Please help with the following prelim problem. Thanks !.
Express the integral as a complex integral of a meromorphic function, where $\rho > 0$ and $a$ is complex valued
$$ \int_{\left\vert z\right\vert\ =\ \rho}\ \left\vert z - a\right\vert^{-2} \,\left\vert{\rm d}z\right\vert $$
Because we are integrating along a circle, $|dz|=\rho d\phi$, where $\phi \in [0,2 \pi)$. If we write $a=a_r+i a_i$, then we write $z=\rho e^{i \phi}$ and we have
$$\int_{|z|=\rho} \frac{|dz|}{|z-a|^2} = \rho \int_0^{2 \pi} \frac{d\phi}{\rho^2+|a|^2 - 2 \rho (a_r \cos{\phi} + a_i \sin{\phi})}$$
We can convert this to a complex integral over a meromorphic function by letting $\zeta = e^{i \phi}$ so that $d\phi = d\zeta/(i \zeta)$. Using $\cos{\phi}=(\zeta + \zeta^{-1})/2$ and $\sin{\phi} = (\zeta - \zeta^{-1})/(2 i)$, we get, after some algebra
$$i \rho \oint_{|\zeta|=1} \frac{d\zeta}{\rho \bar{a} \zeta^2 - (\rho^2+|a|^2) \zeta + \rho a}$$
BONUS
The integral is relatively straightforward to evaluate via the residue theorem. I leave it to the reader to look after the details; I get
$$\int_{|z|=\rho} \frac{|dz|}{|z-a|^2} = \frac{2 \pi \rho}{|\rho^2-|a|^2|}$$