# Thread: Laurent Series

1. ## Laurent Series

Find the laurent series on 2<|z|<3
f(z)=1/[(z^2)*(z-2)*(z+3)]

2. Originally Posted by scubasteve123
Find the laurent series on 2<|z|<3
f(z)=1/[(z^2)*(z-2)*(z+3)]
Around what value of z is the series to be found?

3. There isnt an indicated center. All the question says is "Find Laurent series expansion(in powers of z) that represents the function in the region 2<|z|<3

4. Originally Posted by scubasteve123
There isnt an indicated center. All the question says is "Find Laurent series expansion(in powers of z) that represents the function in the region 2<|z|<3
"(in powers of z)" means you find it around z = 0. So the centre is given. It helps if the whole question is posted.

I suggest writing the given function as $\frac{1}{z^2} \left( \frac{A}{z-2} + \frac{B}{z + 3}\right)$ where the partial fraction decomposition is left for you to finish.

Then expand each of $\frac{1}{z - 2}$ and $\frac{1}{z + 3}$ around z = 0. Then substitute etc.

eg. $\frac{1}{z - 2} = \frac{1}{1 - \frac{2}{z}} = 1 + r + r^2 + r^3 + .....$ where you should recognise an infinite geometric series with $r = \frac{2}{z}$ (note that $\left| \frac{2}{z} \right| < 1$ since $2 < |z| < 3$.