Calculate the following integral $\int_{|z|=1}\frac{z^m}{(z-a)^n}dz$

Question

Given $n,m\in\mathbb{N},|a|\neq1$
Calculate the following integral $\int_{|z|=1}\frac{z^m}{(z-a)^n}dz$

I thought maybe using Cauchy's integral formula and I'm not sure what happens when $a$ is outside the domain

Answer 1

For $|a| <1$ it is $2\pi i/{(n-1)!} f^{n-1}(a)$ where $f(z)=z^{m}$. For $|a|>1$ it is $0$ by Cauchy's Theorem.

Answer 2

If $a$ is not contained in $|z| \leq 1$, then the integral is over a closed curve and has no singularities, so it is equal to $0$. If, instead, $a$ is contained in $|z| < 1$, we have, by Cauchy Integral Theorem (assume that $n \geq 1$): $$ \int_{|z|=1} \frac{z^m}{(z-a)^n} dz = \frac{2 \pi i}{(n-1)!} f^{(n-1)}(a)$$ If $a$ in on the curve itself, $|z|=1$, the integral does not exist in the usual sense. However, it may exist its Cauchy principal value.