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Math Help - Independence of Path

  1. #1
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    Independence of Path

    i am a little confused by the idea of "independence of path" in complex analysis. i understand that certain functions, when integrated over simple paths with the same endpoints a and b yield the same value. however i do not understand how to determine whether the function is independent of the path (without calculating over arbitrary simple paths).

    can anyone help enlighten me? thanks in advance!
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  2. #2
    MHF Contributor chisigma's Avatar
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    A fundamental theorem due to the French mathematician Augustin Cauchy extablishes that if f(z) is analytic in a region inside a closed path \gamma is...

    \displaystyle \int_{\gamma} f(z)\ dz =0 (1)

    Now if f(z) is analytic in the whole complex plane [i.e. is an entire function...] the (1) is valid independently from \gamma , and that means that, if we devide \gamma in two parts, one coming from z_{1} to z_{2} and the other coming back from z_{2} to z_{1} is...

    \displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz + \int_{z_{2}}^{z_{1}} f(z)\ dz =0 (2)

    ... and that means that \displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz depends only from z_{1} and z_{2} and doesn't depend from the path connecting z_{1} and z_{2}...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    A fundamental theorem due to the French mathematician Augustin Cauchy extablishes that if f(z) is analytic in a region inside a closed path \gamma is...

    \displaystyle \int_{\gamma} f(z)\ dz =0 (1)

    Now if f(z) is analytic in the whole complex plane [i.e. is an entire function...] the (1) is valid independently from \gamma , and that means that, if we devide \gamma in two parts, one coming from z_{1} to z_{2} and the other coming back from z_{2} to z_{1} is...

    \displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz + \int_{z_{2}}^{z_{1}} f(z)\ dz =0 (2)

    ... and that means that \displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz depends only from z_{1} and z_{2} and doesn't depend from the path connecting z_{1} and z_{2}...

    Kind regards

    \chi \sigma
    It doesn't need to be entire. Only holomorphic on a path connected subspace.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Drexel28 View Post
    It doesn't need to be entire. Only holomorphic on a path connected subspace.
    In order to avoid misunderstanding a simple example. Susppose that is f(z)=\frac{1}{z}, z_{1}= -i\ \sqrt{2}, z_{2}= i\ \sqrt{2} and \gamma is the 'red circle' in the figure...



    By the residue theorem is...

    \displaystyle \int_{\gamma} \frac{dz}{z} = 2\ \pi\ i (1)

    ... and that means that \displaystyle \int_{z_{1}}^{z_{2}} \frac{dz}{z} will be \pi i if the point z=0 is on the left of the half circle connecting z_{1} to z_{2} and - \pi\ i if it is on the right...

    Kind regards

    \chi \sigma
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  5. #5
    Senior Member roninpro's Avatar
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    I think that Drexel is partially correct. You don't need the function to be entire. However, it must be defined on an open simply-connected domain.

    See Cauchy's integral theorem - Wikipedia, the free encyclopedia.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roninpro View Post
    I think that Drexel is partially correct. You don't need the function to be entire. However, it must be defined on an open simply-connected domain.

    See Cauchy's integral theorem - Wikipedia, the free encyclopedia.
    Ah crap. I always say the wrong hypothesis. You'd think someone who's interested in alg. top. I'd remember haha
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