I need prove that:
$$\int_{0}^{2\pi} \frac{R^{2}-r^{2}}{R^{2}-2Rr\cos \theta +r^{2}} d\theta= 2\pi$$
By deformation theorem, with $0<r<R$.
Professor gave us the hint to use the function $f(z)= \frac{R+z}{z(R-z)}$, and define an adequate $\gamma : [a,b]\rightarrow \mathbb{C}$ circular curve and with deformation theorem, we could find the integral. But I have been able to find the curve $\gamma$. Any advice is very helpful
The integrand is the famous Poisson kernel function! Here's a sketch of how to integrate it using the Residue Theorem:
Show that
$$ \frac{R^2-r^2}{R^2-2Rr\cos\theta +r^2} = \text{Re}\left(\frac{R+re^{i\theta}}{R-re^{i\theta}}\right) $$
and hence
$$ \int_0^{2\pi}\frac{R^2-r^2}{R^2-2Rr\cos\theta +r^2} \,d\theta = \int_0^{2\pi}\text{Re}\left(\frac{R+re^{i\theta}}{R-re^{i\theta}}\right)\,d\theta = \text{Re}\int_0^{2\pi}\frac{R+re^{i\theta}}{R-re^{i\theta}}\,d\theta. $$ View this integral as an integral along the contour $\gamma(\theta) = re^{i\theta}$ and evaluate $$ \int_\gamma \frac{R+z}{iz(R-z)}\,dz $$ using the Residue Theorem.
Hint: Let
$$z=e^{i\theta}\implies dz= i e^{i\theta} d\theta , $$
then the new integral is
$$ \int_{|z|=1} \frac{R^{2}-r^{2}}{R^{2}-Rr(z+1/z) +r^{2}} \frac{dz}{iz}.$$
Now, you need to use residue theorem.
