Cauchy Integral Formula

7/28/2019 Cauchy Integral Formula

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Cauchy Integral Formula Infinite Differentiability Fundamental Theorem of Algebra Maximum Modulus Principle

The Cauchy Integral Formula

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science



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Introduction




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Introduction

1. One of the most important consequences of the Cauchy-Goursat

Integral Theorem is that the value of an analytic function at a

point can be obtained from the values of the analytic function on

a contour surrounding the point




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Introduction




a contour surrounding the point (as long as the function is

defined on a neighborhood of the contour and its inside).




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Introduction






2. The result itself is known as Cauchys Integral Theorem.




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Introduction






2. The result itself is known as Cauchys Integral Theorem.3. Among its consequences is, for example, the Fundamental

Theorem of Algebra, which says that every nonconstant complex

polynomial has at least one complex zero.



C h I t l F l I fi it Diff ti bilit F d t l Th f Al b M i M d l P i i l


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Introduction



C h I t l F l I fi it Diff ti bilit F d t l Th f Al b M i M d l P i i l


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Introduction

4. Once we have introduced series, another consequence is the fact

that every analytic function is locally equal to a power series.





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Introduction



This very powerful result is a cornerstone of complex analysis.





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Introduction




5. In this presentation we will at least be able to prove that analytic

functions have derivatives of any order.





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Introduction




5. In this presentation we will at least be able to prove that analytic

functions have derivatives of any order.

6. ... and the above are only highlights of the consequences ofCauchys Integral Theorem.





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Theorem.





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y g y g p

Theorem. Cauchys Integral Formula.





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y g y g p

Theorem. Cauchys Integral Formula. Let C be a simple closed

positively oriented piecewise smooth curve, and let the function f be

analytic in a neighborhood of C and its interior.





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Theorem. Cauchys Integral Formula. Let C be a simple closed

positively oriented piecewise smooth curve, and let the function f be

analytic in a neighborhood of C and its interior. Then for every z0 in

the interior of C we have that

f(z0) =1

2i

C

f(z)

zz0dz.





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Example.





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Example. Lets first check out if the theorem works for f(z) = 1 andthe circle C(r,z0) of radius r around z0.





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12i

C(r,z0)

1zz0

dz





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12i

C(r,z0)

1zz0

dz = 12i

20

1(z0 + reit)z0

ireit dt





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12i

C(r,z0)

1zz0

dz = 12i

20

1(z0 + reit)z0

ireit dt

=1

2i

20

1

reitireit dt





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12i

C(r,z0)

1zz0

dz = 12i

20

1(z0 + reit)z0

ireit dt

=1

2i

20

1

reitireit dt

=

1

22

0 dt





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12i

C(r,z0)

1zz0

dz = 12i

20

1(z0 + reit)z0

ireit dt

=1

2i

20

1

reitireit dt

=

1

22

0 dt= 1





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12i

C(r,z0)

1zz0

dz = 12i

20

1(z0 + reit)z0

ireit dt

=1

2i

20

1

reitireit dt

=

1

22

0 dt= 1 = f(z0)

Bernd Schroder Louisiana Tech University, College of Engineering and ScienceThe Cauchy Integral Formula



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Proof of Cauchys Integral Formula.




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D

r

-

?O

--

1

]

C

C(r,z0)




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D

r

-

?O

--

1

]

C

C(r,z0)

r




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D

r

-

?O

--

1

]

C

C(r,z0)

rr




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D

r

-

?O

--

1

]

C

C(r,z0)

rr r




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D

r

-

?O

--

1

]

C

C(r,z0)

rr rr




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D

r

-

?O

--

1

]

C

C(r,z0)

rr rrI




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D

r

-

?O

--

1

]

C

C(r,z0)

rr rrIR




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D

r

-

?O

--

1

]

C

C(r,z0)

rr rrIR

C

f(z)

zz0dz =

C(r,z0)

f(z)

zz0dz.




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C

f(z)

z

z0

dz2if(z0)




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C

f(z)

zz0

dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz





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C

f(z)

zz0

dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz





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C

f(z)

zz0

dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)

f(z)f(z0)zz0 d|z|





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C

f(z)

zz0

dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)

f(z)f(z0)zz0 d|z| max

C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|





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C

f(z)

zz0dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)


C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|

= maxC(r,z0)

f(z)f(z0)

2r

1

r





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C

f(z)

zz0dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)


C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|

= maxC(r,z0)

f(z)f(z0)

2r

1

r= 2 max

C(r,z0)

f(z)f(z0)





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C

f(z)

zz0dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)


C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|

= maxC(r,z0)

f(z)f(z0)

2r

1

r= 2 max

C(r,z0)

f(z)f(z0)

and if we choose rsmall enough, we can make the last term arbitrarily

small.




P f f C h I l F l


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C

f(z)

zz0dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)


C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|

= maxC(r,z0)

f(z)f(z0)

2r

1

r= 2 max

C(r,z0)

f(z)f(z0)

and if we choose rsmall enough, we can make the last term arbitrarilysmall. But that means that the original difference must be zero.




P f f C h I t l F l


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C

f(z)

zz0dz2if(z0)

=

C(r,z0)

f(z)

zz0dz

C(r,z0)

f(z0)

zz0dz

=

C(r,z0)

f(z)f(z0)

zz0dz

C(r,z0)


C(r,z0)

f(z)f(z0)

C(r,z0)

1

rd|z|

= maxC(r,z0)

f(z)f(z0)

2r

1

r= 2 max

C(r,z0)

f(z)f(z0)

and if we choose rsmall enough, we can make the last term arbitrarilysmall. But that means that the original difference must be zero.




1

f ( )


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Analyzing f(z0) =1

2i

f(z)

zz0dz.




1

f ( )


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Analyzing f(z0) =1

2i

f(z)

zz0dz.

1. Note that the right side is a function ofz0.




1

f (z)


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Analyzing f(z0) =1

2i

f(z)

zz0dz.

1. Note that the right side is a function ofz0.2. For z fixed,

1

zz0is differentiable in z0.




1

f (z)


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Analyzing f(z0) =1

2i

f(z)

zz0dz.


1


3. So if we could move the derivative into the integral, we could get

a formula for f.




1

f (z)


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Analyzing f(z0) =1

2i

f(z)

zz0dz.


1



a formula for f.

4. And, anticipating that the new integrand will involve1

(zz0)2,

there is no reason to think that the process should stop there.




1

f (z)


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Analyzing f(z0) =1

2i

f(z)

zz0dz.


1



a formula for f.

4. And, anticipating that the new integrand will involve1

(zz0)2,

there is no reason to think that the process should stop there.

5. So for analytic functions, being once differentiable should imply

that we have derivatives of any order.




Warning


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Warning.




Warning. The next result (as motivated on the preceding panel) does


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notwork for functions of a real variable.






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notwork for functions of a real variable. Consider

f(x) = x2; for x > 0,

x2

; for x 0.






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g ( p g p )


f(x) = x2; for x > 0,

x2

; for x 0.Its derivative is 2|x|






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g ( p g p )


f(x) = x2; for x > 0,

x2

; for x 0.Its derivative is 2|x|, which is not

differentiable at 0.




Theorem.


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logo1Bernd Schroder Louisiana Tech University, College of Engineering and Science



Theorem. Cauchys Integral Formula


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Theorem. Cauchys Integral Formula (extended).


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Theorem. Cauchys Integral Formula (extended). Let C be a simple


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closed positively oriented piecewise smooth curve, and let the

function f be analytic in a neighborhood of C and its interior.




Theorem. Cauchys Integral Formula (extended). Let C be a simple


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closed positively oriented piecewise smooth curve, and let the

function f be analytic in a neighborhood of C and its interior. Then

for every z0 in the interior of C and every natural number n we havethat f is n-times differentiable at z0 and its derivative is

f(n)(z0) =n!

2i

C

f(z)

(zz0)n+1dz.




Proof.


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Proof. Induction on n.


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B 0


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Base step, n = 0.





B t 0 Thi i C h I t l F l


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Base step, n = 0. This is Cauchys Integral Formula.





B t 0 Thi i C h I t l F l


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Base step, n = 0. This is Cauchys Integral Formula.Induction step, n (n + 1).





Base step 0 This is Cauchys Integral Formula


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Base step, n = 0. This is Cauchys Integral Formula.Induction step, n (n + 1).

limwz0

f(n)(w)f(n)(z0)wz0





Base step n 0 This is Cauchys Integral Formula


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Base step, n = 0. This is Cauchy s Integral Formula.Induction step, n (n + 1).

limwz0

f(n)(w)f(n)(z0)wz0

= limwz0

1

wz0

n!

2i

C

f(z)

(zw)n+1dz

n!

2i

C

f(z)

(zz0)n+1dz





Base step n = 0 This is Cauchys Integral Formula


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limwz0

f(n)(w)f(n)(z0)wz0

= limwz0

1

wz0

n!

2i

C

f(z)

(zw)n+1dz

n!

2i

C

f(z)

(zz0)n+1dz

= limwz0

n!2i

1wz0

C

f(z)(zw)n+1

f(z)(zz0)n+1

dz







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limwz0

f(n)(w)f(n)(z0)wz0

= limwz0

1

wz0

n!

2i

C

f(z)

(zw)n+1dz

n!

2i

C

f(z)

(zz0)n+1dz

= limwz0

n!2i

1wz0

C

f(z)(zw)n+1

f(z)(zz0)n+1

dz

= limwz0

n!

2i

C(z0,r)

f(z)

1(zw)n+1

1(zz0)n+1

wz0dz







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limwz0

f(n)(w)f(n)(z0)wz0

= limwz0

1

wz0

n!

2i

C

f(z)

(zw)n+1dz

n!

2i

C

f(z)

(zz0)n+1dz

= limwz0

n!2i

1wz0

C

f(z)(zw)n+1

f(z)(zz0)n+1

dz

= limwz0

n!

2i

C(z0,r)

f(z)

1(zw)n+1

1(zz0)n+1

wz0dz

= n!2i

C(z0,r)

f(z)((n + 1)) 1(zz0)n+2

(1) dz




n!

1


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n!

2i

C(z0,r)

f(z)((n + 1))1

(zz0)n+2(1) dz




n!

( )( ( ))1

( )


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n!

2i

C(z0,r)

f(z)((n + 1))1

(zz0)n+2(1) dz

= (n + 1)!2i

C(z0,r)

f(z)(zz0)n+2

dz




n!

f ( )( ( 1))1

( 1) d


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n!

2i

C(z0,r)

f(z)((n + 1))(zz0)n+2

(1) dz

= (n + 1)!2i

C(z0,r)

f(z)(zz0)n+2

dz

=(n + 1)!

2i

C

f(z)

(zz0)n+2dz




n!

f ( )( ( 1))1

( 1) d


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2i

C(z0,r)

f(z)((n + 1))(zz0)n+2

(1) dz

= (n + 1)!2i

C(z0,r)

f(z)(zz0)n+2

dz

=(n + 1)!

2i

C

f(z)

(zz0)n+2dz




Corollary.


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Corollary. Let f be an analytic function on an open domain.


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Corollary. Let f be an analytic function on an open domain. Then all

derivatives of f are analytic on this domain, too.


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derivatives of f are analytic on this domain, too. Moreover, the


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component functions (the real and imaginary parts) have continuous

partial derivatives of all orders throughout the domain.







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Proof.







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Proof. By the preceding result, we see that f has derivatives of all

orders.







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orders. That means all derivatives are differentiable







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orders. That means all derivatives are differentiable and thus all

derivatives are analytic on the domain.






f i ( h l d i i ) h i


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derivatives are analytic on the domain. Therefore real and imaginary

part have






f i ( h l d i i ) h i


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part have (by repeated application of the Cauchy-Riemann equations)






t f ti (th l d i i t ) h ti


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part have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.


The Cauchy Integral Formula Cauchy Integral Formula Infinite Differentiability Fundamental Theorem of Algebra Maximum Modulus Principle





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part have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.



Theorem.


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Theorem. Moreras Theorem.


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Theorem. Moreras Theorem. Let f be a continuous complex

function on an open set so that for every simple closed curve C

t i d i th t h

f ( ) d 0


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contained in the open set we have

C

f(z) dz = 0.





t i d i th t h

f ( ) d 0 Th f i l ti i


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C

f(z) dz = 0. Then f is analytic in

the open set.






f (z) dz 0 Then f is analytic in


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C


the open set.

Proof.






f (z) dz = 0 Then f is analytic in


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C


the open set.

Proof. By theorem from an earlier presentation, f has an

antiderivative F.








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C


the open set.


antiderivative F. By the preceding theorem, all derivatives of F








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C


the open set.



(including f) are analytic.








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C


the open set.



(including f) are analytic.



Theorem. Cauchys Inequality.


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Theorem. Cauchys Inequality. Let f be analytic on a circle CR(z0)of radius R centered at z0 and in the circles interior.


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Theorem. Cauchys Inequality. Let f be analytic on a circle CR(z0)of radius R centered at z0 and in the circles interior. If |f| is bounded

by MR on the circle then for all n we have f(n)(z0)

n!MR


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by MR on the circle, then for all n we have f (z0) Rn .




by MR on the circle, then for all n we have f(n)(z0)

n!MR.


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by MR on the circle, then for all n we have f (z0) Rn .Proof.






n!MR.


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by R on the ci cle, then fo all n we have f (z0) RnProof. f(n)(z0)

=

n!

2i

CR(z0)

f(z)

(zz0)n+1dz






n!MR

R.


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y R , f f (z0)

Rn

Proof. f(n)(z0) =

n!

2i

CR(z0)

f(z)

(zz0)n+1dz

n!

2

CR(z0)

f(z)(zz0)n+1 d|z|






n!MR

Rn.


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y R f f ( 0)

Rn

Proof. f(n)(z0) =

n!

2i

CR(z0)

f(z)

(zz0)n+1dz

n!

2

CR(z0)

f(z)(zz0)n+1 d|z|

n!

2

CR(z0)

MR

Rn+1d|z|






n!MR

Rn.


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y f

f ( )

RnProof. f(n)(z0)

=

n!

2i

CR(z0)

f(z)

(zz0)n+1dz

n!

2

CR(z0)

f(z)(zz0)n+1 d|z|

n!

2

CR(z0)

MR

Rn+1d|z|

n!

22R

MR

Rn+1





by MR on the circle, then for all n we have

f(n)(z0)

n!MR

Rn.


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RnProof. f(n)(z0)

=

n!

2i

CR(z0)

f(z)

(zz0)n+1dz

n!

2

CR(z0)

f(z)(zz0)n+1 d|z|

n!

2

CR(z0)

MR

Rn+1d|z|

n!

22R

MR

Rn+1=

n!MR

Rn





by MR on the circle, then for all n we have

f(n)(z0)

n!MR

Rn.


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RnProof. f(n)(z0)

=

n!

2i

CR(z0)

f(z)

(zz0)n+1dz

n!

2

CR(z0)

f(z)(zz0)n+1 d|z|

n!

2

CR(z0)

MR

Rn+1d|z|

n!

22R

MR

Rn+1=

n!MR

Rn




Theorem.


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Theorem. Liouvilles Theorem.


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Theorem. Liouvilles Theorem. Let f be an entire function.


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Theorem. Liouvilles Theorem. Let f be an entire function. (That is,

f is analytic in C.)


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f is analytic in C.) If f is bounded, then f is constant.


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Proof.


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Proof.






Proof. Let |f | be bounded by B.


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Proof. Let |f| be bounded by B.






Proof. Let |f| be bounded by B. Let R > 0 be arbitrary.


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|f | y > y






Proof. Let |f| be bounded by B. Let R > 0 be arbitrary. By Cauchys


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|f | y y y y

Inequality with n = 1, for every z0 in C we havef(z0)








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|f | y y y y

Inequality with n = 1, for every z0 in C we havef(z0) BR








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| |

Inequality with n = 1, for every z0 in C we havef(z0) BR 0 (R ).








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Because R was arbitrary, we infer that f(z0) = 0.








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Because R was arbitrary, we infer that f(z0) = 0. Because z0 was

arbitrary, we have f

= 0.








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= 0. But f

= 0 implies that f is constant.








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= 0. But f

= 0 implies that f is constant.




Theorem.


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Theorem. Fundamental Theorem of Algebra.


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Theorem. Fundamental Theorem of Algebra. Every nonconstant

complex polynomial has at least one complex zero.


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Proof.


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Proof. Let p be a complex polynomial without zeros.


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Proof. Let p be a complex polynomial without zeros. Then 1p

is

l i i h l l


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analytic in the complex plane.







is

l ti i th l l W l i th t 1 i b d d C


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analytic in the complex plane. We claim that 1p

is bounded on C.







is

l ti i th l l W l i th t 1 i b d d C T


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is bounded on C. To

see this claim, first note that limz

1

p(z)= 0







is

analytic in the complex plane We claim that 1 is bounded on C To


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is bounded on C. To


1

p(z)= 0, because lim

zp(z) = .







is



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is bounded on C. To


1


zp(z) = .

Thus 1p

is bounded outside a circle C(R,0) for sufficiently large R.







is



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analytic in the complex plane. We claim that 1p is bounded on C. To


1


zp(z) = .

Thus 1p


But the only way1

p can be unbounded inside C(R,0) is for thedenominator p(z) to go to zero somewhere







is



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1


zp(z) = .

Thus 1p


But the only way1

p can be unbounded inside C(R,0) is for thedenominator p(z) to go to zero somewhere, which was excluded byhypothesis.







is



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1


zp(z) = .

Thus 1p


But the only way1

p can be unbounded inside C(R,0) is for thedenominator p(z) to go to zero somewhere, which was excluded byhypothesis. Thus 1

pis bounded on C.







is



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1


zp(z) = .

Thus 1p


But the only way1


pis bounded on C.

Now by Liouvilles Theorem we infer that 1p

is constant.







is



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analytic in the complex plane. We claim that p is bounded on C. To


1


zp(z) = .

Thus 1p


But the only way1


pis bounded on C.


is constant. Hence p is

constant.







is

analytic in the complex plane. We claim that 1 is bounded on C. To


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analytic in the complex plane. We claim that p is bounded on C. To


1


zp(z) = .

Thus 1p


But the only way1


pis bounded on C.


is constant. Hence p is

constant.




Lemma.


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Lemma. If f is analytic in some neighborhood|zz0|< and|f(z)| |f(z0)| for all z in that neighborhood


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Lemma. If f is analytic in some neighborhood|zz0|< and|f(z)| |f(z0)| for all z in that neighborhood, then in fact f(z) = f(z0)for all z in that neighborhood.


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Proof.


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Proof. By Cauchys Formula, for all circles C(z0,r) with r< wehave that

f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz


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f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z)

d| |


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2

C(z0,r)

f ( )zz0 d|z|





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z)

d|z|


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2

C(z0,r)

( )zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z|





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z)

d|z|


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2

C(z0,r)

( )zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z| =f(z0)





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z) d|z|


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2

C(z0,r)

zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z| =f(z0)

and the latter integral will be strictly smaller thanf(z0) if there is

even a single point z on C(z0,r) wheref(z)< f(z0).





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z) d|z|


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2

C(z0,r)

zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z| =f(z0)


even a single point z on C(z0,r) wheref(z)< f(z0). Thus for all

r< and all |zz0| = r we must havef(z) = f(z0).





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z) d|z|


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2

C(z0,r)

zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z| =f(z0)



r< and all |zz0| = r we must havef(z) = f(z0). But if|f| is

constant for |zz0|< , then so is f.





f(z0)

=

1

2i

C(z0,r)

f(z)

zz0dz

1 f(z) d|z|


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2

C(z0,r)

zz0 d|z|

1

2

C(z0,r)

f(z0)r

d|z| =f(z0)



r< and all |zz0| = r we must havef(z) = f(z0). But if|f| is

constant for |zz0|< , then so is f.




Theorem.


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Theorem. Maximum Modulus Principle.


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Theorem. Maximum Modulus Principle. Let f be analytic and not

constant on an open domain.


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constant on an open domain. Then f does not assume a maximum

value on the domain.


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value on the domain. That is, there is no z0 in the domain so that

|f(z)| |f(z0)| for all z in the domain.


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value on the domain. That is, there is no z0 in the domain so that

|f(z)| |f(z0)| for all z in the domain.

Proof.


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logo1Bernd Schroder Louisiana Tech University, C

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Cauchy Integral Formula