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If f is analytic on the closed disc

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    If f is analytic on the closed disc, show that for r<1 we have

    [itex]f(re^{i\phi}) = \frac{1}{2\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{1-re^{i(\phi - \theta)}}d\theta[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I tried using cauchy integral formula and end up with

    [itex]f(re^{i\phi}) = \frac{1}{2i\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta} - re^{i(\phi)}}d\theta[/itex]

    i factor the denominator and get

    [itex]f(re^{i\phi}) = \frac{1}{2i\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}(1-re^{i(\phi-\theta)})}d\theta[/itex]

    Don't really know where to go from here.
     
  2. jcsd
  3. Jun 8, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Given ##z = e^{i \theta}##, did you remember to compute the transformation of ##dz##?
     
  4. Jun 8, 2014 #3
    Of course! Thanks a bunch man!
     
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