# Difference between revisions of "Cauchy Integral theorem"

Line 9: | Line 9: | ||

For <math> f(a) </math> a constant, we may rewrite <math> dz = d(z-a) </math>, and noting that <math> z - a = | z - a | e^{i \phi} </math> | For <math> f(a) </math> a constant, we may rewrite <math> dz = d(z-a) </math>, and noting that <math> z - a = | z - a | e^{i \phi} </math> | ||

and <math> d(z-a) = i |z -a | e^{i\phi} \; d\phi </math>, and | and <math> d(z-a) = i |z -a | e^{i\phi} \; d\phi </math>, and | ||

− | taking the contour of integration to be a circle of unit radius <math> C : |z -a= 1 </math>, we may write | + | taking the contour of integration to be a circle of unit radius <math> C : |z -a |= 1 </math>, we may write |

## Latest revision as of 15:59, 17 August 2021

The Cauchy Integral theorem states that for a function which is analytic inside and on a simple closed curve in some region of the complex plane, for a complex number inside

Proof:

For a constant, we may rewrite , and noting that and , and taking the contour of integration to be a circle of unit radius , we may write

.

By Cauchy's Theorem, we may deform the contour into any closed curve that contains the point and the result holds.

For the case when is not constant we may write

.

We must show that the second term on the left is identically zero.

In a vanishingly small neighborhood

.

Because the derivative of an analytic function is also analytic, the integral vanishes identically within a neighborhood of .
By Cauchy's theorem, the contour of integration may be expanded to any closed curve within that contains the point
thus showing that the integral is identically zero.