$f(z)=\displaystyle\frac{1}{\pi}\int_0^1rdr\int_{-\pi}^{\pi}\frac{d\theta}{re^{i\theta}+z}$ and $|z|<1\implies$ $f(z)=\bar{z}$

Question

Given $f(z)=\displaystyle\frac{1}{\pi}\int_0^1rdr\int_{-\pi}^{\pi}\frac{d\theta}{re^{i\theta}+z}$ and $|z|<1$, I want to show that $f(z)=\bar{z}$. This looks so simple but I have not been able to make any progress. Any help is appreciated!

Answer 1

Consider $\int_{-\pi}^{\pi} \frac {d\theta} {re^{i\theta}+z}$. This can be written as $\frac 1 {ir} \int_{\gamma} \frac 1 {\zeta(\zeta +z/r)} d\zeta$ where $\gamma$ is the unit circle (in the anti-clockwise direction). If $|z|>r$ then there is only one pole at 0 and the residue is $\frac r z$. If $|z|<r$ the there are two poles and the residues at these points add up to 0. Hence $f(z)=\frac 1 {\pi} \int_0 ^{|z|} 2\pi r/z dr=\frac {z\overline {z}} z =\overline {z}$.