Exercise :
Let $0<r<R$.
Using the fact that :
$$\frac{1}{2πi}\oint_{|z|=r} \frac{R+z}{(R-z)z} \,dz = 1$$
Show that :
$$\frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{R^2-r^2}{R^2 + r^2 - 2rR\cosθ}dθ =1$$
Attempt :
Let $z=e^{iθ}, θ \in [-\pi,\pi]. $ Then :
$$\cos \theta = \frac{e^{i\theta }+e^{-i\theta}}{2} = \frac{1}{2}\bigg(z+\frac{1}{z}\bigg) $$
$$dz = ie^{i\theta}d\theta \Leftrightarrow dz = izd\theta \Leftrightarrow d\theta = \frac{dz}{iz}$$
Skipping some elementary steps, the integral becomes :
$$\frac{1}{2 \pi i} \oint_{|z|=1}\frac{R^2-r^2}{R^2z + r^2z - rR(z^2+1)}dz$$
Now, from now on, I cannot seem to find a way to continue on the proof.
I'd like to ask first of all if this is the correct way to approach this problem (and such problems as well) and if it is, I'd appreciate a hint or a solution on how to continue on (I think it's something minor that I am not seeing, probably some trick that will lead to using the initial equation or a step for a Residue solution).
Start with the first integral, and extract the real and imaginary parts: multiply the integrand by $1$, in the form $$ \frac{R+z}{R-z} \frac{R-z^*}{R-z^*} = \frac{R^2-zz^*+R(z-z^*)}{R^2-(z+z^*)R+zz^*}. $$ But of course $zz^*=\lvert z \rvert^2=r^2$, $z+z^* = 2\Re(z)$, and $z-z^* = 2i\Im(z)$, so this becomes $$ \frac{R+z}{R-z} = \frac{R^2-r^2+2i\Im(z)}{R^2-2R\Re(z)+r^2}. $$ Now put $z=re^{i\theta}$, so $dz/z = i \, d\theta$. So the integral becomes $$ 1 = \frac{1}{2\pi i}\int_{|z|=r} \frac{R+z}{R-z} \frac{dz}{z} = \int_{-\pi}^{\pi} \frac{R^2-r^2}{R^2-2Rr\cos{\theta}+r^2} d\theta + i\int_{-\pi}^{\pi} \frac{R\sin{\theta}}{R^2-2Rr\cos{\theta}+r^2} d\theta. $$ The second integral clearly vanishes as the numerator is odd and the denominator is even.