I am tasked with evaluating the following integral utilizing the Cauchy theorem: $$\int_{0}^{2\pi} \frac{1}{(2 - \cos t)^{2}}dt$$
I have initiated my calculation as follows:
$$\begin{align*} \int_{0}^{2\pi} \frac{1}{(2 - \cos t)^{2}}dt &= \int_{0}^{2\pi}\frac{1}{4-2\cos t + (\cos t)^{2}}dt \\&= \int_{0}^{2\pi} \frac{dt}{4 - 2\cdot\frac{e^{it} + e^{-it}}{2} + \frac{(e^{it} + e^{-it})^{2}}{4}}\\&=\int_{0}^{2\pi} \frac{4\cdot dt}{16 - 4e^{it} - 4e^{-it} + e^{2it}+ 2e^{0} + e^{-2it}} \\&= \int_{0}^{2\pi} \frac{4e^{it}}{18e^{it} - 4e^{2it} - 4 + e^{3it} + e^{-it}} \cdot \frac{d(e^{it})}{i \cdot e^{it}}\\&= \frac{4}{i} \int_{0}^{2\pi} \frac{d(e^{it})}{18e^{it} - 4e^{2it} - 4 + e^{3it} + e^{-it}} \\&= \frac{4}{i} \oint \frac{dz}{18z - 4z^{2} + z^{3} + z^{-1} - 4} \end{align*}$$
To advance to the next stage of the calculations, I must determine the roots of the equation $18z - 4z^{2} + z^{3} + z^{-3} - 4 = 0$
Unfortunately, I'm uncertain about the method to find these roots. It's possible that I'm overlooking a crucial step, and there might be an alternative approach to simplify the equation and make it more amenable to root-finding
For $z=e^{it}$ you have that $2-\cos t=2-\frac{z+\bar z}{2}=\frac{z^2-4z+1}{-2z}$ and $dt=\tfrac{dz}{iz}$, therefore
$$ \int_{0}^{2\pi}\frac{dt}{(2-\cos t)^2}=\oint \frac{4z^2}{(z^2-4z+1)^2}\cdot \frac{dz}{iz}=-4i\oint \frac{z\,d z}{(z-\alpha )^2(z-\beta )^2} $$
for $\alpha =2-\sqrt{3}$ and $\beta =2+\sqrt{3}$. Now observe that $|\alpha |<1$ and $|\beta |>1$ so the unique pole on the unit disc of the integrand is at $\alpha $ with multiplicity two.