Find the laurent series expansion for the following function: $$f(z)=\frac{e^z}{z^3-z^4}$$
What I've done is:
\begin{align} f(z) & = \frac{e^z}{z^3-z^4} \\ & = \frac{e^z}{z^3(1-z)} \\ & = \frac{e^z}{z^3} .\frac{1}{1-z} \\ & = \frac{e^z}{z^3}\Big(1+z+z^2+\dots\Big) \\ & = e^z\Big(\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+\dots\Big) \\ & = \Big(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{60}+\dots \Big)\Big(\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+\dots\Big) \\ \end{align}
Am I supposed to multiply this out? Surely there's a better way?
Here's how I would do it. First of all, I would compute the Taylor expansion of $\dfrac{e^z}{1-z}$:\begin{align}\frac{e^z}{1-z}&=\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)\left(1+z+z^2+z^3+\cdots\right)\\&=1+2z+\frac52z^2+\cdots+\left(\sum_{k=0}^n\frac1{k!}\right)z^n+\cdots\end{align}And then\begin{align}\frac{e^z}{z^3-z^4}&=\frac1{z^3}\cdot\frac{e^z}{1-z}\\&=\frac1{z^3}\left(1+2z+\frac52z^2+\cdots+\left(\sum_{k=0}^n\frac1{k!}\right)z^n+\cdots\right)\\&=\frac1{z^3}+\frac2{z^2}+\frac5{2z}+\cdots+\left(\sum_{k=0}^{n+3}\frac1{k!}\right)z^n+\cdots\end{align}