Q:

A complex integral (explicit solution?)

Let $r<1$ be a positive number close to 1. How can I compute
$$
\int_0^{2\pi}\frac 1 {e^{i\theta}-r}d\theta?
$$
Can this be computed in a closed form? Thanks!

A:

The answer is $0$. By Cauchy's Theorem $\int_C \frac 1 {1-rz}=0$. This gives $\int_0^{2\pi} \frac {ie^{i\theta}} {1-re^{i}\theta} d\theta=0$. Hence  $\int_0^{2\pi} \frac 1 {e^{-i\theta}-r} d\theta=0$. Can you take it from here?

