# Finding residue of a function

I am learning about residues, and want to know how to find the residue of this function: $$f(z)=\frac{1}{z\sinh^2(z)}$$

I can see that this function has simple poles at $$z=2k\pi i$$ when, $$k \in \Bbb{Z} \backslash \{0\}$$, and that it has a triple pole at $$0$$. But how do I find the residues at these poles? I do not seem to be able to use the regular formula, that $$\text{Res}_{z_0}(f)=\lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z))$$ as I get really complicated derivatives, esepcially for the pole of order $$3$$. I'm guessing I want to expand the Laurent series, but I'm struggling to do this! Any help appreciated-thanks.

Let's deal with the pole at $$z=0$$ first. Find the Laurent expansion: $$\frac{1}{z\sinh^2(z)}=\frac{1}{z^3\left(1+\frac{z^2}{6}+...\right)^2}=\frac{1}{z^3}\frac{1}{1+\frac{z^2}{3}+...}=\frac{1}{z^3}\left(1-\left(\frac{z^2}{3}+...\right)+...\right)=\\ =\frac{1}{z^3}-\frac{1}{3z}+...$$ So $$\mathrm{Res}\left(\frac{1}{z\sinh^2(z)};0\right)=-\frac{1}{3}$$.
The other poles are not simple as you stated, but are of second order. Anyway, write $$z=2\pi ik+w$$. Then $$\sinh^2(2\pi ik+w)=\sinh^2(w)$$. But this is an even function of $$w$$, so the residue of its reciprocal is $$0$$.