Complex Theorems

7/25/2019 Complex Theorems

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Complex Theorems

1 2009-2010

Cauchys Theorem (Homotopy Version).Power series expansion for holomorphic functions (state).Liouvilles Theorem.Argument Principle (state).Rouches Theorem (state).Inverse Function Theorem.

Montels Theorem.Every memomorphic function on the riemann sphere is rational.Riemann Mapping Theorem (state).Mittag-Leflers Theorem.

2 2007-2008

Cauchys Theorem (Homotopy Version).Laurent Series Expansion of a Holomorphic Function (state).Moreras Theorem.

Schwarz Reflection Principle (prove using Morera).Global Maximum Modulus Theorem.Liouvilles Theorem.Mobius Transformations define Meromorphic Functions on the Riemann Sphere..

3 2005-2006

Cauchys Theorem (any version).Cauchy Integral Formula.Power series expansion theorem for holomorphic functions (state).Identity Principle.Weirerstrass Theorem (state).Montels Theorem (state).Riemann Extension Theorem (state).Argument Principle (state).Rouches Theorem (state).Open mapping theorem (deduce from Rouche).Riemann Mapping Theorem (idea of proof).Schwarz Lemma.

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4 Cauchy Theorem (Homotopy Version)

Let 1, 2 : [a, b] . Assume a homotopy between 1, 2 relative to{a, b}Iff: Cis holomorphic, then

1

f dz=2

f dz

Proof

5 Cauchy Theorem (Homology Version)

Let f: Cbe holomorphic and let c be a 1-cycle which is a boundary in .{c= c : c is a 2-chain in }Then

c

f dz= 0

Proof

We have c= c=

jnj[j ] where c=

jnj[j]. We want

[j]

f dz= 0

Let

j : C

[j ] = [ 1] + [ 2] + [ 3]

: [a, b] C

[] = [1] + [2] + [3]

0(t) =z0

So, we have a homotopy (t, s) = (1 s)(t) +s0(t)(t, s) Now((t, s)) is homotomy in between and 0= constant is null-homotopicSo the homotopy version of Cauchy gives

f dz= 0

[j [f dz= 0

c

f dz=

nj

[j]

f dz= 0

6 Moreras Theorem (Converse of Cauchy)

Assume

1. fcontinuous in

2. triangle :

f dz= 0

Then f is holomorphic in

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Proof

f d= 0

[a,z]

f d= 0

=

[a,z0]

f d

= [z,z0]

f d

=F(z) F(z0)

F(z) F(z0)

z z0=

1

z z0

[z0,z]

f d

= 1

z z0

[z0,z]

f(z0)d+ 1

z z0

[z0,z]

(f() f(z0))d

Where I=f() f(z0)

1z z0 I sup|f() f(z0)|

z z0 |z z0|

0 as z z0

So

F(z) F(z0)

z z0

1

z z0

z,z0

f(z0)d

=f(z0)

f(z) =F(z)F holomorphicLocally expands in Taylor Seriesfexpands in power seriesf holomorphic

7 Liouvilles Theorem

Every function fwhich is holomorphic in C and is bounded is constant, i.e. forf O(C)and |f(z)| c, for some constant c, then f=const.

Proof

f is holomorphic in C fis holomorphic in some ball = BR(0). The Cauchy Estimatefor fgives us

|f(n)(0)| n!c

Rn

where c= sup |f|We already know thatc : |f(z)| c so

c c

|f(n)(0)| n!cRn

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for all R. Now we take the limit as R

limR

|f(n)(0)|= 0

for all n >0. For n= 0 we have|f(0)| c

If we consider the power series expansion off(z),

f(z) =n=0

cnzn

=n=0

f(n)(0)

n! zn

=f(0)

So f(z) = const.

8 Identity Principle

Let f, g be holomorphic in with connected and f(z) =g(z)z subset AA has unit points in

fg in

Proof

Let h= f g h|A= 0. We want h 0.To show this use factorisation lemma: z0 unit points ofA. Assume d= ordh(z0)< infh(z) = (z z0)

dg(z), g(z0)= 0 where g(z0) has no zeros near z0z0 is isolated zero ofhz0 cannot be unit point ofA, h|A 0d = infh 0 in neighbourhood ofz0

Use open closed arguement. Set U={a : h 0 in some neighbourhood ofa} U open

To showU closed: let z0 be unit set ofU z0 Uby the first part of the proofU is opan and closed,U = U=

h 0

9 Rouches Theorem

Let D and F, fbe holomorphic in with |f(z)|< |F(z)| on D = . Then # of zerosofF in D = # of zeros ofF+f in D

Proof

F+tfhas no zeros on gamma

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Rightarrow(t) NtF(z)+tf(z)F(z)+tf(z)

is continuous in (z, t) x{0, 1}

(t) is continuous = constant# of zeros of F in D = # of zeros of F + f in D

10 Montels TheoremLet fn be holomorphic in .Assume k compact C(k), |fn(z)| C(k), z k,n

Proof

Choose a disk BR(z0) with z0 being any fixed pointWe want to prove that (|fn| BR(z0)) is equicontinuous.Let

bn(z) = 1

2i BR+(z0)fn()

( z)d

with such that BR(z0)

fn(z) fn(w) = 1

2i

BR+(z0)

fn()d

1

z

1

w

where w BR(z0)Now:

1

z

1

w=

z w

( z)(=w)

BR+(z0) | z|, | w|

And so z w( z)( w) |z w|2

Also

|fn(z) fn(w)| 2(R+) 1

2 supBR+

|fn|z w|

2 |, n

= R+

2 sup |fn||z w|

Take k= BR+(z0)(fn|K) is uniformly bounded by assumption C(K)> 0; sup |fn| C(K), n |fn(z) fn(w)|

R+2

C(K)|z w|Let R+

2 C(K) =M >0, say, >0, take = 2M

Then |z w|< |fn(z) fn(w)| M = 2

<

fn|Br(z0) equicontinuous

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11 Schwarz Reflection Principle

Let f: + I Cwhere I R+ is an interval and we have the following assumptions:

1. f O(+) C(+ I)

2. f(I) R

Then fhas a holomorphic extension f: + I

Cand, moreover, it is given by

f(z) =

f(x) ifz + I

f()z ifz

Proof

Firstly, we need to check that f O(). We have

z zf(z) f()z

where z zand f(z) f()zare both anti-holomorphic maps and z f(z) is a holomor-phic map. This implies that f()zis holomorphic in z Now we wish to show that f is holomorphic at I. We have

limzx0I

f(z) = f(x0)

since f= f|+ I where z x0 Iand fC(+ I).z , z x0

z +, z x0 =x0f(z) f(x0), by our second assumption

f()z f(x0) =f(x0) f(z) f(x0) =f(x0) fC( I) f is continuous in + I

Now we use the Removability Lemma for = I + and f. The lemma states that iffC() O(\R) f O(). Therefore, by the lemma, we have

fC()f O(+ )

where + = \R. By the lemma, f is holomophic in . This proves the reflectionprinciple.

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