TITLE: Finding all Laurent expansions of $f=\frac{1}{z}$
QUESTION [5 upvotes]: I have the following homework problem I need help with:

Let $G=\{z\in\mathbb{C}:\, z\neq0\}$ and define $f:\, G\to G$ by
  $f(z)=\frac{1}{z}$.
Find all possible Laurent expansions of $f$ which are not Taylor
  expansions and for each such expansion specify in what set the Laurent
  series converges.

What I tried:
For the point $z=z_{0}$ there are Laurent expansions in $$E_{1}:=0<|z|<R$$
and in $$E_{2}:=|z|>R$$
Where $R>0$ is real.
I am unsure about in which set the Laurent series converges to $f$
, but it seems that in both cases the Laurent series of $f$ is simply
$\frac{1}{z}$ and it converges in all points of $E_{1}$or $E_{2}$
(according to the case).
Am I correct ?
For $z_{0}\ne0$ the Laurent expansions which is not Taylor is in
$$E_{1}:=0\leq|z|<R$$ where $R>|z_{0}|$ and $$E_{2}:=|z|>R$$ where
$R<|z_{0}|$.
I am not sure about where I should of used $<$ or $\leq$ in the
description of $E_{1},E_{2}$.
I am also having difficulty finding the Laurent series at $E_{i}$
and telling where it is convergent.
Can someone please help me out with the case $z\neq z_0$ and tell me if I did correctly the first case ($z=z_0$) ?

REPLY [7 votes]: What you have is a bit confused. When $z_0=0$, then there is only the one expansion with $0<|z_0|$, and as you said, the Laurent expansion is just $f(z) = \dfrac 1z$.
When $z_0\ne 0$, you have two possible regions: (1) $0<|z-z_0|<|z_0|$ and (2) $|z_0|<|z-z_0|<\infty$. You need an expansion in powers of $z-z_0$. So you start like this:
$$\frac1z = \frac1{z_0+(z-z_0)}=\frac1{z_0}\cdot\frac1{1+\frac{z-z_0}{z_0}}\,.$$
When $|z-z_0|<|z_0|$, substitute $u=\frac{z-z_0}{z_0}$ and note that $|u|<1$. Now use geometric series. When $|z-z_0|>|z_0|$, this doesn't work, so you factor out the larger term:
$$\frac1z = \frac1{z_0+(z-z_0)}=\frac1{z-z_0}\cdot\frac1{1+\frac{z_0}{z-z_0}}\,.$$
As before, substitute $u=\frac{z_0}{z-z_0}$, note that $|u|<1$, and proceed.