prove that ${1 \over 2\pi} \int_0^{2\pi} {\lvert {re^{i\theta} \over (1 - re^{i\theta}) ^2}\rvert} d\theta = {r \over 1- r^2}$, where $0 \lt r \lt1$.

Question

Let $z = e^{i\theta}$, and I get $${1 \over 2\pi i} \int_{\lvert z \rvert = 1} {r \over \lvert r^2 z^2 - 2rz + 1\rvert}{dz \over z}$$ Then I get stuck.

Answer 1

Hint:) From $$|1 - re^{i\theta}|^2=(1 - re^{i\theta})\overline{(1 - re^{i\theta})}=1+r^2-2r\cos\theta$$ we find $${1 \over 2\pi} \int_0^{2\pi} {\lvert {re^{i\theta} \over (1 - re^{i\theta}) ^2}\rvert} d\theta = {1 \over 2\pi} \int_0^{2\pi} {{r \over 1+r^2-2r\cos\theta}} d\theta$$