I am given the following.
Let $f(z) = \frac{e^z-e^{-z}}{z^4}$. Give a Laurent Series for $f(z)$ in terms of powers of $z$. For which values of $z$ does the Laurent series converge? Justify your answer.
Laurent Series is new to me. I am familiar with Taylor Series and it seems similar. If I am not mistaken, the Laurent series is a the series representation in terms of both positive and negative powers of $(z - z_0)$.
This is my approach thus far.
\begin{eqnarray} f(z) & = & \frac{e^z-e^{-z}}{z^4}\\ & = & \frac{1}{z^4}[e^z - e^{-z}]\\ & = & \frac{2}{z^4}\cdot\frac{e^z - e^{-z}}{2}\\ & = & \frac{2}{z^4}\sinh(z)\\ & = & \frac{2}{z^4}\sum_0^{\infty}\frac{z^{2k+1}}{(2k+1)!}\\ & = & \sum_0^{\infty}\frac{2z^{2k-3}}{(2k+1)!}\\ \end{eqnarray}
I am wondering whether I approached it correct or not. Where do I continue from here? I want to say this is what I would do next, but I am not sure. With a slight change of index (z = 2k - 3), then we have the Laurent series representation
\begin{eqnarray} f(z) & = & \sum_{n = -3}^{-\infty}\frac{2z^n}{(n + 4)!} \end{eqnarray}
This is where I am really stuck because I am sure a negative factor is undefined.
I am sure once I figure out the Laurent series, figuring out what values of $z$ makes the series converge would involve the use of a ratio test of some sort.
Thank you for your time and Thanks in advanced for your feedback.