I am trying to evaluate the integral $$ \frac{1}{2\pi}\int_0^{2\pi} \log|re^{it}-\zeta| dt$$ for $\zeta\in\mathbb{C}$.
My approach so far has been to first assume $r\leq\zeta$, since this implies the integrand is the real part of a holomorphic branch of the logarithm, hence harmonic, which gives $$ \frac{1}{2\pi}\int_0^{2\pi} \log|re^{it}-\zeta| dt = \log|\zeta| $$
For the case $r > \zeta$ I would like to conclude $$\frac{1}{2\pi}\int_0^{2\pi} \log|re^{it}-\zeta| dt = \frac{1}{2\pi}\int_0^{2\pi} \log|re^{it}| dt = \log r $$ but I don't quite see why the first equality in this equation should be true. Any help and hints are appreciated! :)
Ok, so we have the case $\;r>|\zeta|\;$ , but then
$$|re^{it}-\zeta|=|\overline{re^{it}-\zeta}|=|re^{-it}-\overline\zeta|=\frac{|r-\zeta e^{it}|}{|e^{it}|}=|r-\zeta e^{it}|$$
and we can apply the mean value theorem $\;|r-\overline\zeta z |=|\overline\zeta e^{it}+(-r)|\;$ , and we have the mean value of this function of $\;r\;$ around the canonical circle of radius $\;|\zeta|\;$ :
$$\frac1{2\pi}\int_0^{2\pi}\log|r-\overline\zeta e^{ir}|dz=\frac1{2\pi}\int_0^{2\pi}\log|\overline\zeta e^{ir}+(-r)|dz=\log|(-r)|=\log r$$