I have a complex analysis question that I am stuck on.
Let $\sum_{n=0}^{\infty}a_{n}z^{n}$ be the Taylor series expansion at 0 for $(1-z)^{-a}$ where $a>0$. Using the generalized Cauchy integral Formula, obtain an estimate for the coefficient $a_{n}$.
I know the Cauchy integral Formula says for an analytic function $f$ and a regular closed curve $\gamma$ with $z\in\gamma$ then $ f^{(k)}(z)=\frac{k!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{k+1}}dw $. So I see that $a_{n}=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w)}dw$.
Additionally, I see that the distance is 1 between 0 the singularity at $z=1,$ so that $\gamma$ is at most of radius 1. However, I am stuck on how to proceed. I want to plug in for my particular function $(1-z)^{-a}$ but I don't know exactly how to do it. Can an exact expression for $a_n$ be attained or just an estimate?
Any help would be appreciated.