Question

[ Complex Analysis ] Find the laurent series for 1/(1-Z2) about Z0=1.

Answer 1

Recall that for $|z|$, we have

$\dfrac1{1-z}=\displaystyle\sum_{n\ge0}z^n$

Then

$\dfrac1{1-z}=-\dfrac1z\dfrac1{1-\frac1z}=-\dfrac1z\displaystyle\sum_{n\ge0}z^{-n}=-\sum_{n\ge0}z^{-n-1}$

valid for $|z|>1$, so that

$\dfrac1{1-z^2}=-\dfrac1{z^2}\dfrac1{1-\frac1{z^2}}=-\dfrac1{z^2}\displaystyle\sum_{n\ge0}z^{-2n}=-\sum_{n\ge0}z^{-2n-2}$

also valid for $|z|>1$.