# Math Help - Another Laurent expansion

1. ## Another Laurent expansion

Expand $f(z) = \frac{2}{z(z-2)}$ in a Laurent series valid for $|z-2|>0$.

I've tried many things, but the |z-2|>0 thing gets me. Tips?

2. Ok, I did a substitution $w=z-2$ and got

$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?

3. Originally Posted by scorpion007
Ok, I did a substitution $w=z-2$ and got

$f(z)=\begin{cases}\displaystyle -\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^{\color{red}2}, & \text{for}~ |z-2|<2

\\ \displaystyle \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^{\color{red}2}, &\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 $z-2$, or $\frac{1}{z-2}$, respectively.

For $0<|z-2|<2$:

$\frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{2}\cdot\frac{1}{1+\frac{z-2}{2}}$

which gives
$=\frac{1}{z-2}-\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n$

For $2<|z-2|$, on the other hand:

$\frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{z-2}\cdot\frac{1}{1+\frac{2}{z-2}}$

which gives
$=\frac{1}{z-2}-\frac{1}{z-2}\cdot\sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n$

4. 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.

5. I have a related question though -- we found the series for |z-2|>2 and |z-2|< 2, but what about |z-2| = 2?

6. Originally Posted by scorpion007
I have a related question though -- we found the series for |z-2|>2 and |z-2|< 2, but what about |z-2| = 2?
Well, the condition that $|z-2|=2$ is satisfied by $z=0$ 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 $z$.

But it might be that one of the two series that you have already found converges for some values other than $z = 0$ that satisfy $|z-2|=2$.

7. Ah, I see!

Also, would it be correct to say that while there is a pole of order 1 for the first case (|z-2|<2), that there is an essential (unremovable) singularity for the second case? (Since we have infinitely many negative power terms)

8. Originally Posted by scorpion007
Ah, I see!

Also, would it be correct to say that while there is a pole of order 1 for the first case (|z-2|<2), that there is an essential (unremovable) singularity for the second case? (Since we have infinitely many negative power terms)
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.