Contour integral of non analytic function

Question

Let $a,b\in \mathbf{C}$ with $|b|<1$. I want to calculate $$\int_{|z|=1} \frac{|z-a|^2}{|z-b|^2}\frac{dz}{z}\, .$$

I'm not sure what tools I can use here since the function is not analytic so (I think) this rules out Cauchy's integral formula, etc.

Answer 1

Note that when $|z| = 1$, $\overline{z} = 1/z$, so this integral is the same as

$$ \oint_{|z|=1} \frac{(z-a)(1/z - \overline{a})}{(z-b)(1/z-\overline{b})}\; \frac{dz}{z} $$ which can be done with residues.