# Whats a Laurent series? And how do I use one to represent a function?

1. May 10, 2007

### laura_a

1. The problem statement, all variables and given/known data
Write TWO laurent series in powers of z that represent the function

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

In certain domains, and specify the domains

2. Relevant equations

Well thats my prob, not sure what the terms in the Laurent series are

The formula I'm looking at is

\sum^{infty}_{n=0} a_n * (z-c)^n

for a complex function f(z) about a point c and a_n is a constant.

I sort of understand that, but do I use it to represent a function? Thats the part I'm not sure about

3. The attempt at a solution

My lame guess is that sub in the f(z) i'm given into the equation. I've seen one example where the fraction is split into two and then the Laurent expansion is applied. I have the answers (because it's a text book question)

They are

\sum^{\infty}_{n=0} (-1)^{n+1} * z^{2n+1} + 1/z (0 < |z| < 1)

and

\sum^{\infty}_{n=0} [(-1)^{n+1}] / z^{2n+1} (1 < |z| < \infty)

All I need to know is what to I plug in to where and I'll work on the rest :) any suggestions will make my day (night actually, but who's counting)

Thanks
Laura

2. May 10, 2007

### nrqed

Btw, put your TeX code between [ itex ] and [ /itex ] (with no spaces...I left spaces there just so that they would show up in my post.)

First you must identify the poles of your function.

Once you have the poles, pick one and use that point as the center of your Laurent series. Then you must expand everything in powers of (z-c)^n as you said, where c is the location of the pole.
The formula to use is the geometric series
$$\frac{1}{1-q} = \sum_{n=0}^\infty q^n$$
at the condition that q < 1. (in your case, it will be a condition on the magnitude of a complex quantity).

For the domain, you will have to check if you have any restriction from using the geometric series and you must laos make sure that you don't go far enough from your pole that you run into one of the other poles.

Last edited: May 10, 2007
3. May 11, 2007

### HallsofIvy

Staff Emeritus
To answer the question in your title, a Laurent series is a power series that includes negative powers. When your original function is a rational function, like the one you give, $$f(x)= \frac{1}{z(1+z^2)}$$, a good method is to take out one of the factors in the denominator, say 1/z, and expand the remaining function, 1/(1+z2), in a Taylor's series. Here we can write $$\frac{1}{1+z^2}= \frac{1}{1- (-z^2)}$$ and recognize that as a geometric series with "common factor" -z2 (I just noticed that is what nrqed said): $1- z^2+ z^4- z^6+ \cdot\cdot\cdot$. Finally, put the 1/z back in: $z^{-1}- z+ z^3- z^5+ \cdot\cdot\cdot$. Obviously, that cannot converge at z= 0. Further more the original power series only converges for the absolute value "common factor" less than 1. Here that is z2< 1 so the Laurent series converges for 0< |z|< 1.

For a different Laurent series, notice that 1/(1+ z^2) is discontinuous at z= i. Factor out 1/(z- i), leaving $$\frac{1}{z(z+ i)}$$ and find a Taylor's series for that, about z= i. A good method is to separate into two different fractions using "partial fractions" and write each as a geometric series.