Complex integral $\oint_{|z| = 1} \frac{dz}{(z-a)^n (z-b)^n}$, where $|a| < |b| < 1$

Question

I don't know how to start solving the integral below. I have already solved it for $1 < |a| < |b|$, because the result it's pretty trivial (it is $0$), but I can't find the idea to do it for $|a| < |b| < 1$. $$\oint_{|z| = 1} \frac{dz}{(z-a)^n (z-b)^n}$$ where $n \in \mathbb{N}$, $a,b \in \mathbb{C}$ and $|a| < |b| < 1$.

I do not want the solution! I just want the advice on how to start! Thanks!!!

Answer 1

I see from the comments that you don't want to use the residue theorem.

I see that you don't want the full solution but some hints:

$n=1$ is easy because $\dfrac{1}{(z-a)(z-b)}=\dfrac{1}{a-b}\left(\dfrac{1}{z-a}-\dfrac{1}{z-b}\right)$.

Here's a trick:

\begin{align*}\dfrac{\partial}{\partial a}\dfrac{\partial}{\partial b}\oint_{|z| = 1} \dfrac{dz}{(z-a)^n (z-b)^n}&=\oint_{|z| = 1} \dfrac{\partial}{\partial a}\dfrac{\partial}{\partial b}\dfrac{1}{(z-a)^n (z-b)^n}\,\text{d}z\\&=\oint_{|z| = 1}\dfrac{n^2}{(z-a)^{n+1} (z-b)^{n+1}}\,\text{d}z.\end{align*}

What can you conclude from this?