# Laurent series of $\frac1{1-z}$

I had an exam question that has been troubling me. It looks simple, but I cannot seem to be able to figure this out.

How can I find the Laurentian series expansion of $\frac1{1-z}$ in the region $|z+2|<3$?

• If a series in powers of $z+2$ confuses you, change variables to $w=z+2$ do it in powers of $w$. Then at the end change back. – GEdgar Mar 23 '17 at 17:54

$\frac {1}{1-z}$ centered at $z = -2$
$\frac {1}{3-(z+2)}\\ \frac {\frac 13}{1-\frac {(z+2)}{3}}$
$\frac 13 \sum_\limits{i=0}^{\infty} (\frac {z+2}{3})^i$