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Math Help - Integration along curve

  1. #1
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    Integration along curve

    This kind of problem is new to me, and I am having a hard time to approach it it. Can someone help please?
    1) Evaluate the integrals \int_\gamma z^n dz for all integers n, where \gamma is any circle centered at the origin with counterclockwise orientation.
    2) Same question but with \gamma being any circle not containing the origin.
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    This kind of problem is new to me, and I am having a hard time to approach it it. Can someone help please?
    1) Evaluate the integrals \int_\gamma z^n dz for all integers n, where \gamma is any circle centered at the origin with counterclockwise orientation.
    2) Same question but with \gamma being any circle not containing the origin.
    Case 1: n > 0.

    You can appeal to the Cauchy-Goursat Theorem which says that if f(z) is analytic inside and on a closed contour C then \oint_C f(z) \, dz = 0. Alternatively you can easily modify the argument given in case 2: n < 0.


    Case 2: n < 0.

    Let C_{\rho}: \, |z| = \rho, that is, the circle with radius \rho and centre at z = 0. On C_{\rho}:

    z = \rho e^{i \theta}, ~ 0 \leq \theta < 2 \pi.

    dz = i \, \rho e^{i \theta} \, d \theta.

    Then \oint_{C_{\rho}} \frac{1}{z^m} \, dz = \int_{0}^{2 \pi} \frac{i \, \rho e^{i \theta}}{\rho^m e^{i m \theta}} \, d \theta = \frac{i}{\rho^{m-1}} \int_{0}^{2 \pi} e^{i (1 - m) \theta} \, d \theta.

    m = 1 (that is, n = -1): \oint_{C_{\rho}} \frac{1}{z} \, dz = 2 \pi i.

    m \neq 1: \oint_{C_{\rho}} \frac{1}{z^m} \, dz = \frac{1}{\rho^{m-1}(1 - m)} \left[ e^{i(1 - m) \theta} \right]_{0}^{2 \pi} = \frac{1}{\rho^{m-1}(1 - m)} \left( e^{2 \pi i (1 - m)} - 1 \right) = \frac{1}{\rho^{m-1}(1 - m)} (1 - 1) = 0.


    Case 3: n = 0.

    Left for you to do.


    These results readily follow from Cauchy's Integral Theorem (which you might not have met yet).


    It follows from the Cauchy-Goursat Theorem that when the circle does not contain the origin, the integral will equal zero. Alternatively, the case 2 argument is readily modified.
    Last edited by mr fantastic; September 14th 2008 at 03:53 AM. Reason: Small formatting tweak
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    You can also do this without Cauchy's theorem. You just need to find an anti-derivative which exists throught the entire disk. That can be done since we stay away from the origin, thus, \tfrac{z^{n+1}}{n+1} is the anti-derivative for n\not = -1. If n=-1 then it is possible to define \log z in an analytic way so that it does hit the origin and is therefore anaytic on the disk.
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    We're not supposed to use the Cauchy-Goursat's Theorem on this problem because we haven't learned it yet. So when the circle doesn't contain the origin, we still have to do some work to show that the integral equals 0.
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  5. #5
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    Quote Originally Posted by namelessguy View Post
    We're not supposed to use the Cauchy-Goursat's Theorem on this problem because we haven't learned it yet. So when the circle doesn't contain the origin, we still have to do some work to show that the integral equals 0.
    You can do it by finding an anti-derivative.
    That is my above post.
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