<s>[INST] Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$. I need to find the Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$.
I believe I am nearly there but I have a confusion.
Notice $f(z)=\frac{z^2-1}{1-z^3}+\frac{1}{z}$ so if $0<|z|<1$ we have
$$f(z)=\frac{1}{z}+(z^2-1)\sum_{n=0}^{\infty}z^{3n}=\frac{1}{z}+\sum_{n=0}^{\infty}(-z^{3n}+z^{3n+2})$$
First question: Is there any better way to present this series or is this in general acceptable?
Now for $|z|>1$, $f$ is entire so it should have a Taylor Series, however I am getting this:
$$f(z)=\frac{1}{z}+\left(\frac{1+z^2}{z^3}\right)\left(\frac{1}{1-\frac{1}{z^3}}\right)=\frac{1}{z}+\sum_{n=1}^{\infty}\left(z^{-3n}+z^{-3n+2}\right)$$
and this is not a Taylor series...
Edit $1$: For the case $0<|z|<1$ I have tried a different approach and got
$$f(z)=\frac{1-z}{z}\left(\frac{1}{1-z^3}\right)=\sum_{n=0}^{\infty}(z^{3n-1}-z^{3n})$$
It is the same series but a much cleaner approach
 [/INST] If $|z|<1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=(1-z)(1+z^3+z^6+z^9+\cdots)\\&=1-z+z^3-z^4+z^6-z^7+\cdots\end{align}and therefore$$f(z)=z^{-1}-1+z^2-z^3+z^5-z^6+\cdots$$If $|z|>1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=-(1-z)(z^{-3}+z^{-6}+z^{-9}+\cdots)\\&=z^{-2}-z^{-3}+z^{-5}-z^{-6}+\cdots\end{align}and therefore$$f(z)=z^{-3}-z^{-4}+z^{-6}-z^{-7}+\cdots$$
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