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Math Help - Help with Laurent Series

  1. #1
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    Help with Laurent Series

    I'm trying to construct Laurent series around z = 0 and z= \infty for f(z) = 1/{z-2} with z_{0} = 2 the only pole of this function.

    I have f(z) = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n

    with a_{n} = \oint_{\gamma}f(z)/(z-c)^{n+1}dz

    and \oint_{\gamma}g(z)dz = 2i\pi res_{z_{0}}g(z)

    and res_{z_{0}}g(z) = \lim_{z \to \infty}(z - z_{0})g(z)

    so am I correct in assuming that around z=c:

    f(z) = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n = \sum_{n=-\infty}^{\infty} (z - c)^{n}\lim_{z \to z_{0}}f(z)\frac{z - z_{0}}{(z - c)^{n+1}}

    or did I make a mistake?

    And how the hell do I evaluate the function around z=infinity?
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  2. #2
    MHF Contributor

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    I am confused about what you are doing here. f(z)= \frac{1}{z- 2} has a pole at z= 2, as you say, but is analytic at z= 0. A function's "Laurent series" about any point at which it is analytic is just its Taylor series.
    In this case, \frac{1}{2-z}= \frac{1/2}{1- z/2} which is the sum of a geometric series with first term a= 1/2 and common ratio r= z/2.

    As for "at z=infinity", the Laurent series of any function, f(z), about z= \infty is just the Laurent series of f(1/z) about z= 0.

    If f(z)= \frac{1}{2-z} then f(\frac{1}{z})= \frac{1}{2- \frac{1}{z}}= \frac{z}{2z-1}.
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  3. #3
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    Thank you.

    Yeah, I was kinda wondering why only this problem had different values for c and z0 so it makes sense this is a special case. But if I were told to look for the Laurent Series near c=2=z0 would my last equation be correct?
    I looked up the answer using an online calculator (for z0 = 2 and c=0) and it corresponded to what I found using my method above, except for the sign: the calculator found: -1/2 -z/4 -(z^2)/8 etc.. while I found 1/2 z/4 (z^2)/8 etc... is it a coincidence that I got so close?
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