Laurent Series

  • Thread starter FiberOptix
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12
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Main Question or Discussion Point

Hello all,

I've got an exam tomorrow so any quick responses would be appreciated. I'm following the Boas section on Laurent series... Anyway, here's my problem:

In an example Boas starts with f(z) = 12/(z(2-z)(1+z), and then using partial fractions arrives at f(z) = (4/z)(1/(1+z) + (1/2-z)). Fine. So there are three singular points, at z = 0, z = 2, and z = -1. So, we have two circles about z = 0 and should be able to obtain three Laurent series, one valid for 0 < |z| < 1, 1 < |z| < 2, and |z| > 2. I'll skip the details of the rest of this example but she expands the partial fraction in terms of z to obtain f(z) for 0 < |z| < 1, then proceeds to expand in terms of 1/z to obtain f(z) for |z| > 2 and then one of the partial fractions in terms of z and the other one in terms of 1/z to obtain f(z) for 1 < |z| < 2.

I'm a bit confused as to why z, 1/z, and then a combination and also how you know which will correspond to which solution for f(z).

As I said, the exam is tomorrow so any quick responses would be helpful.

Thanks
 

Answers and Replies

12
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Alright, so I now understand it's because of the convergence in particular regions but I'm still not 100% on this.
 

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