My Math Forum Is finding laurent series expansion of f at z_0 using geometric series convenient?

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 December 28th, 2014, 06:50 AM #1 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Is finding laurent series expansion of f at z_0 using geometric series convenient? Hello there, I was taught in my complex analysis course that I can find the laurent series expansion for $f:\Omega \to \mathbb{C}$ by using the formula for geometric series: $$f(z) = \frac{1}{1-g(z)} = \sum_{n=0}^\infty g(z)^n$$ This is useful for series expanded at $z=0$. However, I am interested in finding convenient methods for making laurent expansions at at other points $z_0$. I would like to make series of the form: $$f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n,$$ where $z_0$ may be different for $0$. Is it convenient to use the geometric series construction to do this? If so, how? And if not, which are the convenient methods to do so? I know that the coefficients are given by $a_n = 1/(2\pi i) \oint_C f(z)/(z-z_o)^{n+1} \mathrm{d}z$, but it is tedious to calculate all the coeffients manually. Thank you for your time. Kind regards, Marius Last edited by king.oslo; December 28th, 2014 at 07:44 AM.

 Tags convenient, expansion, finding, geometric, laurent, series

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