analysis chapter5.mt08.update

8/6/2019 Analysis Chapter5.MT08.Update

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5 Applications of Cauchys Integral Theorem(s)

Definition 5.1. For continuous functions f : D C we define|za|=r

f(z) dz =

f(z) dz

where the path : [0, 2] D is given by (t) = a + r eit.

Theorem 5.1 (Cauchys Integral Formula). Suppose f : D C is holomorphic,a D and R > 0 are such that BR(a) D, and that 0 < r < R. Then

f(a) =

1

2i|za|=r

f(z)

z a dz.

Proof.

1

2i

|za|=r

f(z)

z adz =

1

2i

|za|=r

f(a)

z adz +

1

2i

|za|=r

f(z) f(a)

z adz.

Note that

limza

f(z) f(a)

z a= f(a)

because f is holomorphic. Hence for |z a| r,f(z) f(a)z a

is bounded (byM, say). The function

ha(z) =f(z) f(a)

z a

is holomorphic on D \ {a} and hence by the deformation version of CauchysIntegral Theorem for closed paths,

1

2i

|za|=r

ha(z)dz =1

2i

|za|=

ha(z)dz,

for any small positive < r. Hence

12i|za|=r

ha(z)dz

= 12i

|za|=

ha(z)dz

M by the standard estimate for integrals and any small > 0 implying that

1

2i

|za|=r

ha(z)dz = 0.

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Then putting (t) = a + reit for t [0, 2], we see that

12i|za|=r

f(a)z adz = 12i20

f(a)a + reit arieitdt

=1

2if(a) 2i = f(a),

as required.

Example 5.1.

|z|=1

sin z

zdz = 2i sin 0 = 0.

Example 5.2. |z|=1

sin z

z2dz =?

sin z = z z3

3!+

z5

5!

sosin z

z= 1

z2

3!+

z4

5! =

n=0

(1)nz2n

(2n + 1)!.

|z|=1

sin z

z2dz =

|z|=1

n=0(1)

n z2n

(2n+1)!

zdz = 2i

Example 5.3.

|z1|=1

dz

z2 1=

|z1|=1

1z+1

z 1dz = 2i

1

1 + 1= i because

1

z + 1is

holomorphic in B2(1).

Example 5.4.

|z|=3

dz

z2 1=

|z|=3

12

z 1

12

z + 1

dz

=

|z|=3

12

z 1dz

|z|=3

12

z + 1dz

=

|z1|=1

12

z 1dz

|z+1|=1

12

z + 1dz

= 2i

1

2

1

2

= 0.

using the deformation version for closed paths version of the Cauchys Integral For-mula.

OR

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|z|=3

1

z

2

1

dz = |z|=R

1

z

2

1

dz

for any R > 3 and |z|=R

1

z2 1dz

1R2 1 2R 0as R 0. Hence

|z|=3

1

z2 1dz = 0.

Lecture 12

Theorem 5.2 (Taylors Theorem). Let : BR(a) C be holomorphic. Then f has apower series expansion

f(z) =n=0

bn(z a)n

converging for all z BR(a), where

bn =1

2i

|za|=r

f(z)

(z a)n+1dz,

for any 0 < r < R.

Proof. Let z BR(a). Then for any > 0 such that |z a| + < R, so B(z) BR(a),

f(z) =1

2i

|zw|=

f(w)

w zdw,

by Cauchys Integral Formula. Deforming the contour of integration we have

f(z) =1

2i

|za|=r

f(w)

w zdw,

where r > |z a| + .[Note, B(z) Br(a) BR(a) and the deformation may be achieved using

H(s, t) = (1 s)

z + eit

+ s

a + reit

, but when writing out proofs you arenot expected to give explicit deformations.] Now

1

w z=

1

w a + a z=

1

(w a)

1 zawa

so

|wa|=r

f(z)

w zdw =

|wa|=r

f(w)

(w a)

n=0

z a

w a

ndw.

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The set {w : |w a| = r} is compact and hence there exists K > 0 such that|f(w)| K when |w a| = r. Put

Mn = K|z a|n

rn+1.

Then

n=0 Mn < because

z ar < 1. By the Weierstass M-test, the series

n=0

f(w)(z a)n

(w a)n+1

converges uniformly on {w C : |w a| = r}. Hence

|wa|=r

f(w)n=0

(z a)n

(w a)n+1dw =

n=0

|wa|=r

f(w)

(w a)n+1dw

(z a)n.

Thus

f(z) =n=0

1

2i

|wa|=r

f(w)

(w a)n+1dw

(z a)n.

This formula for the coefficients holds for all 0 < r < R. The series is calledthe Taylor series for f about a.

(i) We have now proved that, near a, the function f is given by a power se-ries of positive radius. It follows that its derivative f is given by a powerseries of the same radius. Continuing in the way, f fn are all givenby powers series of the same radius. Thus f has complex derivatives ofall orders and so is infinitely differentiable.

(ii) We have now derived the additional Cauchy Integral formulae for thevalues of the derivatives of f. We have shown the function f has theform

f(z) = n=0 bn(z a)

n near a and since

f(k)(a) = bk k!,

f(k)(a) =k!

2i

|za|=r

f(z)

(z a)k+1dz,

for all k N. These are the Cauchy Integral Formulae. The formulaecan be derived directly, without using the Taylor series. When you havestudied sufficient integration theory you will know that they can also bederived by pulling the complex derivative through the integral in thefirst Cauchy Integral formula.

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1. it is easy to show, using Taylor series, that versions of lHopitals rulehold for limits of quotients of holomorphic functions.

Example 5.5. 1. Find the Taylor series for sin z about z = 2

.

This can be done by noting that the odd derivatives ofsinz vanish when z = 2

and the 2kth derivative sin z is (1)k sin z. Thus the Taylor series for sin about2

is

sin z =k=0

(1)k

2k!(z

2)2k.

In practice, there are sometimes of finding Taylor series from known series with-out repeated differentiation of the given function.

sin z = sin(z

2+

2)

= sin(z

2)cos

2+ cos(z

2)sin

2

= cos(z

2)

=n=0

(1)k

(2k)!(z

2)2k.

(ii) Find the Taylor expansion off(z) = 1z21 about z = 3. From the general theorywe can expect the Taylor series to have radius 2 as the function is holomorphicon C \ {1, 1} and B2(3) is the biggest ball centred at 3 lying in the domain ofthe function.

1

z2 1=

12

z 1

12

z + 1

and f(k)(z) = 12

(1)kk!(z1)k+1

(1)kk!

(z+1)k+1

. So that

f(k)(3) =1

2

(1)kk!1

2k+1

1

4k+1 .

The coefficient of(z 3)k in the Taylor expansion is

1

2(1)k

1

2k+1

1

1

2k+1

.

Thus1

z2 1=

k=0

(1)k1

2k+2

1

1

2k+1

(z 3)k.

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Lecture 13

More applications of Cauchys Integral FormulaTheorem 5.3 (Liouvilles Theorem). Suppose that f : C C is holomorphic andand that there exists M > 0 such that |f(z)| < M for all z C. Then the function fis constant.

Proof. Let a, b C. Then

f(a) f(b) =1

2i

|za|=r1

f(z)

z adz

1

2i

|zb|=r2

f(z)

z bdz

for any r1, r2 > 0. Choose R > max{|a| + r1, |b| + r2}. Then by the deformation

version of Cauchys theorem for closed curves,

1

2i

|za|=r1

f(z)

z adz =

1

2i

|z|=R

f(z)

z adz,

1

2i

|zb|=r2

f(z)

z bdz =

1

2i

|z|=R

f(z)

z bdz.

Hence

f(a) f(b) =1

2i|z|=R

f(z)

z a

f(z)

z b dz

=1

2i

|z|=R

f(z)a b

(z a)(z b)dz,

and so

|f(a) f(b)| 1

2M

|a b|

(R |a|)(R |b|)2R,

for all sufficiently large R. Let R . The inequality then implies that

|f(a) f(b)| = 0

and, as a and b are arbitrary points in C, we have shown that f is the constantfunction.

Theorem 5.4 (Generalised Liouvilles Theorem). Suppose that f : C C isholomorphic and that there exist positive k , M , R such that |f(z)| M|z|k for all|z| > R. Then f is a polynomial of degree not exceeding k,

Proof. Sketch!

f(z) =n=0

bnzn

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|bn| =

1

2i

|z|=r

f(z)

zn+1dz

12

M rk

rn+12r

= O(1

rnk)

as r . Therefore, bn = 0 ifn > k.

Theorem 5.5 (Fundamental Theorem of Algebra). Let f be a polynomial of degreen > 0,

f(z) = bnzn + bn1z

n1 + + b0,

bn, bn1, . . . , b0 C, bn = 0. Then there exists a C such that f(a) = 0.

Proof. Suppose that f(z) = 0 for all z C. Then g(z) =1

f(z)is holomor-

phic in the whole complex plane. Now f(z) = zn(bn + bn11

z+ + b0

1

zn) and

bn + bn11

z+ + b0

1

zn bn as |z| . Hence |f(z)| as |z| and

so g(z) 0 as |z| which implies that g is bounded on C. By LiouvillesTheorem, g (and hence f) is constant which gives a contradiction. Therefore,there exists a C such that f(a) = 0.

Corollary 5.1. There exist a1, a2, . . . , an C

such thatf(z) = bn(z a1)(z a2) (z an).

This follows from the theorem above and the Euclidean algorithm. Thereexists a1 such that f(z) = (z a1)s(z) where s(z) is a polynomial of degreen 1 and by recursion such a1, a2, . . . , an exist.

Theorem 5.6 (Moreras Theorem). Let D be a domain in C. Suppose that f : D C is continuous and such that for each triangle D (interior and boundary oftriangle lie in D),

f(z)dz = 0.

Then f is holomorphic.

Proof. Let a D and let > 0 be such that B(a) D. The ball B(a) is convexand defining

F(z) =

[a,z]

f(w) dw,

for each z BR(a), we obtain a function F : B(a) C such that F = f.

Thus F is holomorphic. By Taylors Theorem, F is given by a power series on

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B(a) and so all its derivatives are also holomorphic. In particular, f = F is

holomorphic on B(a). Since this holds for each a D, f is holomorphic on

D.

Proposition 5.1. Suppose that fn, f : D C, n N are such that all fn areholomorphic and that fn f uniformly on D. Then f is holomorphic.

Proof. The functions fn are holomorphic and hence continuous. As the con-vergence is uniform, the function f is also continuous. Let be a trianglecontained in D. Then

f(z) dz =

limn

fn(z) dz

= limn

fn(z) dz = 0

because the convergence is uniform. It follows from Moreras Theorem that fis holomorphic.

Proposition 5.2. Suppose that f : D C is continuous and that L is a line in Csuch that f : D \ L C is holomorphic. Then f is holomorphic on D.

Proof. Let be a triangle in D. Suppose that L = . Then

f(

z) d

z=1

f(

z) d

z+2

f(

z) d

z,

where 1 and 2 are two polygons sharing a common edge lying along L,the other edges of 1 and 2 being made up of edges or parts of edges of. (See diagram.) Next we move 1 and 2 off the line L. Suppose thatL = {a + rei : r R}. Let 1 = 1 + e

(2), choosing the sign which

moves 1 so that it does not intersect L.(See diagram.) Then1

f(z) dz = 0

and lim01

f(z) dz =1

f(z) dz because f is continuous and hence uni-

formly continuous on compact sets. Similarly,2

f(z) dz = 0 and therefore,

f(z)dz = 0 for any triangle contained in D. It follows from Moreras Theo-rem that f is holomorphic on D.

As a special case we note that if f is holomorphic on B(a) \ {a} and con-tinuous at a then f is holomorphic on B(a).

Lecture 14

Proposition 5.3 (Zeros are isolated). Suppose f : BR C is holomorphic, f(a) =0 but f(z) = 0 for some z BR(a). Then there exists > 0 such that f(z) = 0 forall z B(a) \ {a}.

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Proof. The function f has a Taylor series expansion on BR(a):

f(z) =

n=0

bn(z a)n,

and f(a) = 0 implies that b0 = 0. Since f does not vanish identically, thereexists k N such that b0 = b1 = = bk1 = 0 but bk = 0. Then

f(z) =n=k

bn(z a)n =

n=0

bk+n(z a)k+n.

If

g(z) =

n=0

bk+n(z a)n

,

then g is holomorphic in BR(a) and g(a) = bk = 0. There exists > 0 such that

|g(z) g(a)| < |bk|2 for all |z a| < , and

|g(z)| = |g(z) g(a) + g(a)| | |g(z) g(a)| |g(a)| | |bk| |bk|

2= 0.

Then g(z) = 0 ifz B(a) \ {a}. Certainly, (z a)k = 0 ifz B(a) \ {a}and hence f(z) = 0 for all z B(a) \ {a}.

Theorem 5.7 (Identity Theorem). Let D be a domain in C and f, g : D Cholomorphic functions. Then the following are equivalent:

(i) f g on D;

(ii) {z D : f(z) = g(z)} has an accumulation point in D;

(iii) there exists a D such that f(k)(a) = g(k)(a) for all k N {0}.

Proof. Let h = f g. Then (i) becomes: h 0; (ii) becomes: h1(0) has anaccumulation point in D; (iii) becomes: there exists a D such that h(k)(a) = 0

for all k N {0}.Assume (i) holds. Then h1(0) = D which is a nonempty open set and

hence has an accumulation point in D. (In fact, each point ofD is an accumu-lation point ofD.) So (ii) holds.

Assume (ii) holds. There exists a D which is an accumulation point ofh1(0). Let R > 0 be such that BR(a) D. On BR(a), the holomorphic functionh has a Taylor expansion

h(z) =n=0

bn(z a)n.

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There exists a series of points zn in D such that zn = a, zn a and h(zn) = 0.Then as h is continuous at a and zn a, h(a) = 0. Hence h(a) = b0 = 0.

Now suppose that k N is such that b0 = b1 = = bk1 = 0. Then

h(z) =n=k

bn(z a)n

= (z a)kn=0

bk+n(z a)n.

Let q(z) =

n=0 bk+n(z a)n. Then q(z) is holomorphic on BR(a). For each

m N, zm a = 0 but h(zm) = 0 This implies that q(zm) = 0. Further, zm aas m so q(a) = bk = 0. it follows that all coefficients, b0, b1, b2, . . . vanish

and as bn =h(n)(a)

n!we have shown that (iii) holds.

Assume that (iii) holds. There exists a D such that h(n)(a) = 0 for alln N {0}.

LetA = {z D : h(n)(z) = 0 for all n N {0}}.

By assumption A = . Let w A. Then h(n)(w) = 0 for all n N {0}. Thereexists R > 0 such that BR(w) D and for all z BR(w),

h(z) =

n=0

h(n)(w)

n! (z w)n

.

Therefore, h(z) = 0 for all z BR(w) and we have shown that A is an opensubset ofD.

Each h(n) : D C is a holomorphic and hence continuous function. Let

An = {z D : h(n)(z) = 0} =

h(n)

1(0).

The each An is a closed subset ofD because the pre-image of a closed set by acontinuous map is closed.

Further,

A =

n=0

An

and so A is an intersection of closed sets subsets ofD and hence a closed subsetofD. But D is connected. Therefore A = D and so h(z) = 0 for all z D.

We have now shown (i) (ii) (iii) (i).

Example 5.6. We know that tan2x =2tan x

1 tan2 xfor all x (

4, 4

) and hence the

formula

tan2z =2tan z

1 tan2 z

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holds on any domain containingx (4

, 4

) and on which both tan2z and2tan z

1 tan2 zare defined and holomophic.

Example 5.7. Suppose that f, g : D C are holomorphic and suppose that theproduct f.g 0. Let Zf = {z D : f(z) = 0}, Zg = {z D : g(z) = 0}. Then

D = Zf Zg.

Indeed, let a D and R > 0 be such that

D = {z C : |z a| R} =: BR(a)

is contained in D. Then D is a closed bounded subset ofC and

D = (D Zf) (D Zg).

At least one ofDZf and DZg contains infinitely many points and hence has a limitin D and hence in D. By the Identity Theorem, at least one off and g is identicallyequal to 0. We have now shown that the ring of holomorphic functions on a domain Dis, in algebraic language, an integral domain.

Example 5.8. Suppose that D is a domain which is symmetric across the real axis,so z D z D. Suppose that f : D C and f(z) R if z D R. Let

g(z) = f(z). Then we know that g : D C is holomorphic and that f and g coincide

on the intersection of the real axis and D. Hence f g, that is, f(z) = f(z) for allz D.

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