Laurent series around a singular point

[jamilmalik]

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

Homework Statement

f(z) = \frac{1}{z^2-4} ; z_0=2

Homework Equations

Laurent series
\sum_{-\infty}^{\infty} a_n (z-c)^n

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?

 

[Office_Shredder]

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)

 

[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