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Math Help - Complex Contour Integrals

  1. #1
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    Complex Contour Integrals

    Hello,

    I have been working on complex integration in my own time and have managed to solve a few, but these three seriously have me stumped. I was wondering if anyone would be willing to help me understand these problems.

    a) \int_{\gamma } \frac{z}{(z+2)(z-1)}dz where \gamma is the circle of radius 4 about the origin, traversed twice in the clockwise direction.

    b) \int_{\gamma } \frac{2z^{2}-z+1}{(z-1)^2(z+1)}dz where \gamma proceeds around the boundary of the figure eight formed by two circles of radius 1 with centres 1 and 1 by starting at 0, going once counterclockwise around the right circle followed by going once counterclockwise around the left circle.

    c) \int_{\gamma } \frac{5z-2}{z(z-1)(z-3)}dz where \gamma is the circle of radius 2 about the origin.

    I thank you all in advance! Your help is very much appreciated.
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  2. #2
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    What tools are you allowed to use? Cauchy's integral theorem? Residue theorem?
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  3. #3
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    Yes I've read about those two. I'm just learning this on my own so all tools are helpful. Would those two be helpful in the given scenarios?
    Thanks
    Last edited by JoAdams5000; October 3rd 2010 at 06:17 AM.
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  4. #4
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    I was also wondering if trying to parameterize the curve would be an option as well, but I have a feeling that trying to solve using the basic definition will not produce the most efficient result. Any ideas?
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  5. #5
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    Hey, sorry for the delay.

    For your first post - yes they will be very useful here.

    What is absolutely essential here is that you draw the curves that you are integrating over, so that you know how to parameterize them. This is important not because you will use the parametrizations explicitly, but it will give you an idea of how to use the needed theorems.

    I'll do the first one for you - try to do the others in a similar way.

    Let \displaystyle I = \int_{\gamma } \frac{z}{(z+2)(z-1)}dz where \gamma is as mentioned.
    Since this is not a simple curve (it intersects itself) you will have to reparameterize it as a concatenation of two simple curves \gamma = \gamma_1 * \gamma_1 where \gamma_1 traverses |z|=4 once, clock-wise.

    By properties of concatenation of curves, you have \displaystyle I = \int_{\gamma } \frac{z}{(z+2)(z-1)}dz = \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz + \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \cdot \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz

    Now, using the residue theorem:
     \displaystyle \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \pi i \sum_{k} Res(f(z), z_k) where \displaystyle f(z) = \frac{z}{(z+2)(z-1)} = \frac{z}{z^2 + z -2} and z_k are the poles of f(z) inside \gamma_1 - z_1 = -2, \ z_2 = 1

    Now calculate the residues:
    \displaystyle Res(f(z), -2) = \frac{z}{2z + 1} |_{z = -2} = \frac{-2}{-3} = \frac{2}{3}
    \displaystyle Res(f(z), 1) = \frac{z}{2z + 1} |_{z = 1} = \frac{1}{3}

    And so \displaystyle \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \pi i \left( \frac{1}{3} + \frac{2}{3} \right) = 2 \pi i
    And finally \displaystyle I = 2 \cdot \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 4 \pi i

    If you are not familiar with all of this then you are probably expected to use Cauchy's integral theorem. If you need that I'll also show you how to solve this integral with that.
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  6. #6
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    Thanks!

    I was wondering if you don't mind could you also demonstrate how to do this using Cauchy's Integral theorem? I appreciate it!
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  7. #7
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    I tried solving this problem using CIT and I got these answers.

    a) -4ipi
    b) 4ipi
    c) -4ipi

    Has anyone else tried this?
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