Residue of $\frac{1}{(1+\cos z)^2}$ for $z_0=\pi$

Question

I'm trying to find the residue of $f(z)=\frac{1}{(1+\cos z)^2}$ around $z_0=\pi$. I know that $z=\pi$ is a pole of order 4, so I'm using the formula: $$\frac{1}{(4-1)!}\lim_{z\to\pi}\frac{d^3}{dz^3}\left[(z-\pi)^4\left(\frac{1}{(1+\cos z)^2}\right)\right]$$ But when I try to calculate the limit, it just gets worse. After several attempts, I'm stuck at: $$\frac{1}{6}\lim_{z\to 0}\left[\frac{z^4}{(1-\cos z)^2}\left(\frac{24}{z^2} - \frac{24\sin^3z}{(1-\cos z)^3} + \frac{2\sin z}{(1-\cos z)} + \frac{18\sin z\ \cos z}{(1-\cos z)^2} - \frac{24 \cos z}{z (1-\cos z)} + \frac{72\sin^2 z}{z(1-\cos z)^2} - \frac{72\sin z}{z^2(1-\cos z)}\right)\right]$$ Can anyone give me a hint?

Answer 1

The residue of $f$ at $\pi$ is the residue of $f(z+\pi)$ at $0$. But$$f(z+\pi)=\frac1{\left(1+\cos(z+\pi)\right)^2}=\frac1{\left(1-\cos(z)\right)^2},$$which is an even function. Therefore, its residue at $0$ is $0$.