Computing $\int_{|z|=2} z^n(1 - z)^m\ dz$ My two questions are bolded below.
Hypothesis: Let $\gamma$ denote the circle about the origin of radius $2$.
Goal: Compute
$$
\int_{\gamma} z^n(1 - z)^m\ dz
$$
Attempt:


*

*We have that
$$
\int_{\gamma} z^n(1 - z)^m\ dz = \int_{\gamma} z^n(-1)^m(z-1)^m\ dz
$$

*Take the integral of the inverse of the integrand.  Once we figure out an answer to this question, we can inverse that answer to find the integral of our original integrand. Is this correct reasoning?

*Assuming this is correct reasoning, we have that
$$
\int_\gamma {1 \over z^n(-1)^m(z-1)^m}\ dz = \int_\gamma {{1 \over z^n}(-1)^{(m-1)+1} \over (z-1)^{(m-1)+1}}\ dz =  {2 \pi i \over n!} f^{(m-1)}(1) \text{ s.t. } f(z) = ??
$$
Here is $f(z) = {(-1)^{m} \over z^n}$?  I'm trying to make heavy use of Cauchy's integral formula but think I've computationally confused myself in that pursuit.  How does one finish this computation?
 A: As others have said, if both $m$ and $n$ are non-negative integers, then the function $f(z)=z^n(1-z)^m$ is entire,and the integral is zero.
If $m\ge 0$ and $n<0$, then $$z^n(1-z)^m = \sum_{k=0}^m \binom{m}{k}(-1)^kz^{k+n}$$  so that $$\int_\gamma f(z) \, dz =2\pi i \, \text{res}_{z=0} f(z) = 2\pi i \, \binom{m}{-n-1}(-1)^{-n-1}$$
If $m<0$ and $n\ge 0$, then let $p=z-1$, and
$$z^n(1-m)^m = (-1)^m (p+1)^n p^m =(-1)^m \sum_{k=0}^n \binom{n}{k} p^{m+k}$$
and
$$\int_\gamma f(z)\, dz= \int_{\gamma^\star} f(p) \, dp =2\pi i \, \text{res}_{p=0} f(p) = -2\pi i \, \binom{n}{-m-1}$$
It can be shown that if both $m$ and $n$ are negative, then the residues at $z=1$ and at $z=0$ add to zero, so again the integral is zero.
A: If $n,m$ are positive integers the integrand is holomorphic on the whole $\mathbb C$ hence it has NO singularities and so NO residues. Then you can conclude by Residue Thm that the intergral is zero.
A: The integral is zero if $n,m\in\Bbb Z^+$, by Cauchy's theorem (you are integrating a holomorphic (analytic)  function over a closed curve).
