# Laurent series with tricky denominator

I'm asked to find "each possible Laurent series of $\frac{1}{z^3-2z^2}$ around $z=2$" and find its annulus of convergence. I can take out $\frac{1}{z-2}\cdot \frac{1}{z^2}$ And now guess I need the series for $\frac {1}{z^2}$, and that's where my understanding is not enough for this problem. I can get $$\frac{1}{z}=\frac 12 \sum^{inf} (-1)^n(z-2)^n$$

and answer with the cauchy product of that squared, but I'm guessing that is not a great answer, I have looked around but can't find any examples with singularity of multiplicity >2.

The function $f$ has a simple pole at $z=2$ and a pole of order two at $z=0$. So, it can be expanded around $z=2$ as Laurent series with annulus of convergence $0<|z-2|<2$. On the other hand in the region $|z-2|>2$ we can expand $f$ as another Laurent series.

It is convenient to represent $f$ using partial fraction decomposition

\begin{align*} f(z)&=\frac{1}{z^2}\frac{1}{z-2}\\ &=-\frac{1}{2z^2}-\frac{1}{4z}+\frac{1}{4(z-2)} \end{align*}

Laurent series of $f$ in annulus $0<|z-2|<2$:

\begin{align*} f(z)&=-\frac{1}{2(z-2+2)^2}-\frac{1}{4(z-2+2)}+\frac{1}{4(z-2)}\\ &=-\frac{1}{8}\frac{1}{\left(1+\frac{z-2}{2}\right)^2}-\frac{1}{8}\frac{1}{\left(1+\frac{z-2}{2}\right)}+\frac{1}{4(z-2)}\\ &=-\frac{1}{8}\sum_{n=0}^{\infty}\binom{-2}{n}\left(\frac{z-2}{2}\right)^n-\frac{1}{8}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-2}{2}\right)^n+\frac{1}{4(z-2)}\\ &=-\frac{1}{8}\sum_{n=0}^{\infty}(n+1)(-1)^n\left(\frac{z-2}{2}\right)^n-\frac{1}{8}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-2}{2}\right)^n+\frac{1}{4(z-2)}\\ &=-\frac{1}{8}\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^n(n+2)(z-2)^n+\frac{1}{4(z-2)}\\ \end{align*}



Laurent series of $f$ in region $|z-2|>2$:

\begin{align*} f(z)&=-\frac{1}{2(z-2+2)^2}-\frac{1}{4(z-2+2)}+\frac{1}{4(z-2)}\\ &=-\frac{1}{2(z-2)^2\left(1+\frac{2}{z-2}\right)^2}-\frac{1}{4(z-2)\left(1+\frac{2}{z-2}\right)}+\frac{1}{4(z-2)}\\ &=-\frac{1}{2(z-2)^2}\sum_{n=0}^{\infty}\binom{-2}{n}\left(\frac{2}{z-2}\right)^n-\frac{1}{4(z-2)}\sum_{n=0}^{\infty}(-1)^n\left(\frac{2}{z-2}\right)^n+\frac{1}{4(z-2)}\\ &=-\frac{1}{2}\sum_{n=0}^{\infty}(n+1)(-1)^n\frac{2^n}{(z-2)^{n+2}}-\frac{1}{4}\sum_{n=0}^{\infty}(-1)^n\frac{2^n}{(z-2)^{n+1}}+\frac{1}{4(z-2)}\\ &=-\frac{1}{2}\sum_{n=2}^{\infty}(n-1)(-1)^n\frac{2^{n-2}}{(z-2)^{n}}-\frac{1}{4}\sum_{n=1}^{\infty}(-1)^n\frac{2^{n-1}}{(z-2)^{n}}+\frac{1}{4(z-2)}\\ &=-\frac{1}{8}\sum_{n=2}^{\infty}(n-1)(-2)^n\frac{1}{(z-2)^{n}}+\frac{1}{8}\sum_{n=1}^{\infty}(-2)^n\frac{1}{(z-2)^{n}}\\ &=-\frac{1}{8}\sum_{n=2}^{\infty}(n-2)(-2)^n\frac{1}{(z-2)^{n}} \end{align*}