# Laurant Series

• April 23rd 2010, 12:43 PM
Negativ
Laurant Series
Forgive me if this takes a while to post, attempting to get the LaTeX right.

Attempting to find a Laurent series expansion for the function:

$f (z)=\frac{4z^2}{1-z^2}$

Over 2 regions:

$0<|z|<1$

$1<|z|< {\infty}$

I have seen similar problems like:

$f(z)=\frac{z+2}{z-2}$ where you can simplify by taking

$f(z)=\frac{z+(-2+2)+2}{z-2}=\frac{z-2}{z-2}+\frac{4}{z-2}=1+\frac{2}{\frac{z}{2}-1}=1-\frac{2}{1-\frac{z}{2}}$

Which is solved here

I can't seem to find a way to simplify my equation in a similar manner.

I know that (obviously):

$f(z)=\frac{4z^2}{(1+z)(1-z)}$

Not sure where to go from there. If I knew a Maclaurin series for something like $\frac{z}{1+z}$ and $\frac{z}{1-z}$ or how to find one, that would make things much simpler as I could just multiply the series together...?

• April 23rd 2010, 01:46 PM
Laurent
Quote:

Originally Posted by Negativ
Forgive me if this takes a while to post, attempting to get the LaTeX right.

Attempting to find a Laurant series expansion for the function:

$f (z)=\frac{4z^2}{1-z^2}$

Over 2 regions:

$0<|z|<1$

$1<|z|< {\infty}$

Consider the following: for $|z|<1$, we have $|z^2|<1$ hence $\frac{1}{1-z^2}=\sum_{n=0}^\infty z^{2k}$ and $\frac{z}{1-z^2}=\sum_{n=0}^\infty z^{2k+1}=z+z^3+z^5+\cdots$. I let you consider the other case.

By the way it is Laurent series, with an "e"!
• April 23rd 2010, 03:14 PM
Negativ
I can see, judging by your username, that I was wrong. Haha. Thanks for the reply. I'll see what I can work out.
• April 23rd 2010, 07:40 PM
chisigma
For $|z|>1$ is...

$f(z)= \frac{4\cdot z^{2}}{1-z^{2}} = \frac{-4}{1-(\frac{1}{z})^{2}} = -4 - \frac{4}{z^{2}} - \frac{4}{z^{4}} - \dots$ (1)

For $|z|<1$ is...

$f(z)= \frac{4\cdot z^{2}}{1-z^{2}} = 4\cdot z^{2} \cdot (1 + z^{2} + z^{4} + \dots)$ (2)

Kind regards

$\chi$ $\sigma$