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Contour Integrals

  1. Apr 23, 2006 #1
    Find [tex] \int_{C} 3(z-i)^2 dz [/tex] where C is the circle |z-i|=4 traversed once clockwise

    well i know it is zero but i just want to prove it.. kind of

    so we can parametrize [tex] z(t) = i + 4e^{it}, \ 0\leq t \leq 2 \pi [/tex]

    [tex] \int_{C} 3(z-i)^2 dz = \int_{0}^{2\pi} 3(i + 4e^{it}-i)^2 (4ie^{it}) dt [/tex]

    is the setup good?

    Compute [itex] \int_{\Gamma} \overline{z} dz [/tex] where Gamma is the circle |z|=2 tranversed once counterclockwise

    [tex] z(t) = 2e^{it} [/tex]
    [tex] \int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt [/tex]

    is this correct??
    Thank you for the help!
    Last edited: Apr 23, 2006
  2. jcsd
  3. Apr 23, 2006 #2


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    Homework Helper

    This parameterization is going clockwise, not counterclockwise. Otherwise it looks fine.

    Check what [tex]\overline{2e^{it}}[/tex] is again.
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