Integral of Logarithm with complex terms

Question

Is there any way to evaluate this integral? $$ \int_0^{1}\left(\int_{0}^{2\pi} \log(z-r e^{i\theta}) \; d\theta\right)r \; dr $$ with $z\in \mathbb{C}$

Answer 1

The basic idea if $|z|$ is large enough is that $\ln(z-re^{i\theta})$ is a power series in $e^{i\theta}$, but only the constant term doesn't vanish under $\int_0^{2\pi}d\theta$, and for that term it's as if $e^{i\theta}=0$, so $\int_0^{2\pi}\ln(z-re^{i\theta})d\theta=2\pi\ln z$ if $|z|>r$, making the integral $2\pi\ln z\cdot\int_0^1rdr=\pi\ln z$ if $|z|>1$. But if $|z|<1$, we need greater care. Use$$\ln(z-re^{i\theta})=i(\theta+(2n+1)\pi)+\ln r+\ln(1-\tfrac{z}{r}e^{-i\theta}),$$where the choice of $n\in\Bbb Z$ is a matter of the logarithm's branch. Again, the $\theta$-dependent logarithm admits a trivial behaviour under integration, and if we take $n=-1$ (as per @TomP's comment) the term $i(\theta+(2n+1)\pi)$ also integrates to $0$. If $|z|<1$,$$\int_0^{|z|}2\pi r\ln zdr+\int_{|z|}^12\pi r\ln rdr=\pi|z|^2\ln\frac{z}{|z|}-\tfrac12\pi+\tfrac12\pi|z|^2.$$This reduces to $\pi\ln z$ if $|z|=1$.