I have to calculate an integral of the following form:$$\oint_{|z|=1} \log^2\left(\frac{|1+hz|^2}{(1-y2)^2}\right)\left(\frac{1}{z-r^{-1}}\right)\,\mathrm dz,$$ where $h,r$ and $y2$ are constants with $0<h,y2<1<r$. Ideally I'd like to use the Cauchy Residue theorem however the log term is not holomorphic due to the modulus function.
Using the fact that $|1+hz|^2 = (1+hz)(1+h\bar{z})$ we can expand the square logarithm as follows: $$\small\oint_{|z|=1} \left[\log^2\left(\frac{1+hz}{(1-y2)^2}\right) + 2\log\left(\frac{1+hz}{(1-y2)^2}\right)\log\left(\frac{1+h\bar{z}}{(1-y2)^2}\right) + \log^2\left(\frac{1+h\bar{z}}{(1-y2)^2}\right)\right]\left(\frac{1}{z-r^{-1}}\right)\,\mathrm dz.$$ This is useful since we can use the Residue theorem to calculate the integral of the $\log^2$ terms. However I'm still left with the product of logs which I have no idea how to integrate. Does anyone know I can integrate the $\log \log$ term?
This is related to the limiting expectation and variance of an eigenvalue statistic, and a solution does exist.