How can I calculate the following complex integral?

Question

How can I calculate the integral? $$ \int_{\left| z \right| = 1} \frac{z^mdz}{(z-a)^n} $$ for $a\in \mathbb C $ and $ \left| a \right| \neq 1 $ and $ m,n\in \mathbb N$

Answer 1

Here's what I would do. Start by making a change of coordinates $w = z-a$. Then use the binomial theorem to expand the numerator, and cancel $n$ copies of $z$. You probably know that $\int_\gamma \frac{1}{z^k} dz = 0$ whenever $k \neq 1$, so then you can kill all terms except the term with exponent $-1$. You probably are aware already that when $k=1$, that integral has value $2 \pi i$, but if you aren't, you should try to prove that.

Answer 2

First, if $|a|>1,$ then the integrand is analytic in $D(0,|a|)$, hence the integral is $0$ by Cauchy's theorem. So suppose $|a|<1.$ Then for any analytic $f$ holomorphic in $D(0,r)$ for some $r>1,$ we have

$$f^{(n-1)}(a) = \frac{(n-1)!}{2\pi i}\int_{|z|=1} \frac{f(z)}{(z-a)^n}\, dz.$$

This is Cauchy's integral formula for $f$ and its derivatives. Apply this with $f(z) = z^m.$