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Determining order of the poles in z/(e^z-1)

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the order of the poles of

    f(z)=z/(e^z-1)


    2. Relevant equations

    The poles are at z=2 π i k k[itex]\in[/itex]Z\0
    (Because at z=0 f(z) has a removable singularity -set f(0)=1)

    3. The attempt at a solution

    I tried using the Taylor series of e^z - [itex]\sum[/itex]z^n/n!
    But I just got
    f(z)=1/[itex]\sum[/itex]z^n/(n+1)!
    and I somehow need to take out a factor of (z-2 π i k)^j for some j>0 out of that...
     
  2. jcsd
  3. Oct 13, 2012 #2
    Given f(z), what is the minumum power of (z-z_0) would I have to multiply the function by so that:

    [tex]\lim_{z\to z_0} (z-z_0)^n f(z) \neq \infty[/tex]

    Take for example at zero:

    [tex]\lim_{z\to 0} z^0 \left(\frac{z}{e^z-1}\right)\neq \infty[/tex]

    thus the order of the pole at zero is zero, i.e., it's removable. Ok, now you try the pole at [itex] 2n\pi i[/itex].
     
  4. Oct 13, 2012 #3
    Right - got them. I got my brain stuck on taylor series, when I needed L'hopitals rule. Thanks.
    They're simple poles with limits z->z_0=z_0, right? :)
     
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