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.
 A: I don't know why you need an infinite series.  Write your function as
$$f(z) = \frac{(\sinh{z})/z}{z^3 (1-z^2)}$$
The numerator is analytic, while the denominator has poles at $z=0, \pm 1$.  The residue at $z=0$ is
$$\frac{1}{2} \lim_{z \rightarrow 0} \frac{d^2}{dz^2}[z^3 f(z)] =  \frac{1}{2} \frac{d^2}{dz^2} \left[\frac{(\sinh{z})/z}{1-z^2} \right ]_{z=0}$$
The algebra is pretty terrible, so I leave it to you; the answer is:
$$\text{Res}_{z=0} f(z) = \frac{7}{6}$$
A: $$\sinh z = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$$
$$\frac{1}{z^4}\sinh z = \frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots$$
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 $\frac{7}{6}$
A: Hint:
$f(z)$ has a pole of the order $3$ (why?) at $z_0=0.$
