Laurent series expansion for $\frac{(z-2)}{(z+1)}$centered at z=-1

a) How do I find the Laurent series expansion for $\frac{(z-2)}{(z+1)}$centered at z=-1 ?

b) How do I find the region for which it converges?

I've tried rewriting it in different ways, but nothing seems to get me anywhere. I don't think I'm supposed to be integrating anything. Any help would be appreciated, but I ask that you keep the explanation on a low enough level that I can understand, and I don't understand much at the moment.

Edit: I saw the question somewhere else, but I didn't find the answer to be very helpful.

• rewrite the function such that you get terms of $z+1$ everywhere, or a constant. Then, use Maclaurin's expansion at $z+1 = 0$, which will be valid for $| \frac 1 {z+1}| <1$ – bikalpa Nov 12 '17 at 3:25
• Why isn't it just $1-(3/(z+1))$? – Oscar Lanzi Nov 12 '17 at 3:26
• @OscarLanzi, yeah in this case its just what you said. I commented procedure for a generic case though. – bikalpa Nov 12 '17 at 3:28
• How do I do that? Right now I've got it written as 1-3/(z+1). What do I do with that? – CluelessIndividual Nov 12 '17 at 3:29
• Actually, I think I've got it... I'm just exceptionally lost on how to find the domain of convergence of this thing – CluelessIndividual Nov 12 '17 at 3:31

The function \begin{align*} f(z)&=\frac{z-2}{z+1}\\ \end{align*} is to expand around the center $z=-1$.
Since there is a simple pole at $z=-1$ we get as region $D$ of convergence \begin{align*} D:&\quad 0<|z+1|\\ \end{align*}
The region $D$ is a punctured disc with center $z=-1$ and radius $\infty$. It admits a representation as principal part of a Laurent series at $z=-1$.
Expansion in $D$:
• @CluelessIndividual: You're welcome! The form is perfectly valid. Recall a Laurent expansion expanded at $z=-1$ has the form $\sum_{j=-\infty}^\infty a_j(z+1)^j$. Here $a_{-1}=-3,a_0=1$ and all other $a_j=0$. – Markus Scheuer Nov 12 '17 at 19:55