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Math Help - Please help. Totally out to lunch.

  1. #1
    Junior Member
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    Please help. Totally out to lunch.

    I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:


    Use the Complex FTC and/or Cauchy-Goursat Theorem to show that \int_\gamma f(z) \,dz\,=0 , where \gamma
    is the circle with centre 0 and radius 1, oriented as you prefer, and when:

    1. f(z)=\frac{z}{z-3}

    I'm not sure how to proceed. I know that \gamma \, : \, e^{it} , t \in [0,2\pi)

    Do I proceed like:

    \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,

    or

    \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,

    Do I just integrate the original equation and forget about polar coordinates?

    Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by MathSucker View Post
    I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:


    Use the Complex FTC and/or Cauchy-Goursat Theorem to show that \int_\gamma f(z) \,dz\,=0 , where \gamma
    is the circle with centre 0 and radius 1, oriented as you prefer, and when:

    1. f(z)=\frac{z}{z-3}

    I'm not sure how to proceed. I know that \gamma \, : \, e^{it} , t \in [0,2\pi)

    Do I proceed like:

    \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,

    or

    \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,

    Do I just integrate the original equation and forget about polar coordinates?

    Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?
    Simply apply the residue theorem.
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  3. #3
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    I think I'll be getting on to that next, but for now I have to use the complex FTC or Cauchy-Goursat...

    ARgggghhh....
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  4. #4
    Super Member
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    Quote Originally Posted by MathSucker View Post
    I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:


    Use the Complex FTC and/or Cauchy-Goursat Theorem to show that \int_\gamma f(z) \,dz\,=0 , where \gamma
    is the circle with centre 0 and radius 1, oriented as you prefer, and when:

    1. f(z)=\frac{z}{z-3}

    I'm not sure how to proceed. I know that \gamma \, : \, e^{it} , t \in [0,2\pi)

    Do I proceed like:

    \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,

    or

    \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,

    Do I just integrate the original equation and forget about polar coordinates?

    Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?
    The CFTC states that if the antiderivative is defined in a domain containing the contour C=\gamma(t),a\leq t \leq b, then:

    \oint f'(z)dz=f(\gamma(b))-f(\gamma(a))

    and so if it's a closed contour and the antiderivative is analytic over the entire contour, then the integral is zero.
    Then \mathop\oint\limits_{|z|=1} \frac{z}{z-3}dz=\mathop\oint\limits_{|z|=1} \left(1+\frac{3}{z-3}\right)dz=(z+3\log(z-3))\biggr|_{z=1}^{z=1}=0 since \log(z-3) is analytic over the unit circle and in fact in any domain not including the point z=3.

    By way of Cauchy-Gousat, \frac{z}{z-3} is analytic outside z=3 so that \mathop\oint\limits_{|z|=1} \frac{z}{z-3}dz=0
    Last edited by shawsend; March 21st 2010 at 12:20 PM. Reason: spelling
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Quote Originally Posted by MathSucker View Post
    I think I'll be getting on to that next, but for now I have to use the complex FTC or Cauchy-Goursat...

    ARgggghhh....
    Well this problem is a direct result of Cauchy-Goursat.

    Cauchy's integral theorem - Wikipedia, the free encyclopedia
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  6. #6
    Junior Member
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    Great!

    Pardon my stupidity, but how do you get \left(1+\frac{3}{z-3} \right) from \frac{z}{z-3} ? My algebra is a little off.
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  7. #7
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    Quote Originally Posted by MathSucker View Post
    Great!

    Pardon my stupidity, but how do you get \left(1+\frac{3}{z-3} \right) from \frac{z}{z-3} ? My algebra is a little off.
    That's just long division. Try and get good at it as it's useful in Complex Analysis for example when computing residues of certain functions:
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