Finding all Laurent expansions of $f=\frac{1}{z}$

Question

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 ($z=z_0$) ?

Answer 1

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.