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Cauchy's integral formula

  1. Oct 16, 2014 #1
    1. The problem statement, all variables and given/known data

    This problem is from Mary L. Boas - Mathematical Methods in the Physical Sciences, Chapter 14, section 3, problem 23.

    [itex]\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}}[/itex]

    2. Relevant equations

    Cauchy's integral formula

    3. The attempt at a solution

    First isolate the singularity:

    [itex]\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}}[/itex] (This should read something like (e^3z/1^4)/(z-ln2)^4) but I can't seem to do it in latex...)

    Then let g(z) be:

    [itex]g(z) = e^{3z}[/itex]

    Since there's a fourth power in the singularity, g(z) must be derived three times:

    [itex]g'(z) = 3e^{3z}[/itex]
    [itex]g''(z) = 9e^{3z}[/itex]
    [itex]g'''(z) = 27e^{3z}[/itex]

    To solve the integral, we multiply g'''(z) by:

    [itex] (2 \pi i)(27e^{(3)(ln2)}) = 432 \pi i[/itex]

    But the book states that the answer is [itex]72 \pi i[/itex], which is exactly 6 times less than my answer. Where did I go wrong?
    Last edited: Oct 16, 2014
  2. jcsd
  3. Oct 16, 2014 #2


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    You mean differentiated?

    To calculate the residue, you have to divide by 1/(n-1)! where n is the order of the pole. It's part of the limit formula for calculating the residue.
  4. Oct 17, 2014 #3
    Thank you very much for your answer.
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