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Problem with cauchy integral

  1. Sep 30, 2007 #1
    I dont understand the cauchy theorem on complex analysis.
    I have this problem and I would like to have some help for it.

    The question is:
    Use the Cauchy Integral Theorem to prove that:

    It is told to have a closed surface . but here it is not specified. So I have nothing to offer as a try but I am still searching:uhh:

    Thank you for your help.
  2. jcsd
  3. Sep 30, 2007 #2


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    Well, my first thought was to look up the "Cauchy Integral Formula"! I didn't actually look them up but If I remember correctly, it is
    [tex] f^(n)(x)= \int \frac{dz}{(z-z_0)^{n+1}}[/tex]
    You might want to "complete the square" to get the deominator in that form.
  4. Sep 30, 2007 #3


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    The contour you want is along the positive real axis and then close with a counterclockwise circle of radius R is the upper half plane and let R->infinity. The integrand can be factored into 1/((z-r1)(z-r2)) where r1 and r2 are the two roots of the quadratic. Only one is in the upper half plane. Apply the Cauchy Integral formula to that. This tells you what the value of the contour is. Now show you can ignore the contribution from the circular part as R->infinity. Since the value of f(z) there is ~1/R^2 and the length of the circular arc is ~R. This is an application of the Cauchy integral formula called the residue theorem.
  5. Oct 1, 2007 #4
    Thank you both for the insight.
    I will try to work it out.
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