Residue of hyperbolic function How would I find the residue at $z_0=0$ of $$f(z)=\frac{\sinh(z)}{z^4(1-z^2)}$$I tried writing it as a series and reach $$\frac{1}{1-z^2}\sum_{n=0}^\infty \frac{z^{2n-3}}{(2n+1)!}$$ and then don't know where to go form there. Any help/hints appreciated.

$$\sinh z = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$$
$$\frac{1}{z^4}\sinh z = \frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots$$
Now,
$$\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$$
So then, multiplying the series (note we only want to the the $z^{-1}$ terms, so no need to multiply every term)
$$\frac{1}{z^4(1-z^2)}\sinh z = \\
\left(1 + z^2 + z^4 + \cdots\right)\left(\frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots\right) = \\
\cdots + \frac{1}{z3!} + \frac{z^2}{z^3} + \cdots =\\
\cdots + \frac{7}{6z} \cdots
$$
So the residue is $\frac{7}{6}$
