1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Classifying singularities of a complex function

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Find and classify the isolated singularities of the following:

    $$ f(z) = \frac {1}{e^z - 1}$$

    2. Relevant equations


    3. The attempt at a solution

    I have the solution for the positions of the singularities, which is: ## z = 2n\pi i## (for ##n = 0, \pm 1, \pm 2, ...##) and this makes sense. But the solution also says these are simple poles. This part I do not quite understand. For example, if I take the derivative of ##f(z)##, then ##f'(z) = -(e^z -1)^{-2} e^z## where ##f'(2n\pi i)## is also ##\infty##. Perhaps my understanding of simple poles is wrong, but I don't see any singularities of order 1 when considering the Laurent series of ##f(z)##.

    Any ideas on why ## z = 2n\pi i## is considered a simple pole here?
     
  2. jcsd
  3. Nov 21, 2016 #2

    fresh_42

    Staff: Mentor

    If ##z_0=1## is a pole of first order of ##f(z)##, what can be said about ##(z-z_0)\cdot f(z)##?
     
  4. Nov 21, 2016 #3

    Ahh that makes sense. When evaluating ##(z-z_0)\cdot f(z)## at ##z = z_0##, the value is finite. I incorrectly took the derivative above and treated this like a zero of order 1 instead of a pole. Thank you for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Classifying singularities of a complex function
  1. Classify singularities (Replies: 1)

Loading...