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Math Help - Complex Line Integrals

  1. #1
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    Complex Line Integrals

    Here I go again...

    All of these line! integrals are on the circle |z|=2, with counterclockwise orientation.

    1. int(z^104)dz
    2. int(z*conjugate(z))d|z|
    3. int(conjugate(z))dz
    4. int(cos(z)/z^3)dz
    5. int(sin(z)/(z^2-((pi)/2)^2)dz

    Unfortunately I cannot use the residue theorem.

    For numbers 4 and 5 I have been trying to write the integrands as power series, but I am having trouble relating those to line integrals.

    For numbers 1, 2 and 3 I feel like I have to split them up into a real and imaginary parts and then do two integrals, but I am having trouble doing this.

    And throughout I am always wondering why the line integral is evaluated at a circle of radius 2, instead of the usual 1. I think some of the functions have singularities in that region, but I'm not sure what that does.
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  2. #2
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    Bump. I could really use a hint for any of these.
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  3. #3
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    Quote Originally Posted by robeuler View Post
    Here I go again...

    All of these line! integrals are on the circle |z|=2, with counterclockwise orientation.

    1. int(z^104)dz
    2. int(z*conjugate(z))d|z|
    3. int(conjugate(z))dz
    4. int(cos(z)/z^3)dz
    5. int(sin(z)/(z^2-((pi)/2)^2)dz

    Unfortunately I cannot use the residue theorem.

    For numbers 4 and 5 I have been trying to write the integrands as power series, but I am having trouble relating those to line integrals.

    For numbers 1, 2 and 3 I feel like I have to split them up into a real and imaginary parts and then do two integrals, but I am having trouble doing this.

    And throughout I am always wondering why the line integral is evaluated at a circle of radius 2, instead of the usual 1. I think some of the functions have singularities in that region, but I'm not sure what that does.
    You can parameterize the circle |z|=2 \iff z=2e^{2\pi i t} \implies dz=4\pi i e^{2\pi i t}dt

    So the first one becomes

    \int_{0}^{1}(2e^{2 \pi i t})^{104}(4 \pi i e^{2 \pi i t}dt)

    2^{106}\pi i \int_{0}^{1}e^{210 \pi i t}dt=

    2^{106}\pi i \left( \frac{1}{210 \pi i }\right)e^{210 \pi i t} \bigg |_{0}^{1}=0

    This should help with the rest good luck
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    You can parameterize the circle |z|=2 \iff z=2e^{2\pi i t} \implies dz=4\pi i e^{2\pi i t}dt

    So the first one becomes

    \int_{0}^{1}(2e^{2 \pi i t})^{104}(4 \pi i e^{2 \pi i t}dt)

    2^{106}\pi i \int_{0}^{1}e^{210 \pi i t}dt=

    2^{106}\pi i \left( \frac{1}{210 \pi i }\right)e^{210 \pi i t} \bigg |_{0}^{1}=0

    This should help with the rest good luck
    Will the parametrization of z conjugate just be negative the parametrization of z? Also: how do evaluate the absolute value of dz? Does the dt matter?

    Thank you so much for your help.
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