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!

Homework Help: Laurent series expansion of Log(1+1/(z-1))

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the Laurent series expansion of [itex]f(z) = \log\left(1+\frac{1}{z-1}\right)[/itex] in powers of [itex]\left(z-1\right)[/itex].

    2. Relevant equations
    The function has a singularity at z = 1, and the nearest other singularity is at z = 0 (where the Log function diverges). So in theory there should be a Laurent series on the punctured disk centered at z = 1 with radius 1.

    3. The attempt at a solution
    I can't seem to rewrite the expression in a form which can be used to find the Laurent series for [itex]0 < |z - 1| < 1 [/itex]. I can find the Laurent series for [itex] |z - 1| > 1 [/itex] by rewriting [itex]f(z) = -\log\left(\frac{1}{1+\frac{1}{z-1}}\right)[/itex] and then expanding using the series expansion of [itex]\log\left(\frac{1}{1-z'}\right)[/itex] and substituting [itex]z' = \frac{-1}{z-1}[/itex]. This does not work in the region 0 < |z-1| < 1, unfortunately.

    Wolfram and the solutions manual did not give any Laurent series for this region either, which leads me to believe that such a series does not exist. But I don't see why, since the function appears to be analytic in the punctured disk, and Laurent's theorem should apply.
  2. jcsd
  3. Oct 20, 2014 #2
    I just realized I made a rookie mistake. The function is not analytic anywhere on the real line segment between 0 and 1, which means there is no punctured disk throughout which the function is analytic. The only region where the Laurent series exists is for |z-1| > 1.
    Sorry for this :P
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted