Long-time lurker, first time questioner.
Here is the problem:
Let $ a,b \in \mathbb{C} $, with $|b|<1$. Calculate the integral
$$ \frac{1}{2\pi i}\int_{|z|=1} \frac{|z-a|^2}{|z-b|^2} \frac{dz}{z}. $$
I have been looking at the following posts, but I am getting stuck dealing with both $a$ and $b$.
How to compute this contour integral with a modulus sign in the integrand,
How do I solve this complex integration problem?.
This is not for a homework set, it is a practice test I've been working through as a refresher/self-assessment of my knowledge of complex variables.
Thank you for your time and help!
Best Answer
The way I would do it is by using that on $|z| = 1$, $\bar{z} = \frac{1}{z}$, so you can rewrite the fraction $\frac{|z-a|^2}{|z-b|^2}$ as $\frac{(z-a)(1-z\bar{a})}{(z-b)(1-z\bar{b})}$ which is analytic, and then use the residue theorem while considering a few special cases like $a=b$, $a=0$, $b=0$, $b\bar{a} = 1$ separately, since outside of such, there are just two distinct simple poles inside the unit disc, namely $0$ and $b$.