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.
Best Answer
For any given function, the Laurent series about a given point is unique. So you can pretty much use any method you want to find it. Using Cauchy integrals in the fashion you have described may work theoretically, but I'd imagine it would be really tedious to calculate.
The solution simply calculated the taylor series of $\sinh(z)$ and went from there.