# Laurent series around a singular point

1. May 8, 2013

### jamilmalik

Find the Laurent series of the following function in a neighborhood of the singularly indicated, and use it to classify the singularity.

1. The problem statement, all variables and given/known data
$f(z) = \frac{1}{z^2-4} ; z_0=2$

2. Relevant equations
Laurent series
$\sum_{-\infty}^{\infty} a_n (z-c)^n$

3. The attempt at a solution
I started by doing a partial fraction decomposition of $f(z) = \frac{1}{z^2-4} ; z_0=2$ which I got to be $f(z) = \frac{-1}{4} \cdot \frac{1}{z+2} + \frac{1}{4} \cdot \frac{1}{z-2}$ .
Then I used $0< \vert z+2 \vert < 2 \Rightarrow \vert \frac{z+2}{2} \vert < 1$ to apply the geometric series and get $\frac{1}{4} \sum_{k=0}^{\infty} (\frac{2}{z})^k$.

I am not sure if I made a mistake somewhere or if this method is not applicable. Can somebody please provide a hint or point out an error on my behalf?

Last edited: May 8, 2013
2. May 8, 2013

### Office_Shredder

Staff Emeritus
You need to use tex and /tex tags to do math typesetting on this forum.

If you are going to apply the geometric series formula, you need to

1) get a power series in z, not negative exponents
2) Get something that has (z-2)s in it since you want the Laurent series around z=2 (which means you can't take the power series around z=0)

3. May 10, 2013

### yakeyglee

Take note of the following.
$$f(z) = \frac{1}{z^2 - 4} = \frac{1}{z-2} \cdot \frac{1}{z+2} = \frac{1}{z - 2} \cdot \frac{\frac{1}{4}}{1 - \left( - \frac{z - 2}{4} \right)} = \frac{1}{z-2} \cdot \frac{1}{4} \sum_{n=0}^\infty \left( - \frac{z - 2}{4} \right)^n = \cdots$$