Laurent series expansion of $f(z)= \frac{\sinh(z)}{z^3}$

Question

Find the Laurent- series expansion of $$f(z)= \frac{\sinh(z)}{z^3}$$

With small manipulations, I came up with below thing,

$$\frac{\sinh(z)}{z^3}=\frac{1}{z^2}+\frac{1}{3!}+\frac{z^2}{5!}+...$$

Is it useful to find Laurent series?

Answer 1

As @jdc told you, the series you determined is a Laurent series. A Laurent series is defined as $$ f(z) = \sum_{n = -\infty}^{\infty}c_n(z-z_0)^n $$ where $c_n$ is determined by the Cauchy integral formula, $$ c_n = \frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz $$ where $C$ is a closed curve in $\mathbb{C}$ taking counter clockwise. A Laurent series has two parts $(1)$ is the principal part $\sum_{n=-\infty}^{-1}$ and $(2)$ the analytic part $\sum_{n=0}^{\infty}$. Since the series you determined has a coefficient, $c_n$, for $n = -1$, you no longer have a Taylor series. \begin{align} f(z) &= \sum_{n=0}^{\infty}\frac{z^{2n+1}}{z^3(2n+1)!}\\ &= \sum_{n=0}^{\infty}\frac{z^{2n-2}}{(2n+1)!}\\ &= \sum_{n=-1}^{\infty}\frac{z^{2(n+1)-2}}{(2(n+1)+1)!}\\ &= \sum_{n=-1}^{\infty}\frac{z^{2n}}{(2n+3)!} \end{align}

Answer 2

Here is an approach based on the power series of the exponential function

$$ f(z) = \frac{1}{2 z^3}( e^z-e^{-z} )= \frac{1}{2 z^3}\sum_{n=0}^{\infty}(1-(-1)^n)\frac{z^n}{n!}= \frac{1}{z^3}\sum_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}= \sum_{n=0}^{\infty}\frac{z^{2n-2}}{(2n+1)!}.$$