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Math Help - Evaluate the Integral. HELP!!

  1. #1
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    Evaluate the Integral. HELP!!

    Evaluate the following integral. The contour is a square centered at the origin with corners at .

    20.


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    This one is different there is a z in the numerator.
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    Quote Originally Posted by lacy1104 View Post
    This one is different there is a z in the numerator.
    My mistake at not loking closer - sorry.

    The answer is 2 \pi i (i) = - 2 \pi.

    You can get this answer using the following formula (which is a special case of Cauchy's Integral Formula):

    \oint_{\Gamma} \frac{f(z)}{z-\alpha} dz = 2\pi i f(\alpha) where \Gamma is any countor containing \alpha.


    Alternatively, you can integrate along each part of the curve in the way I outlined in your other question .....
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    Quote Originally Posted by mr fantastic View Post
    My mistake at not loking closer - sorry.

    The answer is 2 \pi i (i) = - 2 \pi.

    You can get this answer using the following formula (which is a special case of Cauchy's Integral Formula):

    \oint_{\Gamma} \frac{f(z)}{z-\alpha} dz = 2\pi i f(\alpha) where \Gamma is any countor containing \alpha.


    Alternatively, you can integrate along each part of the curve in the way I outlined in your other question .....
    Alternatively, you can integrate directly:

    I'll do the integral along the line segment Im(z) = 2 from z = 2 + 2i to z = -2 + 2i:

    Note that on this line segment, z = x + 2i => dz = dx:


    \int_{x = 2}^{x = -2} \frac{x + 2i}{{\color{red} (x + 2i) - i}} \, dx = \int_{x = 2}^{x = -2} \frac{x + 2i}{x + i} \, dx


    = \int_{x = 2}^{x = -2} 1 + \frac{i}{x + i} \, dx = \int_{x = 2}^{x = -2} 1 \, dx + i \int_{x = 2}^{x = -2}\frac{1}{x + i} \, dx

    = -4 + i \left( \ln(x + i) \bigg{|}_{2}^{-2} \right) = -4 + i (\ln (-2 + i) - \ln (2 + i) )


     = -4 + i \left( \left[ \ln(\sqrt{5}) + \left( \pi - \arctan \left( \frac{1}{2} \right)\right) \, i \right] - \left[ \ln(\sqrt{5}) + \left(\arctan \left( \frac{1}{2} \right)\right)\, i \right] \right)


     = -4 + i (\pi \, i) = -4 - \pi.


    The integrals along the other line segments are done similarly. Then you add together all the bits from each integral.
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    Or you can do completely avoid Cauchy theorem: \frac{z}{z-i} = \frac{(z-i)+i}{z-i} = 1 + i\cdot \frac1{z-i}.
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    Quote Originally Posted by ThePerfectHacker View Post
    Or you can do completely avoid Cauchy theorem: \frac{z}{z-i} = \frac{(z-i)+i}{z-i} = 1 + i\cdot \frac1{z-i}.
    Indeed. And because f(z) = 1 is analytic on and inside the contour, its integral around the contour is zero ..... (Of course, it's a simple matter to integrate it directly: 4 - 4 + 4 - 4 = 0)
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    Quote Originally Posted by mr fantastic View Post
    And because f(z) = 1 is analytic on and inside the contour, its integral around the contour is zero ..... (Of course, it's a simple matter to integrate it directly: 4 - 4 + 4 - 4 = 0)
    It is even easier than that. (z)' = 1. (So it has a primitive. And integrals over piecewise smooth closed curves of a function possesing a primitive is zero. )

    And interesting question to ask on an exam is why then \oint z^{-1} dz \not = 0. How about \log z?
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    Quote Originally Posted by ThePerfectHacker View Post
    It is even easier than that. (z)' = 1. (So it has a primitive. And integrals over piecewise smooth closed curves of a function possesing a primitive is zero. )

    And interesting question to ask on an exam is why then \oint z^{-1} dz \not = 0. How about \log z?
    Just to avoid potential confusion, the contour encloses the special point z = 0. And therein lies the answer .....
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