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A nasty laurent series

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi! I need to find the laurent series of ##e^{1/(1-z)}## to get the residue at ##z=1##. Can somebody help me?

    3. The attempt at a solution


    I tried using the taylor series of e^x around x=1, but I seem to have failed.. what am I to do? I can't use the maclaurin series of e^x, right?
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2


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    You can and should. [itex]e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}[/itex] is a definition of [itex]e^x[/itex]. Setting [itex]x = (1 - z)^{-1}[/itex] is clearly going to give you a sum of powers of [itex](z - 1)[/itex], which is after all what you want.
  4. Nov 13, 2013 #3
    But why can you do that? Shouldn't the taylor series used to find the laurent series be expanded about the same value as the laurent series is being expanded about (in this case z=1)?

    Could you remind me of the rules for stuff?
    Last edited: Nov 13, 2013
  5. Nov 13, 2013 #4


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    Why should it?

    If the power series [itex]g(z) = \sum a_n z^n[/itex] has radius of convergence [itex]R[/itex], then if [itex]|f(z)| < R[/itex] we have [itex]g(f(z)) = \sum a_n f(z)^n[/itex].

    If [itex]f(z)[/itex] is not within the radius of convergence then one generally has to find a different power series for [itex]g[/itex]. But that would point to expanding [itex]g[/itex] in a Taylor series about [itex]f(z_0)[/itex], not [itex]z_0[/itex] itself.

    Conveniently the power series for exp, which is in fact the definition of the exp function on the complex plane, has infinite radius of convergence, so
    \exp (f(z)) = \sum_{n= 0}^{\infty} \frac{f(z)^n}{n!}
    for any [itex]f(z)[/itex].
  6. Nov 13, 2013 #5
    Hmm, indeed. But still, what happens if f(z) has multiple poles? How would you find the residues of them?
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