Integral of $\ln(|z|)$ over a circle Let $r > |z|$, calculate 
$$\frac{1}{2 \pi}
\int_{0}^{2 \pi} \ln (|r e^{it} - z|) dt.
$$
My guess is that this is $\ln(r)$, obviously for $z = 0$ this goes without saying. Is there some argument for why moving the circle over which we integrate shouldn't change the outcome (as long as $z$ stays inside the disk)? I made some attempt to calculate this directly, but to no avail. Help or hints appreciated, thanks in advance.
 A: Fix $r>0$ 
For fixed $t$ we have that $f(z)=\ln (re^{it}-z)$ is analytic in $|z|<r$ since $re^{it}-z \ne 0$ anywhere there. But now integrating a continuous map of holomorphic functions wr to a finite positive measure on a compact space leads to a holomorphic function $g$ on $|z|<r$ (with $\Re g$ being the original integral up to the $\frac{1}{2 \pi}$) since we can put the $\partial_{\bar z}$ inside and get:
$\partial_{\bar z}g(z)=\partial_{\bar z}
\int_{0}^{2 \pi} \ln (r e^{it} - z) dt=\int_{0}^{2 \pi}\partial_{\bar z} \ln (r e^{it} - z) dt=0$ 
(or just expand in Taylor series in $z$ and show that integration and expansion commute by absolute convergence in small closed discs inside $|z| <r$).
But now differentiating we get (allowed as above) $g'(z)=-\int_{0}^{2 \pi} \frac{dt}{r e^{it} - z}=-\int_{0}^{2 \pi}\sum_{k \ge 1}\frac{z^{k-1}}{r^ke^{ikt}}dt=0$ again by switching sums and integrals by absolute convergence. So $g(z)=g(0)$ is constant and in particular:
$\frac{1}{2 \pi}\int_{0}^{2 \pi} \ln (|r e^{it} - z|) dt=\frac{1}{2 \pi}\Re g(z)=\frac{1}{2 \pi}\Re g(0)=\ln r$ so we are done!
