Computing residue of $f(z)=\frac{z-\pi}{\sin^2z}$ at $z = \pi$ I know how to find residuum simple function but now I have function $$f(z)=\frac{z-\pi}{\sin^2z}$$ and I have to calculate residuum in $\pi$ (that is, $ \operatorname{Res}_{z=\pi}f(z)$). When I calculate the limit in $\pi$ it's infinity. So in $\pi$ we have pole function. And I have problem with calculate times the pole functions. Can someone help me?

One may note that it is a simple pole, hence

$$\text{Res}_{z=\pi}\frac{z-\pi}{\sin^2z}=\lim_{z\to\pi}\frac{(z-\pi)^2}{\sin^2z}=1$$