_n f or some n.

(c) Every component of C_∞−K_n contains a component of C_∞−G.

Proof. For each positive integer n, let K_n = {z : |z|< n} ∩ {z :d(z,C−G)≥ 1 n}. SinceKn is bounded and it is intersection of two closed subsets of C. So Kn is compact.

Now consider the set M ={z :|z| < n+ 1} ∩ {z :d(z,C−G)≥ 1

n+ 1} is open. Hence K_n ⊂ M and M ⊂ K_n+1. So K_n ⊂ int (K_n). G is an open set, so G = ∪^∞

n=1

K_n. Then we can get G= ∪^∞

n=1

int(K_n). If K is compact subset of G, then the set int(K_n) form an open cover of K. So K ⊂ K_n for some n. Now we wish to prove that every component of C_∞−K_n contains a component of C_∞−G. The unbounded component of C_∞−K_n

must contain ∞. So the component of C_∞−G which contains ∞. Also the unbounded component contains {z : |z| > n}. So if D is a bounded component ,it contains a point z with d(z,C−G) < 1

n. According to definition this gives a point w in C−G with

|z−w| < 1

n. But then z ∈B (

w,1 n

)

⊂ C_∞−K_n; since disks are connected and z is in the component D of C∞−k_n, B

( w, 1

n )

⊂ D. If D1 is the component of C∞−D that

containsw it follows that D1 ⊂D.

Proposition2.3. IfG= ∪^∞

n=1

Knwhere eachKnis compact andKn ⊂int(Kn+1),define ρ_n(f, g) = sup{d(f(z), g(z)) :z ∈ K} for all functions f andg in C(G,Ω). Also define, ρ(f, g) = Σ^∞_n=1(¹₂)ⁿ_1+ρ^ρn(f,g)

n(f,g). Prove that (C(G,Ω), ρ) is a metric space.

Proof. (i)ρ(f, g) ≥ 0 ,∀f, g ∈ C(G,Ω).Since ρ_n(f, g) = sup{d(f(z), g(z)) : z ∈ K_n} ≥ 0.(ii)G = ∪^∞

n=1

K_n gives that f = g,then ρ(f, g) = 0.(iii) ρ(f, g) = ρ(g, f),

∀f, g∈C(G,Ω). Since

ρ_n(f, g) = sup{d(f(z), g(z)) :z ∈ K_n}

= sup{d(g(z), f(z)) :z ∈ Kn}

= ρ_n(g, f)

(iv) To prove triangle inequality we first establish the following inequality. Let 0≤ α ≤ beta, thenα+αβ ≤β+αβ.Dividing both sides by (1 +α) (1 +β),we obtain

(2.1) α

1 +α ≤ β 1 +β Now, for all f, g, h∈C(G,Ω), we have ,

0≤ sup{d(f(z), g(z)) :z ∈ K_n}

≤ sup{d(f(z), h(z)) :z ∈ K_n}+ sup{d(h(z), g(z)) :z ∈ K_n}

So from above equation(2.1) it is follows that sup{d(f(z), g(z)) :z ∈ K_n} 1 + sup{d(f(z), g(z)) :z ∈ Kn}

≤ sup{d(f(z), h(z)) :z∈ K_n}+ sup{d(h(z), g(z)) :z ∈ K_n} 1 + sup{d(f(z), h(z)) :z ∈ K_n}+ sup{d(h(z), g(z)) :z ∈ K_n}

Multiplying both sides by 2⁻ⁿ and summing w.r.t n, we get,

Σ^∞_n=1(1

2)ⁿ ρ_n(f, g)

1 +ρ_n(f, g) ≤Σ^∞_n=1(1

2)ⁿ ρ_n(f, h)

1 +ρ_n(f, h) + Σ^∞_n=1(1

2)ⁿ ρ_n(h, g) 1 +ρ_n(h, g) i.e.

ρ(f, g)≤ρ(f, h) +ρ(h, g)

Hence, (C(G,Ω), ρ) is a metric space.

Lemma 2.4. Let the metric ρ be defined as in ρ(f, g) = Σ^∞_n=1(¹₂)ⁿ_1+ρ^ρn(f,g)

n(f,g). If ϵ >0 is given then there is a δ >0 and a compact set K ⊂ G such that for f and g in C(G,Ω);

sup{d(f(z), g(z)) : z ∈ K} < δ ⇒ ρ(f, g) < ϵ. Conversely, if δ > 0 and a compact set K are given, there is an ϵ > 0 such that for f and g in C(G,Ω), ρ(f, g) < ϵ ⇒ sup{d(f(z), g(z)) :z ∈ K}< δ.

Proof. Now we will prove sup{d(f(z), g(z)) : z ∈ K} < δ ⇒ ρ(f, g) < ϵ. Let ϵ > 0 is fixed and p be a positive number such that ∑^∞

n=p+1

(1 2

)n

< 1

2ϵ. Put K = K_p. Choose δ > 0 such that 0 ≤ t ≤ δ gives t

1 +t < 1

2ϵ. Suppose f, g ∈ C(G,Ω) such that sup{d(f(z), g(z)) :z ∈K}< δ. Since K_n⊂K_p =K for 1≤n≤p, 0< ρ_n(f, g)< δ.

So

ρ_n(f, g) 1 +ρ_n(f, g) <

(1 2

) ϵ.

Here,

ρ(f, g) =

∑∞ n=1

(1 2

)n(

ρ_n(f, g) 1 +ρ_n(f, g)

)

=

(∑p n=1

(1 2

)n(

ρ_n(f, g) 1 +ρ_n(f, g)

)) +

( ∑_∞

n=p+1

(1 2

)n(

ρ_n(f, g) 1 +ρ_n(f, g)

))

=

∑p n=1

(1 2

)n( ϵ 2

) + ϵ

2 < ϵ 2+ ϵ

2 =ϵ.

Now, we wish to prove that ρ(f, g) < ϵ ⇒sup{d(f(z), g(z)) : z ∈ K} < δ. LetK and δ are given, Since G= ∪^∞

n=1

k_n = ∪^∞

n=1

intK_n and K is compact there is an integer p≥1 such that K ⊂ K_p; this gives ρ_p(f, g) ≥ sup{d(f(z), g(z)) : z ∈ K}. Let ϵ > 0 be chosen so that 0≤s ≤2^pϵ.

⇒ s

1−s < 2^pϵ

1−2^pϵ =δ ⇒ s 1−s < δ

⇒0≤t≤δ ⇒ t

1 +t < s

1 +s = 2^pϵ So ifρ_p(f, g)< ϵ⇒ ρp(f, g)

1 +ρ_p(f, g) < ϵ

⇒ρ_p(f, g)< δ ⇒sup{d(f(z), g(z)) :z ∈K}< δ. This completes the proof.

Proposition 2.5. (a) A set O ⊂ (C(G,Ω), ρ) is open iff for each f in O there is a compact set K and a δ > 0 such that O⊃ {g :d(f(z), g(z))< δ, z∈K}.

(b) A sequence {f_n} in (C(G,Ω), ρ) converges to f iff {f_n} converges to f uniformly on all compact subsets of G.

Proof. (a)Let O is open and f ∈ O, then for some ϵ > 0, {g : ρ(f, g) < ϵ} ⊂ O. According to lemma (3.3) there existδ >0 and a compact set K such that

{g :d(f(z), g(z))< δ;z ∈K} ⊂O.

Conversely, if for eachf ∈O there is a compact setK and δ >0 such that {g :d(f(z), g(z))< δ;z ∈K} ⊂O,

then from the second part of the previous lemma (3.3); we get Ois open.

(b) Given thatf_n in (C(G,Ω)) converges to f. Now we have to prove thatf_n converges to

f uniformly on all compact subsets of G, i.e to prove f_n converges to f ∀f and ∀ z ∈G.

Let for given ϵ >0,

ρ(f_n, f)< ϵ

⇒

∑∞ n=1

(1 2

)n(

ρ_n(f_n, f) 1 +ρn(fn, f)

)

< ϵ

⇒ ρn(fn, f)< ϵ

⇒ sup{d(f_n(z), f(z)), z ∈K_n}< ϵ.

So f_n converges to f ∀ f and ∀ z ∈ G i.e f_n converges to f uniformly on all compact subsets of G. Conversely, given that f_n converges to f uniformly, d(f_n, f) < ϵ∀f, then sup{d(f_n, f) :z ∈k_n},ρ_n(f_n, f)< ϵ⇒ρ(f_n, f)< ϵ. So f_n converges to f.

Proposition 2.6. C(G,Ω) is a complete metric space.

Proof. Let f_n be a cauchy sequence in C(G,Ω). If we restrict our domain of the sequence of functions gives a Cauchy sequence f_n to C(K,Ω) for every compact sets K inG.i.e, for everyδ >0 there is an integerN such that sup{d(fn(z), fm(z)) :z ∈K}< δ for n, m ≥ N. In particular {f_n} is a Cauchy sequence in Ω; so there is a point f(z) in Ω such that f(z) = limfn(z). This gives a function f : G → Ω; it must be shown that f is continuous and ρ(f_n, f) → 0. Let K be compact and fixed δ > 0; choose N so that sup{d(f_n(z), f_m(z)) : z ∈ K} < δ should satisfy for n, m > N. If z is arbitrary in K but fixed then there is an integer m > N so that d(f(z), f_m(z)) < δ.

But then d(f(z), f_n(z)) < 2δ for all n ≥ N. Since N does not depend on z, this gives sup{d(f(z), f_n(z)) :z ∈K} →0 as n→ ∞. Hence,f_n converges tof uniformly on every compact set inG. In particular converges on all closed balls contained inG. Since uniform limit of a sequence of continuous function is continuous, we see that f is continuous at each point of G. Also ρ(f_n, f)→0 according to proposition (3.4(b)).

Definition 2.7 (Normal families). A set F⊂ C(G,Ω) is normal if each sequence in Fhas a subsequence which converges to a function f inC(G,Ω).

Definition 2.8 (Locally bounded). A set F ⊂ H(G) is locally bounded if for each point a in G there are constants M and r > 0 such that for all f in F, |f(z)| ≤ M, for

|z−a|< r i.e sup{f(z) :|z−a|< r, f ∈F}<∞.

Definition 2.9 (Equicontinuous at a point). A setF⊂C(G,Ω) is equicontinuous at a point z₀ ∈Giff for every ϵ >0 such that for|z−z₀|< δ, d(f(z), f(z₀))< ϵ for everyf inF.

Definition 2.10 (Equicontinuous over a set). F is equicontinuous over a set E ∈ G if for every ϵ > 0 there is a δ > 0 such that for z and z₀ in F and |z − z₀| < δ, d(f(z), f(z₀))< ϵf ∀f ∈F.

Theorem2.11 (Arzela-Ascoli theorem). A setF⊂C(G,Ω)is normal iff the following two conditions are satisfied:

(a) For eachz ∈G,{f(z) :f ∈F} has compact closure in Ω.

(b) F is equicontinuous at each point of G.

Definition 2.12 (Meromorphic unction). If G is open and f is a function defined and analytic inG except for poles, then f is a meromorphic function on G.

Theorem 2.13 (Rouche’s Theorem). Suppose f and g are meromorphic in a neigh- borhood of B(a;R) with no zeros or poles on the circle γ = {z : |z −a| = R}. If Zf, Z_g (p_f,p_g) are the number of zeros(poles) of f, 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. From the hypothesis

|f(z)

g(z) + 1|<|f(z) g(z)|+ 1 on γ. If λ = f(z)

g(z) and if λ is a positive real number, then this inequality becomes λ+ 1 < λ+ 1. This is a contradiction, hence the meromorphic function f

g maps γ onto

Ω = C−[0,∞). If l is a branch of the logarithm on Ω, then l

(f(z) g(z)

)

is well-defined primitive for (f /g)^′(f /g)⁻¹ in a neighborhood of γ.Thus

= 1

2πi

∫

γ

(f /g)^′(f /g)⁻¹

= 1

2πi

∫

γ

[f^′ f − g^′

g ]

= (Z_f −P_f)−(Z_g−P_g).

So we haveZ_f −P_f =Z_g−P_g.

Theorem 2.14 (Hurwitz’s Theorem). Let G be a region and suppose the sequence {f_n} in H(G) converges to f. If f is not identical to zero, B(a;R) and f(z) ̸= 0 for

|z−a| =R, then there is an integer N such that for n ≥ N, f and {f_n} have the same number of zeros inB(a;R).

Proof. LetGbe a region and{f_n}inH(G) converges tof. Sincef(z)̸= 0∀|z−a|= R, let δ = inf{|f(z)|:|z−a|=R}>0. But {f_n} →f uniformly on|z|:|z−a|=R.

So there is an integer N such that ifn ≥N and |z−a|=R, then|f(z)−f_n(z)|< 1 2δ <

|f(z)| ≤ |f(z)|+|f_n(z)|. According to Rouches theorem f and {f_n} have same number

of zeros inB(a;R).

Theorem 2.15 (Montel’s Theorem). A family F in H(G) is normal iff F is locally bounded.

Proof. SupposeF is normal but fails to be locally bounded; then there is a compact set K ∈ G such that sup{|f(z)| : z ∈ k, f ∈ F} = ∞, i.e there is a sequence {f_n} in F such that sup{|f(z)|:z ∈k} ≥n. SinceF is normal there is a function f in H(G) and a sequence {f_nk} such that f_nk →f. But this gives that sup{|f_nk(z)−f(z)| :z ∈k} → 0 asK → ∞. If |f(z)| ≤ M for z inK, n_k ≤ sup{|f_nk(z)−f(z)|: z ∈k}+M. Since the right hand side converges toM, so this is a contradiction. So F is locally bounded.

conversely, suppose F is locally bounded. Here we use Arzela-Ascoli theorem to show F

is normal. It can be easily shown that the first condition is satisfied. Now only we have to prove the second condition of this theorem, i.e we have to proveFis equicontinuous at each point ofG. Let fix a point a ∈G and ϵ >0, so according to hypothesis ∃r >0 and M >0 such that B(a;r)⊂G and |f(z)| ≤M ∀z ∈B(a;r) and ∀f ∈F. Let |z−a|< ¹₂r and f ∈F; then using Cauchy’s formula with γ(t) = a+re^it, 0≤t≤2π, we get

|f(a)−f(z)| ≤ 1 2π

∫

γ

f(w)(a−z) (w−a)(w−z)dw

≤ 1

2π|a−z| ∫

γ

f(w)

(w−a)(w−z)dw (2.2)

Atw=a

lim

w→a

f(w)(w−a) (w−a)(w−z)

= M

(1/2)r = 2M (2.3) r

(2.4) Atw=z

lim

w→z

f(w)(w−z) (w−a)(w−z)

= M

(1/2)r = 2M (2.5) r

According to Cauchy’s formula and from (2.3) and (2.5), we get from (2.2)

∫

γ

f(w)

(w−a)(w−z)dw= 2π (2M

r + 2M r

)

= 8M π (2.6) r

Again from (2.2) and from (2.6) we get,

|f(a)−f(z)|= 1

2π|a−z|8M π

r =|a−z|4M r

Let δ = min{_2r¹,_4M^rϵ }. So |a−z| < δ. So |f(a)−f(z)| < ϵ ∀f ∈ F. Hence the second

condition satisfied. Hence it is proved.

CHAPTER 3

SPACES OF MEROMORPHIC FUNCTIONS AND THE RIEMANN MAPPING THEOREM

In this chapter, we put a metric on extended complex plane and defined a Chordal metric. We also focus on spaces of meromorphic function functions and give proof of the celebratedRiemann Mapping Theorem .

1. Spaces of meromorphic functions

Let G be a region and f is a meromorphic function on G, Let M(G) is the set of all continuous functions onG then consider M(G) as a subset of C(G,C∞). In this section we are going to put a metricd on C_∞ as follows. Let z₁, z₂ ∈C,then

d(z1, z2) = 2|z₁−z₂|

[(1 +|z₁|²)(1 +|z₂|²)]¹²; and for eachz inC we define, d(z,∞) = 2

(1 +|z|²)¹².

Theorem 3.1. Let fn be a sequence in M(G) and fn → f in C(G,C∞).Then either f is meromorphic or f ≡ ∞.If each f_n is analytic then either f is analytic or f ≡ ∞.

Proof. Suppose there is a point a in G with f(a) ̸= ∞ and put M = |f(a)|.We know from one proposition, If a is in C and r > 0 then there is a number ρ > 0 such that B_∞(a;ρ) ⊂ B(a;r). According to this proposition we can find a number ρ > 0 such that B_∞(f(a);ρ)⊂ B(f(a);M). But since f_n → f there is an integer n₀ such that d(f_n(a), f(a))< ¹₂ρfor alln ≥n₀. Also {f, f₁, f₂, ...}is compact inC(G,C_∞) so that it is equicontinuous.That is, there is anr >0 such that|z−a|< rimpliesd(f_n(z), f_n(a))< ¹₂ρ.

That gives that d(f_n(z), f(a)) ≤ ρ for |z −a| ≤ r and for n ≥ n₀. But by choice of ρ,

|f_n(z)| ≤ |f_n(z)−f(a)|+|f(a)| ≤ 2M for all z in B(a;r) and n ≥ n₀. But from the formula for the metric d , we have

2

1 + 4M²|f_n(z)−f(z)| ≤d(f_n(z), f(z)

for z inB(a;r) and n≥n₀. Since d(f_n(z), f(z))→0 uniformly for z inB(a;r). Since the sequence f_n is bounded on B(a;r),f_n has no poles and must be analytic near z = a for n≥n₀ . It follows that f is analytic in a disk about a.

Now suppose that there is a point a in G with f(a) = ∞.For function g inC(G,C∞) define ¹_g by (¹_g)(z) = _g(z)¹ if g(z) ̸= 0 or ∞; (¹_g)(z) = 0 if g(z) = ∞; and (¹_g)(z) = ∞ if g(z) = 0.It follows that ¹_g ∈ C(G,C_∞).Also since f_n → f in C(G,C_∞).From chordal metric it is follows that _f¹

n → _f¹ inC(G,C∞).Now each function _f¹

n is meromorphic onG;so it gives a number r > 0 and a integer n₀ such that _f¹

n and _f¹ are analytic on B(a;r) for n ≥n₀ and _f¹

n → _f¹ uniformly on B(a;r).From Hurwitz’s theorem either _f¹ ≡ 0 or _f¹ has

isolated zeros in B(a;r).

Corollary 3.2. M(G)∪ {∞} is a complete metric space.

Corollary 3.3. H(G)∪ {∞} is closed in C(G,C∞).

2. The Riemann Mapping Theorem

The theorem was stated byBernhand Riemannin 1851 in his Ph.D. thesis. According to Riemann Mapping theorem,any two proper simply connected domains in the plane are homeomorphic.Even though class of continuous functions are vastly larger than the class of conformal maps, it is not easy to construct a one-to-one function onto the disc,knowing only that the domains are simply connected.

Definition 3.4 (Conformally equivalent). A region G₁ is conformally equivalent to G₂ if there is analytic functionf : G₁ → C such that f is one-one and f(G₁) =G₂. C is not equivalent to any bounded region. By Liouville’s theorem, iff is entire and bounded

for all values of z inC, then f(z) is constant through out the plane. So constant function can not be one-one .HenceC is not equivalent to any bounded region.

Theorem 3.5 (Riemann Mapping Theorem). : Given any simply connected region Ω which is not the whole plane and the point a ∈ Ω, then there exits a unique analytic function f(z)∈Ω,normalized by the conditions.

(a) f(a) = 0 and f^′(a)>0 (3.1)

(b) f is one−one.

(c) f(Ω) ={z :|z|<1}

Proof. Suppose Ω ̸= C is a simply connected region.Let z₀ ∈ Ω,We have to show that (1)Uniqueness of f having the properties in (3.1).

Let g be another analytic function which is defined by g : Ω → C, which satisfies all the conditions of (3.1), then g(a) = 0 and g^′(a) > 0 implies that a = g⁻¹(0) (f ◦g⁻¹)(0) = f(a) = 0. Suppose f and g are two functions on Ω,then S=f ◦ g⁻¹: D−→Dis analytic, one-one and onto. f◦g⁻¹(0)=f(a)=0. According to Schwarz’s lemma, (f ◦g⁻¹) is a one-one map then there is constant c with |c|= 1 and (f◦g⁻¹)(z) =cz∀z then f(z) = cg(z) since f^′(a)> 0, so cg^′(a) >0 and also we have g^′(a)> 0. So c must be 1.So finally we havef =g, i.e,f is unique. (2) Existence:To motivate the proof of the existance of f,(2) Existence: Consider the family of functions F of all analytic functions f having properties (a) and (b) from (3.1) and satisfying |f(z)|<1 for z in Ω. Now only we have to choose a member of F having property (c) of the equation (3.1). Suppose {K_n} is a sequence of compact subsets of Ω such that Ω = ∪^∞

n=1

K_n and a ∈ K_n for each n, then {f(Kn)} is a sequence of compact subsets of D where D = {z : |z| < 1}.As n becomes larger, {f(K_n)} becomes larger and larger and tries to fill out the disc D. i.e, D = ∪^∞

n=1

f(K_n). In a simply connected region every non vanishing analytic function has an analytic square root.So to prove existence off, we have to prove the following lemma.

In a simply connected region every non vanishing analytic function has an analytic square

root. So to prove existence of f, we have to prove the following lemma.

Lemma 3.6. Let Ω be a region which is not the whole plane and such that every non vanishing analytic function on Ω has an analytic square root. If a ∈Ω, then there is an analytic function f on Ω such that

(a) f(a) = 0 and f^′(a)>0 (b) f is one−one.

(c) f(Ω) =D={z :|z|<1}

Proof. Define F = {f ∈ H(Ω) : f is one−one, a ∈ Ωf(a) = 0, f^′(a) > 0, f(Ω) ⊂ D}. Sincef(Ω) ⊂D, sup{|f(z)|: z ∈ Ω} ≤ 1 for f ∈ F. According to Montel’s theorem Fis normal if it is non-empty, i.e, we have to prove (i)F̸=ϕand (ii) F⁻=F∪ {0}. First assume that equation (i) and (ii) hold. Consider the function f → f^′(a) of H(Ω) → C. This is a continuous function. SinceF⁻ is compact, then there exist f ∈F⁻ with f^′(a)≥ g^′(a)∀g ∈F. BecauseF̸=ϕthenf ∈F. Now only we have to show thatf(Ω) =Di.e, we have to show that (a)f(Ω)⊂Dand (b)D⊂f(Ω). To prove equationf(Ω) =D; letw does not belongs tof(Ω). Then the function f(z)−w

1−wf(z)¯ is analytic in Ω and never vanishes. By hypothesis there is an analytic functionh : Ω→C such that [h(z)]² = f(z)−w

1−wf(z)¯ . Since the M¨obius transformation T(ξ) = ξ−w

1−wξ¯ maps D ontoD, h(Ω) ⊂D. Define g : Ω →C by

g(z) =

(|h^′(a)| h^′(a)

)( h(z)−h(a) 1−h(a)h(z)

) .

Here, it is easy to see that g(a) = 0, and g^′(z) =

(|h^′(a)| h^′(a)

)(h^′(z)(1−h(a)h(a)) [1−h(a)h(z)]²

) .

Hence

g^′(a) =

(|h^′(a)| h^′(a)

)(h^′(a)(1−h(a)h(a)) [1−h(a)h(a)]²

)

= |h^′(a)|

1− |h(a)|² >0.

Now,|h(a)|² =| −w|=|w| and

h(z)² = f(z−w) 1−wf¯ (z). On differentiation of the functionh yields,

2h(z)h^′(z) = (1−wf(z))f¯ ^′(z)−(f(z)−w)( ¯−wf^′(z)) (1−wf¯ (z))²

= f^′(z)−wf¯ (z)f^′(z) +f(z)f^′(z) ¯w−wwf¯ ^′(z) (1−wf(z))¯ ²

= f^′(z)(1− |w|²) (1−wf¯ (z))² . Hence,

2h(a)h^′(a) = f^′(a)(1− |w|²)

(1−wf¯ (a))² =f^′(a)(1− |w|²) (∵ f(a) = 0).

and as a result,

h^′(a) = f^′(a)(1− |w|²) 2h(a) . Now

g^′(a) = |h^′(a)| 1− |h(a)|² >0

⇒g^′(a) = f^′(a)(1− |w|²)

2h(a)(1− |w|) = f^′(a)(1− |w|²) 2√

|w|(1− |w|) = f^′(a)(1 +|w|) 2√

|w| > f^′(a).

This gives that g is in F and contradicts the choice of f. So w ∈ f(Ω). So, D ⊂ f(Ω).

Again f(Ω) ⊂ D. So we get f(Ω) = D. Now we are going to prove the equations (i) and (ii). Since Ω ̸= C, let b ∈ C−Ω and let g be an analytic function on Ω such that [g(z)]² = z −b. If z₁ and z₂ are points in Ω, and g(z₁) = ±g(z₂), then it follows that z1 = z2. In particular g is one to one . According to Open mapping theorem there is a r >0 such that B(a, R)⊂g(Ω). So there is a point z in Ω such that g(z)∈B(−g(a);r) then r > |g(z) +g(a)| = | − g(z)−g(a)|. Since B(a, R) ⊂ g(Ω), so there is a w in Ω with g(w) = −g(z); but B(a, R) ⊂ g(Ω) shows that w = z which gives g(z) = 0.

But then z −b = [g(b)]² = 0 implies that b is in Ω, a contradiction. Hence g(Ω)∩ {ξ :

|ξ+g(a)| < r} = ϕ. Let U be the disk {ξ : |ξ +g(a)| < r} = B(−g(a);r). There is a M¨obius transformationT such that T(C_∞−U⁻) =D. Let g₁ =T◦g; theng₁(Ω)⊂D. If

α= g(a), then let g₂(z) =ϕ_α◦g₁(z); so we will have that g₂(Ω)⊂ D and g₂ is analytic, but we also have that g₂(a) = 0. Now there is a complex number c with |c| = 1, such that g₃(z) = cg₂(z) has positive derivative at z = a and is therefore in F. Here f is g3 = c(ϕα ◦T ◦g). So from this we conclude that the set F is nonempty. Suppose {fn} is a sequence in F and f_n → f in H(Ω). Clearly f(a) = 0 and since f_n^′(a) → f^′(a), then it follows that f^′(a) ≥ 0. Let z₁ be an arbitrary element of Ω and put ξ = f(z₁);

let ξ_n = f_n(z₁). Again let z₂ ∈ Ω, z₁ ̸= z₂ and let K be a closed disk centered at z₂ such that z₁ ∈/ K. Then f_n(z) → ξ_n never vanishes on disk K. Since f_n is one- one.

But f_n(z)−ξn→ f(z)−ξ uniformly on K. According to Hurwitz’s theorem gives that f(z)−ξ never vanishes on K or f(z) ≡ ξ. If f(z) ≡ ξ on K, then f is the constant function ξ throughout Ω; since f(a) = o we have that f(z) ≡ 0. Otherwise we get that f(z₁) ̸= f(z₂) for z₁ ̸= z₂; that is f is one-one. But if f is one-one then f^′ can never vanish. Sof^′(a)>0 andf is inF. This proves the equation (ii) completely, which proves

the existence of f inF.

Now from this above lemma we conclude that f exists with properties in equation (3.1). This completes the proof of the Riemann mapping theorem.

CHAPTER 4

ANALYTIC CONTINUATION AND MONODROMY THEOREM

In this chapter, we define Analytic continuation and some interesting examples. We also focus on Analytic continuation along a path and give proof of celebrated Schwarz Reflection Principle and Monodromy Theorem.

1. Analytic Continuation

Definition4.1. Analytic Continuation is a technique to extend the domain of a given analytic function. Supposef is an analytic function defined on an open subset U of the complex plane C. If V is a larger open subset of C containing U and F is an analytic function defined onV such thatF(z)=f(z), then F is called an analytic continuation of f.

Example: Consider the first function f₁ is defined by the equation

f1(z) =

∑∞ n=0

zⁿ

= lim

n→∞[1 +z+z²+· · ·+zⁿ]

= lim

n→∞

1−zⁿ⁺¹ 1−z

= 1

1−z − lim

n→∞

zⁿ⁺¹ 1−z

= 1

1−z whenever |z|<1.

If |z| > 1, then lim |z|ⁿ = ∞ and the series diverges. Now the function f₂(z) = ₁₋¹_z is defined and analytic in C− {1}. We see that f₂(z) = f₁(z), for all |z| < 1. Hence the functionf₂ is analytic continuation of f₁.

2. Schwarz Reflection Principle

The theorem was stated byH.A.Schwarz. This article is about the reflection principle in complex analysis. In mathematics, Schwarz reflection principle is a way to extend the domain of definition of an analytic function of a complex variable F, which is defined on the upper half-plane and has well-defined and real number boundary values on the real axis.In that case,the putative extension ofF to the rest of the complex plane isF(z). IfG is a region andG^∗ ={z :z ∈G}and if f is an analytic function onG thenf^∗ :G^∗ −→C defined by f^∗(z) = f(z) is also analytic. Now suppose that G = G^∗; that is, G is symmetric with respect to the real axis. Theng(z) = f(z)−f(z) is analytic onG. Since G is connected it must be that G contains an open interval of the real line. Suppose f(x) is real for all x in G∪ ℜ. But G∪ ℜ has a limit point in G so that f(z) = f(z),

∀z ∈ G. The fact that f must satisfy this equation is used to extend a function defined onG∪ {z :Imz≥0} to all ofG.

IfG is symmetric region (i.e., G=G^∗) then let G₊ ={z ∈G :Imz > 0}, G₋ ={z ∈G: Imz <0}, and G₀ ={z ∈G:Imz = 0}.

Schwarz Reflection Principle: Let G be a region such that f : G₊∪G₀ −→ C is a continuous function which is analytic on G₊ and if f(x) is real for x in G₀, then there is an analytic function g :G−→C such that g(z) = f(z) for z in G₊∪G₀.

Proof. For z inG₋ defineg(z)=f(z) and for z in G₊∪G₀ letg(z) =g(z). It is easy to see that g : G −→ C is continuous; it must be shown that g is analytic. It is trivial thatg is analytic onG₊∪G₋, so fix a pointx₀ inG₀ and letR >0 withB(x₀;R)⊂G. It is sufficient to show that: g is analytic onB(x₀;R). To do this apply Morera’s Theorem.

Let T= [a,b,c,a] be a triangle in B(x₀;R). To show that : ∫

T f = 0 it is sufficient to show that∫

P f = 0, whenever P is a triangle or a quadrilateral lying entirely inG₊∪G₀ orG₋∪G₀. Assume that T ⊂G₊∪G₀ and [a,b]⊂G₀.

Let△designateT together with its inside; theng(z) =f(z) for all z in△. By hypothesis fis continuous onG₊∪G₀and so,f is uniformly continuous on△. (since△=T∪inside⊂ G₊∪G₀). So, ifϵ >0 there is aδ >0 such that when z and z^′ ∈ △ and|z−z^′|< δ then

|f(z)−f(z^′)|< ϵ. Now, suppose αand β on the line segment [c,a] and [b,c] respectively, so that |α −a| < δ and |β −b| < δ. Let T₁=[α, β,c, α] and Q= [a,b, β, α,a]. Then

∫

T f = ∫

T1f +∫

Qf, but T and its inside are contained in G₊ and f is analytic there;

hence (4.1)

∫

T

f =

∫

Q

f

since ∫

T1f = 0 by C.I.T, But, if 0≤t ≤1, then

|[tβ+ (1−t)α]−[tb+ (1−t)a]|< δ so that,

|f(tβ+ (1−t)α)−f(tb+ (1−t)a)|< ϵ IfM =max{|f(z)|:z ∈ △} and l = the perimeter of T, then

|

∫

[a,b]

f +

∫

[β,α]

f| = |(b−a)

∫ 1 0

f(tb+ (1−t)a)dt

−(β−α)

∫ 1 0

f(tβ+ (1−t)α)dt|

≤ |b−a||

∫ 1 0

[f(tb+ (1−t)a)

− f(tβ+ (1−t)α)]dt|+

|(b−a)−(β−α)|

|

∫ 1 0

f(tβ+ (1−t)α)dt|

≤ ϵ|b−a|+M|(b−β) + (α−a)|

≤ ϵl+ 2M δ.

Also,

(4.2) |

∫

[α,a]

f| ≤M|a−α| ≤M δ.

and

(4.3) |

∫

[b,β]

f| ≤M δ.

Combining these last two inequalities with (4.1) and (4.2) gives that

|

∫

T

f| ≤ϵl+ 4M δ.

Since it is possible to chooseδ < ϵ and since ϵ is arbitrary, it is follows that

|

∫

T

f|= 0

Thusf must be analytic. This completes the proof.

3. Analytic Continuation Along a Path

Let us begin this section by recalling the definition of a function. We use the somewhat imprecise statement that a function is a triple (f, G,Ω), whereG and Ω are sets andf is a rule which assigns to each element ofG a unique element of Ω. If we enlarge the range Ω to a set Ω₁ then (f, G,Ω₁) is a different function. However this point should not be emphasized here; we do emphasize that a change in a domain results in a new function.

Indeed, the purpose of analytic continuation is to enlarge the domain.

Definition 4.2. Germ: A function element is a pair (f, G) where G is a region and f is analytic function on G. For a given function elements (f, G) define the germ off at ato be the collection of all function elements (g, D) such that a∈D and f(z) = g(z) for all z in a neighborhood ofa. Denote the germ by [f]_a. Notice that [f]_a is a collection of all function elements and it is not a function element itself.

Definition 4.3. Let γ[[0,1] −→ C be a path and suppose that for each t in [0,1]

there is a function element (f_t,Dt) such that : (a) γ(t)∈Dt;

(b) for each t in [0, 1] there is a δ > 0 such that |s −t| < δ implies γ(s) ∈ Dt and [f_s]_γ(s)=[f_t]_γ(s). Then (f_t,Dt) is the analytic continuation of (f₀,D0) along the path γ; or (f₁,D1) is obtained from (f₀,D0) by analytic continuation along γ.

Proposition4.4. Letγ : [0,1]→Cbe a path fromatoband let{(f_t, D_t) : 0≤t≤1} and {(g_t, B_t) : 0 ≤ t ≤ 1} be analytic continuations along γ such that [f₀]_a = [g₀]_a.Then [f₁]_b = [g₁]_b.

Proof. This proposition will be proved by showing that the set T = {t ∈ [0,1] : [f_t]_γ(t) = [g_t]_γ(t)}is both open and closed in [0,1] so that, in particular, 1∈T. The easiest part of the proof is to show thatT is open. So fix t inT and assumet ̸= 0 or 1. (Ift= 1 the proof is complete; if t = 0 then the argument about to be given will also show that [a, a+δ)⊂T for someδ > 0.) By the definition of analytic continuation there is aδ >0 such that for all|s−t|< δ, γ(s)∈D_t∩B_t and

[f_s]_γ(s) = [f_t]_γ(s) [gs]γ(s) = [gt]γ(s).

But since t∈ T, f_t(z) =g_t(z) for all z inD_t∩B_t. So it follows from the above equation that [f_s]_γ(s) = [g_s]_γ(s) whenever |s−t| < δ. That is, (t−δ, t+δ)⊂ T and so T is open.

To show that T is closed let t be a limit point of T, and again choose δ > 0 so that γ(s)∈D_t∩B_t and the above equation is satisfied when ever|s−t|< δ. Sincetis a limit point ofT there is a pointsinT with |s−t|< δ; soG=Dt∩Bt∩Ds∩Bs containsγ(s) and therefore, is a non-empty open set. Thus,f_s(z) = g_s(z) for allz ∈G by definition of T. But, according to the above equation f_s(z) = f_t(z) and g_s(z) = g_t(z) for all z in G.

Sof_t(z) = g_t(z) for all z inG and, becauseGhas a limit point inD_t∩B_t, this gives that [f_t]_γ(t) = [g_t]_γ(t). That is, t ∈T and so T is closed.

Definition 4.5. If γ : [0,1]→ C is a path from a to b and {(f_t, D_t) : 0 ≤ t ≤ 1} is an analytic continuation alongγ then the germ [f₁]_b is the analytic continuation of [f₀]_a alongγ.

Definition 4.6. If (f, G) is a function element then the complete analytic function obtained from (f, G) is the collection F of all germs [g]_b for which there is a point a inG and a pathγ from a tob such that [g]b is the analytic continuation of [f]a along γ.

Definition 4.7. Unrestricted Analytic Continuation Let (f, D) be a function element and let Gbe a region which contains D; then (f, D) admits unrestricted analytic contin- uation inGif for any pathγ inGwith initial point inDthere is an analytic continuation of (f, D) alongγ.

Theorem 4.8 (Monodromy Theorem). Let (f, D)be a function element and let G be a region containing D such that (f, D) admits unrestricted continuation in G.

Let a ∈ D, b ∈ G and let γ₀ and γ₁ be paths in G from a to b; let {(f_t, D_t) : 0 ≤t ≤ 1} and{(gt, Dt) : 0≤t≤1} be analytic continuations of (f, D) along γ0 and γ1 respectively.

If γ₀ and γ₁ are FEP(fixed-end-point) homotopic in G then [f₁]_b = [g₁]_b.

Bibliography

[1] John B. Conway: Functions of One complex Variable, Second edition, Narosa Publishing House, New Delhi.

[2] T. W. Gamelin: Complex Analysis, Springer International Edition.