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Math Help - Cauchy Theorem Integration Part 2

  1. #1
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    Cauchy Theorem Integration Part 2



    My book seems to do something like this here. My only question is that if we integrate with respect to theta...will it still work?

    The reason I chose to do it this way is precisely because it is so obviously clear that this is 2 cos \theta. But otherwise, it doesn't make a lot of sense to me this way.

    Thanks for your help, in advance.
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  2. #2
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    After more thinking, I tried what seemed a little more natural:

    \int_\gamma \! (z + \dfrac{1}{z}) \, dz = \int_\gamma \! z \, dz + \int_\gamma \! z^{-1} \, dz = \dfrac{z^2}{2}\bigg|_a^b + Log \, z \big|_a^b

     = \dfrac{(6+2i)^2 - (-4+i)^2}{2} + ln|6+2i| - ln|-4+i| + i[arctan(1/3) - arctan(-1/4)]

    This gives me a completely different answer.
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  3. #3
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    Quote Originally Posted by davismj View Post
    After more thinking, I tried what seemed a little more natural:

    \int_\gamma \! (z + \dfrac{1}{z}) \, dz = \int_\gamma \! z \, dz + \int_\gamma \! z^{-1} \, dz = \dfrac{z^2}{2}\bigg|_a^b + Log \, z \big|_a^b

     = \dfrac{(6+2i)^2 - (-4+i)^2}{2} + ln|6+2i| - ln|-4+i| + i[arctan(1/3) - arctan(-1/4)]

    This gives me a completely different answer.
    The \arctan part is not correct. It depends on the sign of the real and imaginary term. You need the principal argument of the complex number -4+i:

    \int_{\gamma} z+1/z dz=17/2+16i+Log[6+2i]-Log[-4+i]

    =(17/2+16i)+\ln|6+2i|-\ln|-4+i|+i\left(Arg(6+2i)-Arg(-4+i)\right)

    Also, you can compute this parametrically with a straight-line between the points:

    z=x(t)+iy(t)=\{t,i(1/10(t+4)+1)\}

    then we'd have:

    \int_{-4}^6 (z(t)+1/z(t))(1+1/10 t) dt\approx 8.9+13.4 i
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