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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.

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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 $0$.

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