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Residues of ((log(z))^2)/(1+z^2)^2

  1. Mar 20, 2010 #1
    1. The problem statement, all variables and given/known data This is actually on Wikipedia, but it doesn't show how you actually calculate the residues.

    I want to calculate the residues for ((log(z))^2)/(1+z^2)^2. Wikipedia claims the sum of the residues is (-(pi/4) + (i*(pi)^2)/16 - (pi/4) - (i*(pi)^2)/16) = -pi/2

    2. Relevant equations


    3. The attempt at a solution The singularities are at i and -i. Multiplying first by (z-i)^2 and taking the limit z ---> i, I get (log(i)^2)/4 = (log(e^(i*(pi/2))^2)/4 = -(pi^2)/16.

    Then, multiplying by (z+i)^2 and taking the limit z ---> -i I get (log(e^-i*(pi/2))^2)/4 = -(pi^2)/16 again. Any help would be very much appreciated.
  2. jcsd
  3. Mar 20, 2010 #2


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    The poles are double poles. Look at http://en.wikipedia.org/wiki/Residue_(complex_analysis [Broken]) under "Limit formula for higher order poles". There's a derivative involved here as well as a limit.
    Last edited by a moderator: May 4, 2017
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