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Complex Integration over a Closed Curve

  1. Feb 12, 2013 #1
    (a) Suppose [itex]\kappa[/itex] is a clockwise circle of radius [itex]R[/itex] centered at a complex number [itex]\mathcal{z}[/itex]0. Evaluate: [tex]K_m := \oint_{\kappa}{dz(z-z_0)^m} [/tex]
    for any integer [itex] m = 0, \pm{1},\pm{2}, ,... [/itex]Show that

    [itex] K_m = -2\pi i[/itex] if [itex] m = -2;[/itex] else :[itex] K_m = 0 [/itex] if [itex] m = 0, \pm{1}, \pm{2}, \pm{3},....[/itex]

    Note the minus sign here: [itex]\kappa[/itex] is clockwise.



    I am not allowed to use or assume the validity of the residue theorem, but I can use Cauchy's integral theorem without proof.

    I was trying to parameterize [itex]K_m[/itex] using

    [itex]z(\tau) = c + re^{i\tau} , \tau \in [a,b][/itex] with [itex] a \equiv {\theta_a}[/itex] and [itex] b \equiv \theta_b,[/itex] if [itex] \theta_a < \theta_b, [/itex]

    But I'm just stuck on how to set this up at this point. Any ideas?
     
  2. jcsd
  3. Feb 12, 2013 #2

    haruspex

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    That's a good start. And c = ? What does it give you for dz and (z-z0)?
     
  4. Feb 12, 2013 #3
    "c" is the complex number around which Km would be centered in the form of [itex]re^{i\tau}[/itex]

    So could I assume a clockwise circle centered at 0, making both c and z0 = 0, which makes would make my equation: [tex] \oint_\kappa z^m dz [/tex] ???
     
  5. Feb 12, 2013 #4

    haruspex

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    You don't need to make them both 0. z0 is a given, so you can't decide what it is. But you can choose c. You need to use the form [itex]z\left(\tau\right)=c+re^{i\tau}[/itex] to substitute for z in the integral. This is supposed to represent the curve of interest (circle of radius R centered at z0) as tau varies and c remains constant. What value of c will give you that?
    What does that substitution give you for dz and for z-z0?
     
  6. Feb 12, 2013 #5
    C = 0

    Tau in the exponent is negative because of the clockwise direction.

    So [itex] z(\tau) = re^{-i\tau} [/itex], [itex]\tau \in [0, 2\pi] [/itex]

    I would plug this in for Z and its respective derivative to replace dz with d[itex]\tau[/itex]
    But what would I do with z_0?
     
    Last edited: Feb 12, 2013
  7. Feb 12, 2013 #6

    haruspex

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    The integration path is a circle radius R, centred at z0. z(t) = c+reit describes a circle radius r, centred where?
     
  8. Feb 12, 2013 #7
    Oh.. z0 is c, which is just some arbitrary constant that I can leave in the equation, where my [itex] (z - z_0)^m dz [/itex] becomes [itex] (re^{i\tau} - c)^m rie^{i\tau} d\tau [/itex]
     
  9. Feb 12, 2013 #8

    haruspex

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    Yes.
    No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?
     
  10. Feb 12, 2013 #9
    **ignore-wrong post**
     
    Last edited: Feb 12, 2013
  11. Feb 12, 2013 #10
    [tex] (c + re^{-i\tau} - c)^m (-rie^{-i\tau}) d\tau [/tex]
    [tex] = (re^{-i\tau})^m (-rie^{-i\tau}) d\tau [/tex]

    How's that?


    where: [itex] z - z_0 \equiv c + re^{-i\tau} - c [/itex]
     
  12. Feb 12, 2013 #11

    haruspex

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    Yes. Now simplify and integrate.
     
  13. Feb 12, 2013 #12
    So here's my integration, which I'm pretty sure is correct. I just need help with one more thing.
    [tex] (re^{-i\tau})^m(-rie^{-i\tau})d\tau = -ir^{m+1}e^{-(m+1)i\tau} d\tau [/tex]
    [tex] \int_{0}^{2\pi} -ir^{m+1}e^{-(m+1)i\tau} d\tau= \frac{r^{m+1}e^{-(m+1)i\tau}}{m+1} |_{0}^{2\pi}[/tex]

    And so for all m [itex]\neq[/itex] -1, this result is always 0. Now, how do I prove the case [itex] K_m = -2\pi i[/itex] if m = -1 ?
     
    Last edited: Feb 13, 2013
  14. Feb 13, 2013 #13

    haruspex

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    Go back to the last point before you did the integration. The integration step you performed, using the usual rule for integrating xm.dx, is only valid when m is not -1. So put m=-1 there and see how you would integrate what results.
     
  15. Feb 13, 2013 #14
    ahhh, I see I see. Ok thank you for all your help hauspex! This one is done.
    sorry it took so long for that one part to click. it's been a long day.


    I might need some more help tomorrow with another :D
     
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