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Math Help - Complex Integration

  1. #1
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    Complex Integration

    I'm stuck on a question of complex integration; there doesn't seem to be any way to integrate the function, so I imagine there must be a 'trick' somewhere, but I can't work it out Here's the problem:

    Compute
    \oint \frac{log(z)}{z^2 + 9}dz
    around |z-4i|=3.

    I parametrized the path as
    \gamma (t) = 4i + 3e^{it},\ 0 \leq t \leq 2\pi
    but then I end up trying to integrate
    \int_0^{2\pi} \frac{log(4i+3e^{it})3ie^{it}}{(4i+3e^{it})^2 + 9}dt
    which I can't work out. Thanks for any help
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  2. #2
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    Why not use the residue theorem?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    Why not use the residue theorem?
    I thought there have to be poles inside the contour. f(z) has poles at z=\pm 3, which aren't inside \gamma (t).
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  4. #4
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    It actually has poles at z= \pm 3i

    :P

    EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    It actually has poles at z= \pm 3i

    :P

    EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.
    Damn.. Such a stupid mistake Thanks for pointing it out!
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  6. #6
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    Alright, so how do I go about calculating the residue? Don't I still need to integrate f(x) to work out Res(f, z_1) where z_1 = 3i?
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  7. #7
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    No.
    Since z=3i is a simple pole, you know that

    Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    No.
    Since z=3i is a simple pole, you know that

    Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)
    How do we know that z = 3i is a simple pole?
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  9. #9
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    How do we know that z = 3i is a simple pole?
    Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.
    Man, I wish the definition I have was that simple... Why can't my lecturer's explain it like that?! Thanks everyone
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  11. #11
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    Well, you gotta make sure the numerator doesn't also go to zero or do something else equally strange at the same location. The precision necessary to correctly define a simple pole means you're probably going to be doing something similar to what your lecturers have. However, this is an intuitive way to think of poles, and it works in quite a few cases.
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  12. #12
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    In a slightly related question, compute
    \oint \frac{\sin (2z)}{z^{15}}
    around |z| = 2.

    There's obviously a pole at z=0 but is that a simple pole, as the numerator will be zero at that point...
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  13. #13
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    The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?
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  14. #14
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    Quote Originally Posted by Ackbeet View Post
    The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?
    Cool, that makes sense. So I can work out the residue at z=0, by using
    b_{1} = \frac{1}{2\pi i} \int_\gamma f(\zeta )d\zeta
    right? I did that and after a bit of manipulation, arrived at
    \frac{1}{2^{15} \pi } \int_{0}^{2\pi }\frac{\sin{4e^{it}}}{e^{14it}}
    which I can't integrate. I've got no idea how to do it..
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  15. #15
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    I would just evaluate the residue using this formula.
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