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Math Help - Contour Integration

  1. #1
    Super Member Aryth's Avatar
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    Contour Integration

    Here's the problem:

    The function f(z) is holomorphic everywhere on a closed contour \Gamma, and also within \Gamma except at a finite set of points where f has poles.

    (Recall that a pole of order n at z = \alpha occurs where f(z) is of the form \frac{h(z)}{(z - \alpha)^n}, where h(z) is regular at \alpha).

    Show that:

    \oint_r f(z)dz = 2\pi i*\{\text{sum of the residues at these poles}\},

    where the residue at the pole \alpha is \frac{h^{(n-1)}(\alpha)}{(n-1)!}.

    Any and all help is appreciated.
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  2. #2
    Super Member wingless's Avatar
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    f(z) is holomorphic everywhere except its poles. And the contour \Gamma doesn't have any singularities / poles on it.

    Now, think of the Laurent Series of f(z) on the point a.

    f(z) = \sum_{n= -\infty}^{\infty} a_n (z-a)^n

    f(z) = ... +  a_{-2} (z-a)^{-2} + a_{-1} (z-a)^{-1} + a_0 (z-a)^{0} + a_1 (z-a)^{1} + ...

    Integrate both sides by the contour \Gamma,

    \oint_{\Gamma} f(z)~dz =  ... ~+~  \oint_{\Gamma} a_{-2} (z-a)^{-2}~dz ~+~ \color{red}\oint_{\Gamma} a_{-1} (z-a)^{-1}~dz \color{black}~+ ~\oint_{\Gamma} a_0 (z-a)^{0}~dz ~+ ~...

    Cauchy's integral theorem tell us that \oint_{\Gamma} (z-a)^n~dz = 0 for any n except -1.

    So it means all integrals except the red one will vanish (because they are 0).

    Now we have,
    \oint_{\Gamma} f(z)~dz =a_{-1} \oint_{\Gamma}  (z-a)^{-1}~dz

    Let \Gamma be a circle of radius 1 with center a in the complex plane.

    z = e^{it}+a

    This will give,

    \oint_{\Gamma} f(z)~dz =a_{-1} 2\pi i

    The -1st coefficient of Laurent Series of f(z) at a point is called the Residue of that function at that point.

    So we can generalize this for contours with multiple poles,

    \oint_{\Gamma} f(z)~dz =  2\pi i \cdot \sum \text{Res}_{f(z)=...}
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  3. #3
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    Good job wingless, proud of you.

    It depends how formal you want to see a proof. If you are okay with a simple geometric proof by induction you can read this. It seems to me you are satisfied with that. I can provide a more formal proof if you want to see it?
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  4. #4
    Super Member Aryth's Avatar
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    Oh yeah. I'd love to see the rigorous proof. The proof wingless provided was sufficient but I'm never against extras.
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