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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
     
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