residue integral for specific range of a

Question

I know for $\int_0^{2\pi} \frac{d\theta}{1+a cos\theta} $ where $-1<a<1$, we take a c: unit circle as the contour and change the integrand into a rational function and then apply the residue theorem.
Now I reached to $\int_c \frac{-2i}{az^2+z+a}dz\ $
the singularities of integrand then are: $z=\frac{-1+\sqrt{}1-4a^2}{2}$ and $z=\frac{-1-\sqrt{}1-4a^2}{2}$. I am stuck here. what is the next step? I appreciate any help.

Answer 1

$$\int_0^{2\pi} \frac{d\theta}{1+a\cos \theta} = \frac{2}{i} \oint \frac{dz}{az^2 +2z+a}$$ where the contour is the unit circle taken counter-clockwise.

The poles occur at $z_{1,2}=-\frac{1}{a} \pm \frac{\sqrt{1-a^2}}{a}.$

The pole $z_{1}=\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}$ is inside the circle. Compute the residue there!

$$\int_0^{2\pi} \frac{d\theta}{1+a\cos \theta} = \frac{2}{ia} \cdot 2\pi i \cdot \textrm{Res}_{z=z_1}\left[ \frac{1}{(z-z_1)(z-z_2)}\right]= \frac{4\pi}{a} \cdot \frac{1}{(z_1-z_2)}=\frac{2\pi}{\sqrt{1-a^2}}$$