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Math Help - Cauchy's Integral Formula

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    Cauchy's Integral Formula

    Hey guys, I was just wondering how I would apply Cauchy's Integral Formula to something like this:



    I would be able to do it if there wasn't a constant of 1/2i there but that constant makes me uncertain.
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    Quote Originally Posted by Alexrey View Post
    Hey guys, I was just wondering how I would apply Cauchy's Integral Formula to something like this:



    I would be able to do it if there wasn't a constant of 1/2i there but that constant makes me uncertain.
    I have a question for you. What do you mean by this notation

    \gamma(0,1)

    Is this a circle of radius 1 centered at the origin?

    If so the Cauchy integral formula does not apply because the integrand is discontinous at z=1.
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    My bad, yes, gamma (0,1) is an open disk of radius 1 centered at the origin. Surely you could apply Cauchy's integral formula:


    with f(w) = z and w-a = z-1, therefore since f(w) is holomorphic in gamma(0, 1) we can apply Cauchy's Integral Formula, with a=1?
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    Quote Originally Posted by Alexrey View Post
    My bad, yes, gamma (0,1) is an open disk of radius 1 centered at the origin. Surely you could apply Cauchy's integral formula:


    with f(w) = z and w-a = z-1, therefore since f(w) is holomorphic in gamma(0, 1) we can apply Cauchy's Integral Formula, with a=1?
    a must be an interior point of

    \gamma

    Cauchy's integral formula - Wikipedia, the free encyclopedia

    The problem is that there is a pole on the countour. Also something else is fishy you sure that it is an open disk?
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    Ah I see, I screwed that one up. Regarding gamma being an open disk, I thought that it was open since I read in my notes somewhere that the interior of gamma was open and therefore presumed that the whole disk was open as well. What can I do if the singularity is ON the contour? Will residues work (I haven't gone over this section yet).

    Okay lets rephrase the question to gamma(0,2). I would be able to use Cauchy's Integral Formula then right, since a is inside the contour.
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    Quote Originally Posted by Alexrey View Post
    Ah I see, I screwed that one up. Regarding gamma being an open disk, I thought that it was open since I read in my notes somewhere that the interior of gamma was open and therefore presumed that the whole disk was open as well. What can I do if the singularity is ON the contour? Will residues work (I haven't gone over this section yet).

    Okay lets rephrase the question to gamma(0,2). I would be able to use Cauchy's Integral Formula then right, since a is inside the contour.
    Yes, this would give

    f(1)=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{f(z)}{z-1}dz, \quad f(z)=z so we get

    1=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{z}{z-1}dz \iff \int_{\gamma (0,2)}\frac{z}{z-1}dz=2\pi i

    So finaly

    \frac{1}{2i}\int_{\gamma (0,2)}\frac{z}{z-1}dz=\frac{1}{2i}(2\pi i)=\pi
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    Awesome, thanks so much.
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