I want to solve this integral:
Let $f$ be a holomorphic function in $\Omega$ such that $\bar D (0,1) \subset \Omega$ and let $a\in D(0,1)$. Compute $$\int_{\vert z\vert =1}\! \frac{f(z)}{\vert z-a \vert ^2}\, \mathrm d z$$
If the modulus wasn't in the denominator, I know this integral would be $2\pi \mathrm{i}\,f'(a)$ because of Cauchy Integral Formula for derivatives. But I don't know how to solve this actual problem.
In the unit circle, $\bar{z} = z^{-1}$, and thus $\lvert z - a\rvert^2 = (z - a)(\bar{z}-\bar{a}) = (z - a)(z^{-1} - \bar{a})$ in the unit circle. Hence
$$\int_{\lvert z\rvert = 1} \frac{f(z)}{\lvert z - a\rvert^2}\, dz = \int_{\lvert z\rvert = 1} \frac{f(z)}{(z - a)(z^{-1}-\bar{a})}\, dz = \int_{\lvert z\rvert = 1} \frac{zf(z)}{(z - a)(1 - \bar{a}z)}\, dz$$
We can decompose
$$\frac{z}{(z - a)(1 - \bar{a}z)} = \frac{1}{1 - \lvert a\rvert^2}\left(\frac{1}{z-a} + \frac{\bar{a}}{1 - \bar{a}z}\right)$$
This allows us to write
$$\int_{\lvert z \rvert = 1} \frac{zf(z)}{(z-a)(1-\bar{a}z)}\, dz = \frac{1}{1-\lvert a\rvert^2}\left(\int_{\lvert z\rvert = 1} \frac{f(z)}{z-a}\, dz + \bar{a}\int_{\lvert z \rvert = 1} \frac{f(z)}{1 - \bar{a}z}\, dz\right)$$
You'll be able to take it from here.