I'm starting to learn about Laurent series. The way I understand it is that it is the same as a Taylor expansion, but with negative terms in addition to the positive terms.
I may be wrong, but isn't the general goal to get it into the form $\frac{1}{1-z}$? And then from there the expansion is $\sum_{n=0}^{\infty} z^n$ and $-\sum_{n=1}^{\infty} \frac{1}{z^n}$
I found some other notes online that kind of point in a current direction as to how to go about solving this problem. Below is my attempt.
We know that $$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$$
So if we change $z$ to $\frac{2}{z-1}$, we get
$$ \sum_{n=0}^\infty \frac{(\frac{2}{z-1})^n}{n!}$$
Taylor series of $$e^x=\sum\limits_{k=o}^{\infty}\frac{x^k}{k!}$$
Set $x=2/(z-1)$
$$e^{2/(z-1)}=\sum\limits_{k=o}^{\infty}\frac{\left(\frac{2}{z-1}\right)^k}{k!}$$
The goal when you want laurent series is to get this form: $\sum C_n z^{-n}$ the power should be negative.