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    asder's Avatar
    asder Posts: 2, Reputation: 1
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    #1

    Nov 20, 2005, 04:10 AM
    Contour Integral
    is there anybody there who is interested in complex calculus?
    Is the theorem right?
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    When you integrate f on a closed loop, you may or may not get 0 as the total integral. It depends a lot on what f is, and what the loop is. If f has an anti-derivative on the loop, then the integral is guaranteed to be 0 by the fundamental theorem of calculus. If f is analytic on the loop and inside the loop, then Cauchy's theorem guarantees that the integral is 0. In all other cases it is very unlikely that the integral vanishes completely.
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    z=2exp([email protected])
    f(z)[email protected] then the antiderivative is exist is 1/z and by calculation the integral yields "2pi*i". But according to the theorem above shouldn't it be 0.
    Where am I wrong?
    CroCivic91's Avatar
    CroCivic91 Posts: 729, Reputation: 23
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    #2

    Nov 20, 2005, 04:04 PM
    I remember this vaguely from my university. I believe Log(z) is not defined on the whole set C. I believe there is a line going from (0, 0) under the angle "fi" (for different Log(z) functions or something) that Log(z) is not defined for points on that line. Am I correct? If true, and if your loop intersects that line, I believe that might be causing the problem. Then again, I remember that an integral is not affected if a function is not defined in a single point, but I don't know if that means anything. Is that any help?

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