Compute the following residue:
\begin{equation} \textrm{res}_{z=0}\frac{(\sin(z))^2}{(\sinh(z))^5} \end{equation}
I thought of using the power series expansions, and trying to find the $\frac{1}{z}$ coefficient as follows:
\begin{equation} \frac{\sin^2(z)}{\sinh^5(z)}=\frac{z^2(1-\frac{z^2}{6}+O(z^4))^2}{z^5(1+\frac{z^2}{6}+O(z^4))^5} \end{equation}
But it seems very hard to find the $\frac{1}{z}$ term from this - any ideas?
You're on the right track, but you need to count the number of terms you need from each series expansion.
It should be evident that you have an odd function, so the Laurent series has only odd powers, and the dominating term around $z=0$ will be $z^{-3}$. Therefore:
$\dfrac{\sin^2(z)}{\sinh^5(z)}\approx z^{-3}+az^{-1}$
and you will use two terms from the numerator and denominator.
Thus render
$\dfrac{\sin^2(z)}{\sinh^5(z)}\approx \dfrac{(z^{-3})(1-z^2/6)^2}{(1+z^2/6)^5}\approx(z^{-3})(1-z^2/6)^2(1-z^2/6)^5$
Using the Binomial Theorem and again keeping two terms, you get the next step
$\approx(z^{-3})(1-z^2/3)(1-5z^2/6),$
and you can carry out this multiplication to get the term proportional to $z^{-1}$.