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Laurent series of 1/(e^z - 1)

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Obtain the first few terms of the Laurent series for the following function in the specified domain:

    [itex]\frac{1}{e^z-1}[/itex] for [itex]0 < |z| < 2\pi.[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I've attempted a few approaches, but haven't really gotten anywhere. For instance, using a Maclaurin series for [itex]e^z[/itex] yields [itex]\frac{1}{e^z-1} = \frac{1}{z} \cdot \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+\cdots}[/itex]

    Of course, [itex]\frac{1}{z}[/itex] can be written as [itex]\frac{1}{2\pi - (-z + 2\pi)} =\frac{1}{2\pi} \cdot \frac{1}{1-\frac{-z+2\pi}{2\pi}}[/itex], and since the latter can be written as a geometric series in the given annulus, we have [itex]\frac{1}{z} = \sum_{k = 0}^{\infty} (-1)^k (\frac{z-2\pi}{2\pi})^k.[/itex]

    Actually, after further contemplation, I think I could set [itex]g(z) = \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+\cdots}[/itex] and differentiate that a few (infinitely many) times to find its Maclaurin series. I think termwise differentiation should be allowed in the annulus in question, since g should be analytic there. Then multiplication by 1/z yields the Laurent series?
  2. jcsd
  3. Feb 7, 2012 #2
    Do you know the Laurent series of [itex]\frac{1}{z-1}[/itex]??
  4. Feb 7, 2012 #3
    Regular Maclaurin (Taylor at z=0) series for [itex]|z|<1[/itex], i.e. [itex]-1-z-z^2-z^3-\cdots[/itex], and [itex]\sum_{k = 0}^{\infty} \frac{1}{z^{k+1}}[/itex] for [itex]|z|>1[/itex]?

    So, [itex]\frac{1}{e^z - 1} = \sum_{k = 0}^{\infty} \frac{1}{e^{z(k+1)}}[/itex]?
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