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Math Help - Laurent expansions

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by conmeo View Post
    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).
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