Computing $\int_{|z|=2} z^n(1 - z)^m\ dz$ for $n,m\in\Bbb Z$

Question

Computing $\int_{|z|=2} z^n(1 - z)^m\ dz$

I need help for this question when $m, n$ are negative integers.

Thanks.

Answer 1

The answer depends on the sum $n+m$. If $n+m<-1$ Then the integral is zero. \begin{gather} f(z)=z^n(1-z)^m\\ I=-2\pi i\underset{z=\infty}{\rm res}f(z)=2\pi i c_{-1} \end{gather} There are no poles outside the contour. Here, $c_{-1}$ is the coefficient of $f(z)$ expansion in the vicinity of the infinity. The leading behavior of $f(z)$ is \begin{gather} f(z)=(-1)^mz^{n+m}\ \ z\rightarrow\infty. \end{gather} Therefore $c_{-1}=0$ unless $m+n=-1$.

If $m+n=-1$ then: \begin{gather} c_{-1}=(-1)^m\ \Rightarrow\ I=2\pi i(-1)^m \end{gather} P.S. We have some videos with explanations about contour integrals. Check out our page: https://www.youtube.com/channel/UCcOPzqeuB8GTt1zfh6JKbSA

Answer 2

You said $n,m\lt0$. By Cauchy's integral formula, I get $2\pi i (1+1)=4\pi i$, when $n=m=-1$.

For other negative values of the exponents, it appears to be more complicated, and I don't see how to do it with CIF. Perhaps you could use the residue theorem.