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[Complex Analysis] Help with Cauchy Integral Problem

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integral,

    [tex]I = \int_{0}^{2\pi} \frac{d \theta}{(1-2acos \theta + a^2)^2}, \ 0 < a < 1 [/tex]

    For such, transform the integral above into a complex integral of the form ∫Rₐ(z)dz, where Rₐ(z) is a rational function of z. This will be obtained through the substitution z = e^iθ. Therefore, solve the complex integral through Cauchy's Integral Formula for the first derivative of an analytic function.

    2. Relevant equations

    Cauchy's Integral Formula

    [tex] f^{(n)} (z_0) = \frac{n!}{2 \pi i} \oint_{\gamma} \frac{f(z)}{(z - z_0)^{(n+1)}}dz [/tex]

    3. The attempt at a solution

    Writting dθ in terms of dz

    [tex]z = e^{i \theta} [/tex]

    [tex]dz = ie^{i \theta} d\theta [/tex]

    [tex]\frac{1}{iz}dz = d\theta [/tex]

    Writting

    [tex] \frac{1}{(1-2acos \theta + a^2)^2} [/tex]

    in terms of z, known that

    [tex]\frac{1}{2}\left( z + \frac{1}{z}\right) = cos \theta [/tex]

    Therefore

    [tex] \frac{1}{(1-a\left( z + \frac{1}{z}\right) + a^2)^2} [/tex]

    Thus

    [tex]I = \oint_{\gamma} \frac{dz}{iz(1-a\left( z + \frac{1}{z}\right) + a^2)^2} [/tex]

    Where do I go from here? Any clues on how to express the denominator in terms of (z-z0)² ?
     
    Last edited: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2
    Try multiplying by z/z. You will have a factor of ##z^2## in the denominator. Thus,

    ##z^2 \cdot (1-a(z+z^{-1})+a^2)^2 = [z \cdot (1-a(z+z^{-1})+a^2)]^2##

    It should be apparent from there!

    Edit: When you finally do this, it should become apparent why ##0 < a < 1## is required.
     
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