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Complex integration problem

  1. Jun 22, 2007 #1

    malawi_glenn

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

    evaluate

    [tex] \int_{c} | z - 1 | |dz| [/tex]

    where c is the positive oriented unit circle.


    2. Relevant equations



    3. The attempt at a solution

    [tex] | z - 1 | = \left[ ( z-1)( \overline{z} - 1 ) \right] ^{1/2} = \left[ |z|^{2} - z - \overline{z} +1 \right] ^{1/2} [/tex]

    [tex] c : z(t) = e^{it} ; 0 \leqslant t \leqslant 2\pi [/tex]

    [tex] |dz| = dt[/tex]

    [tex] |z| = 1 [/tex]

    [tex] \int_{c} | z - 1 | |dz| = \int_{0} ^{2 \pi} (2 - e^{it} - e^{-it}) ^{1/2} dt[/tex]

    [tex] \int_{0} ^{2 \pi} \sqrt{2-2 \cos t } dt [/tex]

    Is this right so far? What to do next? :S
     
    Last edited: Jun 22, 2007
  2. jcsd
  3. Jun 22, 2007 #2

    D H

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    Use some appropriate half-angle formula.
     
  4. Jun 22, 2007 #3

    malawi_glenn

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    you mean I should express 2cos(t) as something else?
     
  5. Jun 22, 2007 #4

    mjsd

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    try [tex]2 \sin^2 (x) = 1 - \cos (2x)[/tex]
     
  6. Jun 22, 2007 #5

    malawi_glenn

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    aha ok, thanx a lot! cheers
     
  7. Jun 22, 2007 #6

    D H

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    Yep. That's the half-angle formula I was talking about.

    [tex]\sin\left(\frac x 2\right) = \sqrt{1-\cos 2x}[/tex]
     
  8. Jun 22, 2007 #7

    malawi_glenn

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    got the right answer now, thanx alot !
     
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