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Math Help - Integral question

  1. #1
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    Integral question

    The question
    Evaluate \oint_{|z| = 4} \frac{z^3 \ dz}{(z - 3)^3(z - 2)^2} using the Cauchy Integral Formula, where the circle is traversed once anti-clockwise.

    My attempt
    z = 3, z = 2 are both singularities in the contour, so:

    Rewrite the integrand such that one of the singularities is in the denominator:
    \frac{\frac{z^3}{(z - 3)^3}}{(z - 2)^2}

    Thus by CIF, f(z) = \frac{z^3}{(z - 3)^3}
    This singularity is of pole 2, so we must use \frac{2\pi i}{1!} f^{1}(2):
    2\pi i (\frac{3z^3(z - 3)^2) - 3z^2(z - 3)^3}{(z - 3)^6})
    = 72\pi i

    Now for the other singularity:
    \frac{\frac{z^3}{(z - 2)^2}}{(z - 3)^3}

    Thus by CIF, f(z) = \frac{z^3}{(z - 2)^2}
    This singularity is of pole 3, so we must use \frac{2\pi i}{2!} f^{2}(3):
    \pi i(\frac{6z^3-6z^2(z - 2) - 6z^2(z - 2) + 6z(z - 2)^2}{(z - 2)^4})
    = 72\pi i

    Add them together, and we get 144\pi i. However, the answer is actually 0. I think I'm supposed to be getting -72\pi i for the second one, but I can't work out where I've gone wrong. :/

    Any help would be greatly appreciated!
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  2. #2
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    Re: Integral question

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  3. #3
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    Re: Integral question

    My lecturers notes say that using partial fractions is an alternative method to the one I've done. But looking now, he was using singularities of pole 1, so perhaps it's different?

    Hmm, actually, just realised that my functions are not analytic in the contour, which breaks it. >_<
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Integral question

    The procedure has been described in...

    http://www.mathhelpforum.com/math-he...la-183122.html

    Here inside the contour f(z) has a second order pole in z=2 and a third order pole in z=3, so that what You have to do is compute the residues...

    r_{z=2} = \lim_{z \rightarrow 2} \frac{d}{d z} \{(z-2)^{2}\ f(z)\}

    r_{z=3} = \lim_{z \rightarrow 3} \frac{1}{2}\ \frac{d^{2}}{d z^{2}} \{(z-3)^{3}\ f(z)\}

    ... and the integral is...

    \int_{c} f(z)\ dz = 2 \pi i (r_{z=2} + r_{z=3})

    Kind regards

    \chi \sigma
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  5. #5
    Aki
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    Re: Integral question

    Quote Originally Posted by Glitch View Post
    This singularity is of pole 2, so we must use \frac{2\pi i}{1!} f^{1}(2):
    2\pi i (\frac{3z^3(z - 3)^2) - 3z^2(z - 3)^3}{(z - 3)^6})
    = 72\pi i
    There is an error in the derivation.
    The correct derivative is
    f'(z)=\frac{3z^2 (z-3)^3 - z^3 \cdot 3(z-3)^2}{(z-3)^6}=-\frac{9z^2}{(z-3)^4}.
    So, f'(2)=-36.

    The derivation formula for f(z)/g(z)is
    \frac{d}{dz}\frac{f(z)}{g(z)}=\frac{f'(z)g(z)-f(z)g'(z)}{g(z)^2}.
    In your equation, the numerator is replaced by f(z)g'(z)-f'(z)g(z).
    It is interesting that you got the correct result for the second integral.
    Does double use of the wrong derivation formula derive correct result ?

    Sorry for my poor English.
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  6. #6
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    Re: Integral question

    Huh, I did do it the wrong way around. That's quite worrying actually. :/
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