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

Question

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 $|z+2|<3$?

Answer 1

Well, try the following:

\begin{align*} \frac{1}{1-z} &= \frac{1}{3 - (z+2)} \\ &= \frac{1}{3} \frac{1}{1- (z+2)/3} \\ &= \frac{1}{3} \sum_{n=0}^\infty \left(\frac{z+2}{3} \right)^n \end{align*}

This is a standard trick by the way, in the end you only need to rewrite it in order to make use of the geometric series.

Answer 2

$\frac {1}{1-z}$ centered at $z = -2$

$\frac {1}{3-(z+2)}\\ \frac {\frac 13}{1-\frac {(z+2)}{3}}$

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