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Cauchy Integral

  1. May 5, 2009 #1
    A. 1. The problem statement, all variables and given/known data

    f is analytic inside and on a simple closed contour C and z0 isn't on C. Show:

    [tex]\int f'(z)dz/ (z- zo)[/tex] = [tex]\int f(z)dz/ (z- zo)^2 [/tex]


    3. The attempt at a solution

    [tex]\int f'(z)dz/ (z- zo)[/tex] = 2[tex]\pi[/tex]i f'(zo)


    [tex]\int f(z)dz/ (z- zo)^2 [/tex] = [tex]\int [f(z)dz/(z- zo)]/(z-zo) [/tex] = 2[tex]\pi[/tex]i [f(zo)/(zo-zo)] (I got stuck here!?)


    1. The problem statement, all variables and given/known data

    B. C is the unit circle z = e^(i[tex]\theta[/tex]). For any real constant a:

    [tex]\int ((e^a)^z) /z) dz[/tex] = 2[tex]\pi[/tex] i (1)

    Derive the following (the integral is from 0 to [tex]\pi[/tex]):

    [tex]\int [(e^a)^(cos\theta)) * cos (a sin\theta)] d\theta [/tex] = [tex]\pi[/tex] (2)

    3. The attempt at a solution

    First, I translated (1) in terms of [tex]\theta[/tex] and got a very long term. Then I substituted the term to (2). But what I got was a huge and complicated mess. Did I miss out something obvious? Thanks!
     
  2. jcsd
  3. May 5, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    For the first one you have a second order pole at z=z0. How do you find the residue at a second order pole. It's not the way you are trying to do it. For the second, yes, write the integral 1) as a an integral dtheta. Then take the imaginary part and compare with 1). Then argue that the integral from 0 to pi is 1/2 the integral from 0 to 2pi by showing it's symmetrical around theta=pi.
     
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