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Math Help - Help me fix this residue integration

  1. #1
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    Help me fix this residue integration

    \int_{0}^{2\pi} \frac{d\theta}{37 - 12cos\theta} = \oint_{C} \frac{dz/iz}{37 - 12(\frac{z}{2} + \frac{1}{2z})}
    = \oint_{C} \frac{dz/iz}{37 - 6z - \frac{6}{z}}
    = \oint_{C} \frac{dz}{i(37z - 6z^{2} - 6)}
    = \oint_{C} \frac{dz}{-i(6z^{2} - 37z + 6)}
    = \frac{-1}{i} \oint_{C} \frac{dz}{(6z - 1)(z - 6)}

    z_{1} = \frac{1}{6}, z_{2} = 6

    This is the part which I'm not really sure. I'm guessing the z_{1} is outside because it's less than unity, but I start having trouble when it comes to identifying more complicated poles. What is the right way?

    Res = \left|\frac{1}{6z - 1} \right|_{z = 6} = \frac{1}{35}
    \Rightarrow 2\pii(\frac{-1}{i})(\frac{1}{35}) = \frac{-2\pi}{35}

    The answer is not a negative, so it may be a careless mistake, but I've pored through the equations without spotting an error.

    PS: Is \\ the tag for line break? I can't seem to get it to work.
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  2. #2
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    You need to compute this:

    \displaystyle-\frac{1}{i}(2\pi i)\text{Res}[\frac{1}{(6z-1)(z-6)},z\to 1/6]=-\frac{2\pi}{6}\left(\frac{1}{z-6}\right)\Big|_{z\to 1/6}=-\frac{2\pi}{6}\left(\frac{1}{1/6-36/6}\right)=-\frac{\pi}{3}\left(-\frac{6}{35}\right)=\frac{2\pi}{35}.
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  3. #3
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    Why is z = 1/6 chosen?

    What about this one? Picking the other pole just changes the residue sign. The answer is 2pi/3.

    \int_{0}^{2\pi} \frac{d\theta}{5 - 4sin\theta}
    = \oint_{C} \frac{dz/iz}{5 - \frac{2z}{i} - \frac{2}{iz}}
    = - \oint_{C} \frac{dz}{2z^{2} - 5iz - 2}
    = - \oint_{C} \frac{dz}{(4z - 8i)(49 - 2i}
    z_{1} = 2i, z_{2} = \frac{1}{2}
    Res = \left|\frac{1}{4z - 2i} \right|_{z = 2i} = \frac{1}{6i}
    \Rightarrow 2\pi i(\frac{1}{6i})(-1) = \frac{-\pi}{3}
    Last edited by geft; November 9th 2010 at 12:41 AM.
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  4. #4
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    You pick z = 1/6, because that pole is inside the contour, and the other one is not. Only poles inside a contour have any effect whatsoever on the value of the integral. The implied contour over which you're integrating is the unit circle. So the correct poles at which to evaluate the residues are the ones inside that contour. Make sense?
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  5. #5
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    How do you check if the pole is inside the contour? In my previous question, you told me to plot the circle, but the function here seems to be quadratic.
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  6. #6
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    Do you know what the contour is? Whether the integrand is quadratic or not has nothing to do with the contour.

    The original problem was to integrate \int_{0}^{2\pi} \frac{d\theta}{37 - 12cos\theta}
    which you changed into the contour integral \oint_{C} \frac{dz/iz}{37 - 12(\frac{z}{2} + \frac{1}{2z})}. How did you do that?
    What was "z" in terms of " \theta". As \theta goes from 0 to 2\pi, how does z change?
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