Laurent sreies expansion of the function $f(z)=z^{-1}\sinh(z^{-1})$ about the point $0$.
I thought I was meant to use this:
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty \frac{b_n}{(z-z_0)^n}$$
$r_1\lt |z-z_0|\lt r_2$
Where $$a_n = \frac{1}{2\pi i} \int_c \frac{f(\xi)}{(\xi - z_0)^{n+1}} d\xi, b_n = \frac{1}{2\pi i} \int_c \frac{f(\xi)}{(\xi - z_0)^{-(n+1)}} d\xi$$
But the answer didn't do any of this? why??
Given answer method:
For $f(z)=\sinh z$, we have $$f^{(n)}(z) = \left\{ \begin{array}{cc}\sinh z&z\text{ even}\\ \cosh z& z\text{ odd} \end{array}\right.$$
$$\implies f^{(n)} = 1, n \text{ odd}. =0, n \text{ even}$$
Gives Maclaurin series for $\sinh z$ is $$z + \frac{z^3}{3!} + \frac{z^5}{5!}+\cdots$$ Laurent series for $\sinh z^{-1}$ is $$\frac{1}{z} + \frac{1}{3!z^3} + \frac{1}{5!z^5}+\cdots$$
Laurent series for $z^{-1}\sinh(z^{-1})$ is $$\frac{1}{z^2}+\frac{1}{3!z^4}+\frac{1}{5!z^6}+\cdots$$
Note that $z^{-1}\sinh(z^{-1})$ is analytic for $z\ne 0$, so the origin is an isolated singularity from the Laurent series, there are arbitrary large negative powers of $z$, so this is essential.
