The Cauchy Integral Formula2.The result itself is known as Cauchy’s Integral Theorem. 3.Among its consequences is, for example, the Fundamental Theorem of Algebra, which says that

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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

1. One of the most important consequences of the Cauchy-GoursatIntegral Theorem is that the value of an analytic function at apoint can be obtained from the values of the analytic function ona contour surrounding the point (as long as the function isdefined on a neighborhood of the contour and its inside).

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

Theorem of Algebra, which says that every nonconstant complexpolynomial has at least one complex zero.



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Introduction1. One of the most important consequences of the Cauchy-Goursat

Integral Theorem is that the value of an analytic function at apoint can be obtained from the values of the analytic function ona contour surrounding the point

(as long as the function isdefined on a neighborhood of the contour and its inside).





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Integral Theorem is that the value of an analytic function at apoint can be obtained from the values of the analytic function ona contour surrounding the point (as long as the function isdefined on a neighborhood of the contour and its inside).





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2. The result itself is known as Cauchy’s Integral Theorem.

3. Among its consequences is, for example, the FundamentalTheorem of Algebra, which says that every nonconstant complexpolynomial has at least one complex zero.



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Introduction

4. Once we have introduced series, another consequence is the factthat every analytic function is locally equal to a power series.This very powerful result is a cornerstone of complex analysis.

5. In this presentation we will at least be able to prove that analyticfunctions have derivatives of any order.

6. ... and the above are only highlights of the consequences ofCauchy’s Integral Theorem.



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Introduction4. Once we have introduced series, another consequence is the fact

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

This very powerful result is a cornerstone of complex analysis.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 of

Cauchy’s Integral Theorem.



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that every analytic function is locally equal to a power series.This very powerful result is a cornerstone of complex analysis.





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Theorem.

Cauchy’s Integral Formula. Let C be a simple closedpositively oriented piecewise smooth curve, and let the function f beanalytic in a neighborhood of C and its interior. Then for every z0 inthe interior of C we have that

f (z0) =1

2πi

∫C

f (z)z− z0

dz.



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Theorem. Cauchy’s Integral Formula.

Let C be a simple closedpositively oriented piecewise smooth curve, and let the function f beanalytic in a neighborhood of C and its interior. Then for every z0 inthe interior of C we have that

f (z0) =1

2πi

∫C

f (z)z− z0

dz.



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Theorem. Cauchy’s Integral Formula. Let C be a simple closedpositively oriented piecewise smooth curve, and let the function f beanalytic in a neighborhood of C and its interior.

Then for every z0 inthe interior of C we have that

f (z0) =1

2πi

∫C

f (z)z− z0

dz.



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Theorem. Cauchy’s Integral Formula. Let C be a simple closedpositively oriented piecewise smooth curve, and let the function f beanalytic in a neighborhood of C and its interior. Then for every z0 inthe interior of C we have that

f (z0) =1

2πi

∫C

f (z)z− z0

dz.



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Example.

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

12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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

12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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12πi

∫C(r,z0)

1z− z0

dz

=1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt

= 1 = f (z0)



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12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1

= f (z0)



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12πi

∫C(r,z0)

1z− z0

dz =1

2πi

∫ 2π

0

1(z0 + reit)− z0

ireit dt

=1

2πi

∫ 2π

0

1reit ireit dt

=1

2π

∫ 2π

0dt = 1 = f (z0)



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Proof of Cauchy’s Integral Formula.

D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

r

rrrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rr

rrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrr

rIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrr

I

R

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrrI

R

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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D

r-�

?O

�

�

�

--

1

]

�

C

C(r,z0)

rrrrIR

∫C

f (z)z− z0

dz =∫

C(r,z0)

f (z)z− z0

dz.



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∣∣∣∣∫C

f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣

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



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Proof of Cauchy’s Integral Formula.∣∣∣∣∫C

f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z|

≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣

and if we choose r small enough, we can make the last term arbitrarilysmall.

But that means that the original difference must be zero.



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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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f (z)z− z0

dz−2πif (z0)∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)z− z0

dz−∫

C(r,z0)

f (z0)z− z0

dz∣∣∣∣

=∣∣∣∣∫C(r,z0)

f (z)− f (z0)z− z0

dz∣∣∣∣

≤∫

C(r,z0)

∣∣∣∣ f (z)− f (z0)z− z0

∣∣∣∣ d|z| ≤ maxC(r,z0)

∣∣f (z)− f (z0)∣∣∫

C(r,z0)

1r

d|z|

= maxC(r,z0)

∣∣f (z)− f (z0)∣∣2πr

1r

= 2π maxC(r,z0)

∣∣f (z)− f (z0)∣∣




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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.

1. Note that the right side is a function of z0.

2. For z fixed,1

z− z0is differentiable in z0.

3. So if we could move the derivative into the integral, we could geta formula for f ′.

4. And, anticipating that the new integrand will involve1

(z− z0)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.



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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.


2. For z fixed,1




(z− z0)2 ,





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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.


2. For z fixed,1




(z− z0)2 ,





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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.


2. For z fixed,1




(z− z0)2 ,





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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.


2. For z fixed,1




(z− z0)2 ,

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

5. So for analytic functions, being once differentiable should implythat we have derivatives of any order.



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Analyzing f (z0) =1

2πi

∫γ

f (z)z− z0

dz.


2. For z fixed,1




(z− z0)2 ,





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Warning.

The next result (as motivated on the preceding panel) doesnot work for functions of a real variable. Consider

f (x) ={

x2; for x > 0,−x2; for x≤ 0.

Its derivative is 2|x|, which is not

differentiable at 0.



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Warning. The next result (as motivated on the preceding panel) doesnot work for functions of a real variable.

Consider

f (x) ={

x2; for x > 0,−x2; for x≤ 0.





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Warning. The next result (as motivated on the preceding panel) doesnot work for functions of a real variable. Consider

f (x) ={

x2; for x > 0,−x2; for x≤ 0.





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f (x) ={

x2; for x > 0,−x2; for x≤ 0.

Its derivative is 2|x|

, which is not




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f (x) ={

x2; for x > 0,−x2; for x≤ 0.





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Theorem.

Cauchy’s Integral Formula (extended). Let C be a simpleclosed positively oriented piecewise smooth curve, and let thefunction f be analytic in a neighborhood of C and its interior. Thenfor 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!

2πi

∫C

f (z)(z− z0)n+1 dz.



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Theorem. Cauchy’s Integral Formula

(extended). Let C be a simpleclosed positively oriented piecewise smooth curve, and let thefunction f be analytic in a neighborhood of C and its interior. Thenfor 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!

2πi

∫C

f (z)(z− z0)n+1 dz.



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Theorem. Cauchy’s Integral Formula (extended).

Let C be a simpleclosed positively oriented piecewise smooth curve, and let thefunction f be analytic in a neighborhood of C and its interior. Thenfor 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!

2πi

∫C

f (z)(z− z0)n+1 dz.



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Theorem. Cauchy’s Integral Formula (extended). Let C be a simpleclosed positively oriented piecewise smooth curve, and let thefunction f be analytic in a neighborhood of C and its interior.

Thenfor 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!

2πi

∫C

f (z)(z− z0)n+1 dz.



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Theorem. Cauchy’s Integral Formula (extended). Let C be a simpleclosed positively oriented piecewise smooth curve, and let thefunction f be analytic in a neighborhood of C and its interior. Thenfor 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!

2πi

∫C

f (z)(z− z0)n+1 dz.



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Proof.

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

limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C

f (z)(z−w)n+1 dz− n!

2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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Proof. Induction on n.

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

limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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Proof. Induction on n.Base step, n = 0.

This is Cauchy’s Integral Formula.Induction step, n→ (n+1).

limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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Proof. Induction on n.Base step, n = 0. This is Cauchy’s Integral Formula.

Induction step, n→ (n+1).

limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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

limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)

= limw→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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limw→z0

f (n)(w)− f (n)(z0)w− z0

= limw→z0

1w− z0

(n!

2πi

∫C


2πi

∫C

f (z)(z− z0)n+1 dz

)= lim

w→z0

n!2πi

1w− z0

∫C

f (z)(z−w)n+1 −

f (z)(z− z0)n+1 dz

= limw→z0

n!2πi

∫C(z0,r)

f (z)1

(z−w)n+1 − 1(z−z0)n+1

w− z0dz

=n!

2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz



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n!2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz

=(n+1)!

2πi

∫C(z0,r)

f (z)(z− z0)n+2 dz

=(n+1)!

2πi

∫C

f (z)(z− z0)n+2 dz



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n!2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz

=(n+1)!

2πi

∫C(z0,r)

f (z)(z− z0)n+2 dz

=(n+1)!

2πi

∫C

f (z)(z− z0)n+2 dz



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n!2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz

=(n+1)!

2πi

∫C(z0,r)

f (z)(z− z0)n+2 dz

=(n+1)!

2πi

∫C

f (z)(z− z0)n+2 dz



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n!2πi

∫C(z0,r)

f (z)(−(n+1))1

(z− z0)n+2 (−1) dz

=(n+1)!

2πi

∫C(z0,r)

f (z)(z− z0)n+2 dz

=(n+1)!

2πi

∫C

f (z)(z− z0)n+2 dz



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Corollary.

Let f be an analytic function on an open domain. Then allderivatives of f are analytic on this domain, too. Moreover, thecomponent functions (the real and imaginary parts) have continuouspartial derivatives of all orders throughout the domain.

Proof. By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.



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

Then allderivatives of f are analytic on this domain, too. Moreover, thecomponent functions (the real and imaginary parts) have continuouspartial derivatives of all orders throughout the domain.




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Corollary. Let f be an analytic function on an open domain. Then allderivatives of f are analytic on this domain, too.

Moreover, thecomponent functions (the real and imaginary parts) have continuouspartial derivatives of all orders throughout the domain.




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Corollary. Let f be an analytic function on an open domain. Then allderivatives of f are analytic on this domain, too. Moreover, thecomponent functions (the real and imaginary parts) have continuouspartial derivatives of all orders throughout the domain.




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Proof.

By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.



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

That means all derivatives are differentiable and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.



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Proof. By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable

and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have (by repeated application of the Cauchy-Riemann equations)continuous partial derivatives of all orders throughout the domain.



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Proof. By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable and thus allderivatives are analytic on the domain.

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



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Proof. By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have

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



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Proof. By the preceding result, we see that f has derivatives of allorders. That means all derivatives are differentiable and thus allderivatives are analytic on the domain. Therefore real and imaginarypart have (by repeated application of the Cauchy-Riemann equations)

continuous partial derivatives of all orders throughout the domain.



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Theorem.

Morera’s Theorem. Let f be a continuous complexfunction on an open set so that for every simple closed curve C

contained in the open set we have∫

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

the open set.

Proof. By theorem from an earlier presentation, f has anantiderivative F. By the preceding theorem, all derivatives of F(including f ) are analytic.



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Theorem. Morera’s Theorem.

Let f be a continuous complexfunction on an open set so that for every simple closed curve C



the open set.




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Theorem. Morera’s Theorem. Let f be a continuous complexfunction on an open set so that for every simple closed curve C


Cf (z) dz = 0.

Then f is analytic in

the open set.




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the open set.




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the open set.

Proof.

By theorem from an earlier presentation, f has anantiderivative F. By the preceding theorem, all derivatives of F(including f ) are analytic.



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the open set.

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

By the preceding theorem, all derivatives of F(including f ) are analytic.



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the open set.

Proof. By theorem from an earlier presentation, f has anantiderivative F. By the preceding theorem, all derivatives of F

(including f ) are analytic.



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the open set.




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the open set.




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Theorem. Cauchy’s Inequality.

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

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

∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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

If |f | is bounded


∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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


∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof.

∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣

≤ n!2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1

=n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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∣∣∣≤ n!MR

Rn .

Proof. ∣∣∣f (n)(z0)∣∣∣ =

∣∣∣∣ n!2πi

∫CR(z0)

f (z)(z− z0)n+1 dz

∣∣∣∣≤ n!

2π

∫CR(z0)

∣∣∣∣ f (z)(z− z0)n+1

∣∣∣∣ d|z|

≤ n!2π

∫CR(z0)

MR

Rn+1 d|z|

≤ n!2π

2πRMR

Rn+1 =n!MR

Rn



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Theorem.

Liouville’s Theorem. Let f be an entire function. (That is,f is analytic in C.) If f is bounded, then f is constant.

Proof. Let |f | be bounded by B. Let R > 0 be arbitrary. By Cauchy’sInequality with n = 1, for every z0 in C we have∣∣f ′(z0)

∣∣ ≤ BR→ 0 (R→ ∞).

Because R was arbitrary, we infer that f ′(z0) = 0. Because z0 wasarbitrary, we have f ′ = 0. But f ′ = 0 implies that f is constant.



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Theorem. Liouville’s Theorem.

Let f be an entire function. (That is,f is analytic in C.) If f is bounded, then f is constant.


∣∣ ≤ BR→ 0 (R→ ∞).




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Theorem. Liouville’s Theorem. Let f be an entire function.

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


∣∣ ≤ BR→ 0 (R→ ∞).




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Theorem. Liouville’s Theorem. Let f be an entire function. (That is,f is analytic in C.)

If f is bounded, then f is constant.


∣∣ ≤ BR→ 0 (R→ ∞).




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Theorem. Liouville’s Theorem. Let f be an entire function. (That is,f is analytic in C.) If f is bounded, then f is constant.


∣∣ ≤ BR→ 0 (R→ ∞).




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Proof.

Let |f | be bounded by B. Let R > 0 be arbitrary. By Cauchy’sInequality with n = 1, for every z0 in C we have∣∣f ′(z0)

∣∣ ≤ BR→ 0 (R→ ∞).




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Proof. Let |f | be bounded by B.

Let R > 0 be arbitrary. By Cauchy’sInequality with n = 1, for every z0 in C we have∣∣f ′(z0)

∣∣ ≤ BR→ 0 (R→ ∞).




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

By Cauchy’sInequality with n = 1, for every z0 in C we have∣∣f ′(z0)

∣∣ ≤ BR→ 0 (R→ ∞).




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∣∣

≤ BR→ 0 (R→ ∞).




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∣∣ ≤ BR

→ 0 (R→ ∞).




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∣∣ ≤ BR→ 0 (R→ ∞).




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∣∣ ≤ BR→ 0 (R→ ∞).

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

Because z0 wasarbitrary, we have f ′ = 0. But f ′ = 0 implies that f is constant.



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∣∣ ≤ BR→ 0 (R→ ∞).

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

But f ′ = 0 implies that f is constant.



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∣∣ ≤ BR→ 0 (R→ ∞).




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∣∣ ≤ BR→ 0 (R→ ∞).




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Theorem.

Fundamental Theorem of Algebra. Every nonconstantcomplex polynomial has at least one complex zero.

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

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

see this claim, first note that limz→∞

1p(z)

= 0, because limz→∞

p(z) = ∞.

Thus 1p is bounded outside a circle C(R,0) for sufficiently large R.

But the only way 1p can be unbounded inside C(R,0) is for the

denominator p(z) to go to zero somewhere, which was excluded byhypothesis. Thus 1

p is bounded on C.Now by Liouville’s Theorem we infer that 1

p is constant. Hence p isconstant.



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Theorem. Fundamental Theorem of Algebra.

Every nonconstantcomplex polynomial has at least one complex zero.




1p(z)


p(z) = ∞.








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Theorem. Fundamental Theorem of Algebra. Every nonconstantcomplex polynomial has at least one complex zero.




1p(z)


p(z) = ∞.








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Proof.

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



1p(z)


p(z) = ∞.








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Proof. Let p be a complex polynomial without zeros.

Then 1p is



1p(z)


p(z) = ∞.








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analytic in the complex plane.

We claim that 1p is bounded on C. To


1p(z)


p(z) = ∞.








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

To


1p(z)


p(z) = ∞.








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1p(z)

= 0

, because limz→∞

p(z) = ∞.








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1p(z)


p(z) = ∞.








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1p(z)


p(z) = ∞.








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1p(z)


p(z) = ∞.



denominator p(z) to go to zero somewhere

, which was excluded byhypothesis. Thus 1





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1p(z)


p(z) = ∞.



denominator p(z) to go to zero somewhere, which was excluded byhypothesis.

Thus 1p is bounded on C.

Now by Liouville’s Theorem we infer that 1p is constant. Hence p is

constant.



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1p(z)


p(z) = ∞.




p is bounded on C.

Now by Liouville’s Theorem we infer that 1p is constant. Hence p is

constant.



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1p(z)


p(z) = ∞.





p is constant.

Hence p isconstant.



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1p(z)


p(z) = ∞.








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1p(z)


p(z) = ∞.








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Lemma.

If f is analytic in some neighborhood |z− z0|< ε 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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Lemma. If f is analytic in some neighborhood |z− z0|< ε 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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Lemma. If f is analytic in some neighborhood |z− z0|< ε 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.

By Cauchy’s Formula, for all circles C(z0,r) with r < ε wehave that ∣∣f (z0)

∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣

and the latter integral will be strictly smaller than∣∣f (z0)

∣∣ if there iseven a single point z on C(z0,r) where

∣∣f (z)∣∣< ∣∣f (z0)∣∣. Thus for all

r < ε and all |z− z0|= r we must have∣∣f (z)∣∣= ∣∣f (z0)

∣∣. But if |f | isconstant for |z− z0|< ε , then so is f .



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Proof. By Cauchy’s Formula, for all circles C(z0,r) with r < ε wehave that ∣∣f (z0)

∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣








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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣








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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|

=∣∣f (z0)

∣∣and the latter integral will be strictly smaller than

∣∣f (z0)∣∣ if there is

even a single point z on C(z0,r) where∣∣f (z)∣∣< ∣∣f (z0)

∣∣. Thus for allr < ε and all |z− z0|= r we must have

∣∣f (z)∣∣= ∣∣f (z0)∣∣. But if |f | is

constant for |z− z0|< ε , then so is f .



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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣








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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣



∣∣f (z)∣∣< ∣∣f (z0)∣∣.

Thus for allr < ε and all |z− z0|= r we must have

∣∣f (z)∣∣= ∣∣f (z0)∣∣. But if |f | is

constant for |z− z0|< ε , then so is f .



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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣





∣∣.

But if |f | isconstant for |z− z0|< ε , then so is f .



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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣








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∣∣ =∣∣∣∣ 12πi

∫C(z0,r)

f (z)z− z0

dz∣∣∣∣

≤ 12π

∫C(z0,r)

∣∣∣∣ f (z)z− z0

∣∣∣∣ d|z|

≤ 12π

∫C(z0,r)

∣∣f (z0)∣∣

rd|z|=

∣∣f (z0)∣∣








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Theorem.

Maximum Modulus Principle. Let f be analytic and notconstant on an open domain. Then f does not assume a maximumvalue 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. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Theorem. Maximum Modulus Principle.

Let f be analytic and notconstant on an open domain. Then f does not assume a maximumvalue 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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Theorem. Maximum Modulus Principle. Let f be analytic and notconstant on an open domain.

Then f does not assume a maximumvalue 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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Theorem. Maximum Modulus Principle. Let f be analytic and notconstant on an open domain. Then f does not assume a maximumvalue 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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Theorem. Maximum Modulus Principle. Let f be analytic and notconstant on an open domain. Then f does not assume a maximumvalue 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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Proof.

Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist.

By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0.

Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain.

Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain.

Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain.

Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0).

Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0.

Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.

Repeat until we have that f (z1) = f (z0). This would prove that f isconstant, a contradiction.



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Proof. Suppose for a contradiction that such a z0 does exist. By thepreceding lemma, f would be constant on a disk around z0. Moreover,we could choose the radius of the disk arbitrarily large, as long as itstays inside the domain. Now let z1 be in the domain. Then there is anarc C from z0 to z1 and the arc has a positive (minimum) distancefrom the boundary of the domain. Now there is a disk of maximumradius around z0 on which f is equal to f (z0). Thus the point on C thatis in this disk and closest to z1 (on the arc C) has the same propertiesas z0. Continue with a disk of maximum radius around this new point.Repeat until we have that f (z1) = f (z0).

This would prove that f isconstant, a contradiction.



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Corollary.

Let f be continuous on a closed and bounded region andlet it be analytic and nonconstant in the interior of the region. Thenthe largest value of |f | will be assumed at some point on the boundaryof the region and it will not be reached in the interior.

Proof. Direct consequence of the fact that continuous functions willassume a maximum on closed and bounded regions and the MaximumModulus Principle.



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Corollary. Let f be continuous on a closed and bounded region andlet it be analytic and nonconstant in the interior of the region.

Thenthe largest value of |f | will be assumed at some point on the boundaryof the region and it will not be reached in the interior.




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Corollary. Let f be continuous on a closed and bounded region andlet it be analytic and nonconstant in the interior of the region. Thenthe largest value of |f | will be assumed at some point on the boundaryof the region and it will not be reached in the interior.




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Proof.

Direct consequence of the fact that continuous functions willassume a maximum on closed and bounded regions and the MaximumModulus Principle.



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Documents

The Cauchy Integral Formula2.The result itself is known as Cauchy’s Integral Theorem. 3.Among its consequences is, for example, the Fundamental Theorem of Algebra, which says that