I am trying to find the residues of the function, $$f(z)=\frac{z(z-\pi)^2}{\sin^2(z)}.$$
My attempt:
I have considered three singularities: $z_0=0,\pi,k\pi \ (k\in\mathbb{Z}, \ k\neq 0,1).$
For $z_0=0$, I've determined this is a simple pole with Res$(f,0)=\pi^2$.
For $z_0=\pi$, I've determined this is a removable singularity and hence Res$(f,\pi)=0.$
For $z_o=k\pi$, I've determined this is a pole of order $2$, but cannot determine it's residue. I know it is possible using the formula, $$\text{Res}(f,k\pi)=\lim_{z\to k\pi}\frac{\partial}{\partial z}\left(\frac{z(z-k\pi)^2(z-\pi)^2}{\sin^2(z)}\right),$$ but I would like to use a more simpler approach through a series (but I am not sure of how to do so). I know I need to compute a Laurent series, but for this type of function, I am unsure.
Some advice would be really appreciated.