# Math Help - Complex integral theorem

1. ## Complex integral theorem

This is a theorem from my complex analysis book, and one of the exercises is to prove it.

When n = 1, the result follows directly from Cauchys integral formula. When n is 0 or negative, the integral is just a regular polynomial. A regular polymoial is analytical everywhere, and an integral around a close curve around it is zero.

These two cases I get, but why is this integral 0 when n is a positive integer greater than or equal to 2?

Thanks!

2. ## Re: Complex integral theorem

First, your two cases are "n= 1" and " $n\ne 0$" which does not make sense. I assume you meant " $n\ne 1$".

You also don't state what the curve is. Obviously, if the curve does not have $z_0$ in its interior, the the integrand is analytic there so the integral is 0. So I presume that C is a curve having $z_0$ in its interior. In that case, we can transform the curve into a circle with center at $z_0$ without changing the integral because the integrand is analytic at every point between the given curve and such a circle.

So we can assume that C is the circle with center at $z_0$ and radius R. In that case $z= z_0+ Re^{i\theta}$ where $\theta$ goes from 0 to $2\pi$. Of course then $dz= Ri e^{i\theta}d\theta$, $z- z_0= Re^{i\theta}$, and the integral becomes $\int_0^{2\pi}\frac{Ri e^{i\theta}d\theta}{R^ne^{ni\theta}}= R^{1-n}i\int_0^{2\pi}e^{i(1-n}\theta}d\theta$.

IF $n\ne 1$, then the integral is $\frac{R^{1- n}}{1-n}\left[e^{i(1-n)\theta\right]_0^{2\pi}$. But the exponential is 1 at both 0 and $2\pi i$ so the integral is 0.

IF $n= 1$,then the integral is $i\int_0^{2\pi} d\theta= 2\pi i$

3. ## Re: Complex integral theorem

Your assumptions are correct, I should learn to be more presise. Thank you very much, I didn't think of it that way!