Expand $\displaystyle f(z) = \frac{2}{z(z-2)}$ in a Laurent series valid for $\displaystyle |z-2|>0$.
I've tried many things, but the |z-2|>0 thing gets me. Tips?
Ok, I did a substitution $\displaystyle w=z-2$ and got
$\displaystyle f(z)=\begin{cases}\displaystyle -\frac{1}{z-2}\sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n, & \text{for}~ |z-2|<2
\\ \displaystyle \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n, &\text{for}~|z-2|>2
\end{cases}$
Does that look right?
To me it looks somewhat right, but not quite. Let me see: I'd do a partial fraction decomposition first, then massage the thing into something I can expand in terms of $\displaystyle z-2$, or $\displaystyle \frac{1}{z-2}$, respectively.
For $\displaystyle 0<|z-2|<2$:
$\displaystyle \frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{2}\cdot\frac{1}{1+\frac{z-2}{2}}$
which gives
$\displaystyle =\frac{1}{z-2}-\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n$
For $\displaystyle 2<|z-2|$, on the other hand:
$\displaystyle \frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{z-2}\cdot\frac{1}{1+\frac{2}{z-2}}$
which gives
$\displaystyle =\frac{1}{z-2}-\frac{1}{z-2}\cdot\sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n$
Thanks, I edited my post, I had 2 typos. The powers of two were supposed to be n in the series. I'll look over your working and see why it's different.
EDIT: It seems that your answer is equivalent! Cool
Well, the condition that $\displaystyle |z-2|=2$ is satisfied by $\displaystyle z=0$ for which the original function has a pole. Hence you cannot expect to get a Laurent-Series expansion that converges for all such values of $\displaystyle z$.
But it might be that one of the two series that you have already found converges for some values other than $\displaystyle z = 0$ that satisfy $\displaystyle |z-2|=2$.
Let me put it this way: you can never have an essential singularity inside the disc of convergence of a power series, or inside the annulus of convergence of a Laurent series.
But you can (and typically do have such a singularity) right on the boundrary of that disc / that annulus of convergence. Stronger: what limits the size of the disc / annulus of convergence is the existence of such a singularity on its boundary.
Convergence or divergence of those power / Laurent series expansions for points that lie right on the boundary of the disc / annulus of convergence is a separate problem that might need some further trickery to decide.