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Laurent series for f(z) = 1/(exp(z)-1)^2 ?

  1. Sep 10, 2008 #1
    Laurent series for f(z) = 1/(exp(z)-1)^2 ??

    1. The problem statement, all variables and given/known data

    Determine the Laurent series and residue for f(z) = [tex]\frac{1}{(e^{z} - 1)^{2}}[/tex].

    2. Relevant equations

    We know that the Taylor series expansion of e[tex]^{z}[/tex] is = 1 + z + (z^2)/2! + ...

    3. The attempt at a solution

    I am soooo confused. It's almost like a geometric series, except the denominator is squared. I know I need to use the Taylor series expansion, but I don't know where. Do I just invert the Taylor series expansion??? How do I deal with the fact that it's squared?

    I almost thought of writing it into the geometric series anyway, then squaring the terms (obviously this isn't really accurate, as I'd be missing tons of cross-terms, but if we only need a few terms of the series anyways...). Can anyone help to explain what's going on?

    I understood the example in our book (there's only 1 example :frown:) but it had two different poles, and I was able to expand the Taylor series around the "other" pole for each Laurent series. Here our only pole is 0 (though of order 2) and I don't know how to proceed.
    Any help would be so appreciated!! :cool:
  2. jcsd
  3. Sep 10, 2008 #2


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    Re: Laurent series for f(z) = 1/(exp(z)-1)^2 ??

    You expand around z=0. exp(z)-1=z+z^2/2!+...=z(1+z/2!+...). Write that as z(1+a) where a=z/2!+z^2/3!+... So 1 over that squared is (1/z^2)*(1/(1+a)^2). The series expansion for 1/(1+a)^2 is 1-2a+3a^2+... As you said, you only need the first few terms. Now start throwing away terms that you know won't contribute to the terms you need. E.g. a^2 starts with a z^2 term.
  4. Sep 10, 2008 #3
    Re: Laurent series for f(z) = 1/(exp(z)-1)^2 ??

    oh,oh, that is SOOO Nice!!!

    IT WORKED! :rofl:

    I think this has helped me a LOT. You use the Taylor series to simplify your expression into a form that you can expand in geometric/other series, because it's a lot simpler than doing the nasty line integral for the coefficients? Basically we just put this into a different looking form, and then it all stood out.

    I even got the same answer as Maple, up to the first 5 terms! :biggrin:

    Thank you Dick!!
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