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Simple Cauchy's integral theorem problem

  1. Sep 3, 2007 #1
    Integrate f(z) counterclockwise around the unit circle indicating whether Cauchy's integral theorem applies, ( show details of your work).
    (A) [tex] z(t) = \cos t + \iota \sin t =\exp{\iota t} \mbox{ for } 0 \leq \ t \ \leq 2\pi \\[/tex]
    So that counterclockwise integration corresponds to an increase of t from 0 to 2[tex] \pi [/tex]
    (B) [tex] \frac{dz(t)}{dt} = \iota \exp{\iota t} \\[/tex]
    (C) f[z(t)] = x(t) = cos(t). Therefore
    (D) [tex] \oint_C \Re z dz = \int_0^{2\pi}\cost \iota\exp{\iota t} \\ [/tex]
    Integrating by parts I get [tex] \iota \int_0^{2\pi} \cos t \exp{\iota t} = \frac{\exp{2\pi\iota}-1}{2} \\ [/tex]. I could be wrong about x(t)=cos(t). Cauchy's integral theorem does not apply as f(z) = Re(z) = cos(t) is not analytic.
     
    Last edited: Sep 4, 2007
  2. jcsd
  3. Sep 3, 2007 #2

    HallsofIvy

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    Are you saying that you are given that f(z)= Re(z)? If so, then you are correct.
     
  4. Sep 4, 2007 #3
    Yes I am saying that f(z)=Re(z), and the answer in the 'book is [tex] \pi\iota [/tex]. Thanks for the reply. In fact [tex] \exp{2\pi\iota} = 1 [/tex], hence the integral is zero.
     
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