# Why does $\frac{1}{2}\int_0^{2\pi}\frac{d\theta}{a+\cos \theta}=-i\int_{|z|=1}\frac{dz}{z^2+2az+1}$?

I'm reading an example of how to calculate $\int_0^\pi\frac{d\theta}{z+\cos\theta}$ for $a>1$ with residues. Since $\cos\theta$ takes the same values on $(0,\pi)$ as $(\pi,2\pi)$, the integral equals $$\frac{1}{2}\int_0^{2\pi}\frac{d\theta}{z+\cos\theta}.$$ Letting $z=e^{i\theta}$, I find $-i\frac{dz}{z}=d\theta$, and so the integral is $$-i\int_{|z|=1}\frac{dz}{2za+2z\cos\theta}$$

How does this convert to $-i\int_{|z|=1}\frac{dz}{z^2+2az+1}$?

I know I should find $z^2+1=2z\cos\theta$, but I'm not seeing it.

-

## 1 Answer

$\cos(\theta)= \frac{z + \overline{z}}{2} =\frac{z+\frac{1}{z}}{2} = \frac{z^2+1}{2z} \Rightarrow 2z \cos(\theta) =z^2+1$.

-