Proof of Cauchy integral formula

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SUMMARY

The discussion centers on proving the Cauchy integral formula for analytic functions in the context of a differential equations class. The proof involves integrating over a larger contour C that encloses a smaller contour C_r, which wraps around a circle of radius r around z_0. The deformation principle is crucial, as it states that the path of integration does not affect the integral's value for analytic functions. The proof for non-analytic functions is similar, except the second integral vanishes for analytic functions.

PREREQUISITES
  • Understanding of contour integrals
  • Familiarity with the Cauchy integral formula
  • Knowledge of the deformation principle in complex analysis
  • Basic concepts of analytic functions
NEXT STEPS
  • Study the properties of analytic functions in complex analysis
  • Learn about the deformation principle and its applications in contour integration
  • Explore the maximum modulus principle in complex analysis
  • Review the Cauchy integral formula and its proofs for different types of functions
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Students in differential equations, mathematicians studying complex analysis, and anyone interested in understanding the Cauchy integral formula and its implications for analytic functions.

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Homework Statement


For an quiz for a diff eq class, I need to prove the Cauchy integral formula.
The assignment says prove the formula for analytic functions. Is the proof significantly different when the function is not analytic?

Homework Equations


Basically, a proof I found online says that I take the integral of a larger contour C that encloses the contour C_r which is the contour that wraps around the cirlce of radius r around z_0. By the deformation principle, if r is sufficiently small, then the integral of f(z) around C and C_r are equal. I've tried searching for 'deformation principle' but none was of much help.


3. Questions
http://www.math.colostate.edu/~achter/419f06/help/cauchy.pdf

What is the deformation principle? I don't know much about contour integrals, so if someone can relate it to just regular integrals, that'd be great.

http://www2.latech.edu/~schroder/slides/comp_var/14_Cauchy_Integral_Formula.pdf Pg 43

What is max C(r, z_0) supposed to mean? I think the this proof uses the same principles as in the first one, but doesn't explicitly use the delta-epsilon limit.

Thank you
 
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Cauchy's formula for smooth functions is

[tex]f(a) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) dz}{z-a} + \frac{1}{2\pi i}\iint_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dz\wedge d\bar{z}}{z-a}[/tex]

The proof is not that much different than for analytic functions other than for analytic functions the second integral is zero.

The deformation principle is that path between the limits of integration does not matter for analytic functions. For real integration this is also true (at least if the function is non pathological and the path does over emphasize a region).
For example if integrating from 1 to 4 is written I(1,4) we have
I(1,4)=I(1,2)+I(2,4)=I(1,3)+I(3,2)+I(2,4) and so on
For complex integrals there is more freedom because for each segment the path can be curved and twisted in endless ways.

max C(r, z_0) is the maximum modulus
the largest absolute value the function can obtain on the circle
 

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