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Math Help - how to find these contour integrals?

  1. #1
    Senior Member
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    how to find these contour integrals?

     \int_{C}{z \ \ bar \ \ dz} , C is along the circle |z|=1
    and

    another case, C along the circle |z-1|=1.

     \int_{C}{\sqrt{z} dz} . C is |z|=1, and |z-1|=1


    i think the first part is the same for both cases which is 2pi*i.

    and the second part is 0 for both cases by Cauchy Integral Theorem.

    But i am not sure if that can be applied. Because sqrt z is only analytic on C\ { x axis } . Therefore, out circles are not connected?
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  2. #2
    MHF Contributor

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    No deep theorems required here. Just parameterize the curve and integrate.

    On the circle |z|= 1, z= e^{i\theta} so z*= e^{-i\theta} and dz= ie^{i\theta}d\theta. Of course to cover the entire circle \theta must go from 0 to 2\pi. The integral is \int_0^{2\pi}(e^{-i\theta})(ie^{i\theta}d\theta= i\int_0^{2\pi}d\theta

    On the circle |z-1|= 1, z= 1+ e^{i\theta} so z*= 1+ e^{-i\theta} and dz= -ie^{-i\theta}.

    If z= e^{i\theta} \sqrt{z}= z^{1/2}= e^{i\theta/2}.

    On z= 1+ e^{i\theta} \sqrt{z}= \sqrt{1+ e^{i\theta}}.
    Last edited by mr fantastic; February 20th 2009 at 08:59 PM. Reason: Fixed all the latex tags
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  3. #3
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    yeah

    yeah. i did those steps too. but then i found the last case is a bit difficult to calculate.
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