How to find Complex Fourier series of $\frac {r\cos(t)}{1+r^2-2r\cos(t)}$

Question

I want to find Complex Fourier series of $$\frac {r\cos(t)}{1+r^2-2r\cos(t)},|r|<1$$ these are my works : $$|r|\lt 1 \Rightarrow \sum_{n=0}^\infty r^{n}e^{int}=\frac {1}{1-re^{it}}=\frac{1}{1-re^{it}}\frac{1-re^{-it}}{1-re^{-it}}=\frac {1-r\cos(t)+i\sin(t)}{1+r^2-2r\cos(t)}$$ but I have no idea how to continue my solution

Answer 1

Since$$\frac{r\cos t}{(1-re^{it})(1-re^{-it})}=\frac{1}{1-r^2}\left(\frac{1+r^2}{2}\left(\frac{1}{1-re^{it}}+\frac{1}{1-re^{-it}}\right)-1\right),$$the $e^{int}$ coefficient for $n\in\Bbb Z$ is$$\frac{\frac{1+r^2}{2}(1+\delta_{n0})r^{|n|}-\delta_{n0}}{1-r^2}.$$