# Laurent expansions

• December 1st 2009, 10:04 PM
conmeo
Laurent expansions
Find all Laurent expansions centered at z0 = 0 for f(z) = 1/[z^2(z^3 + 1)] and find the region of convergence for each expansion.

For this problem, I start with rewrite f(z) = 1/(z^2) *1/ [1-(-z^3)] then use geometric series to write laurent series representation for 1/ [1-(-z^3)]. I am stuck after that. Please help. Thank you.
• December 2nd 2009, 01:14 AM
mr fantastic
Quote:

Originally Posted by conmeo
Find all Laurent expansions centered at z0 = 0 for f(z) = 1/[z^2(z^3 + 1)] and find the region of convergence for each expansion.

For this problem, I start with rewrite f(z) = 1/(z^2) *1/ [1-(-z^3)] then use geometric series to write laurent series representation for 1/ [1-(-z^3)]. I am stuck after that. Please help. Thank you.

There are two cases:

1. If $|-z^3| < 1 \Rightarrow |z| < 1$ then $\frac{1}{1 - (-z^3)} = -z^3 + (-z^3)^2 + (-z^3)^3 + .... = -z^3 + z^6 - z^9 + ....$.

2. If $|-z^3| > 1 \Rightarrow \frac{1}{|-z^3|} < 1 \Rightarrow |z| > 1$ then you need to consider the following:

$\frac{1}{1 + z^3} = \frac{1}{z^3} \cdot \frac{1}{\frac{1}{z^3} + 1} = \frac{1}{z^3} \cdot \frac{1}{1 - \left(-\frac{1}{z^3}\right)}$ $= \frac{1}{z^3} \left( -\frac{1}{z^3} + \left(-\frac{1}{z^3}\right)^2 + \left(- \frac{1}{z^3}\right)^3 + .... \right) = \frac{1}{z^3} \left( -\frac{1}{z^3} + \frac{1}{z^6} - \frac{1}{z^9} + .... \right)$.