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Math Help - Contour Integration

  1. #1
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    Contour Integration

    \int_c\frac{{e}^{z } }{ ({{z}^{2 }-9 })^{2 } } dz where c is the positively oriented circle |z+2|=2 So the formula is \frac{2\pi i {f}^{(n-1) } (z)}{(n-1)! }  f(z) = {e}^{ z} and n = 2 So my answer is 2\pi i {e}^{-3 } But according to my teacher it is over 27
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  2. #2
    MHF Contributor Amer's Avatar
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    f(z) is not e^z in this question

    \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}
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  3. #3
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    \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}
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  4. #4
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    Quote Originally Posted by adam_leeds View Post
    \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}
    Sorry still dont get it.
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  5. #5
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    g(z)=\frac{e^z}{(z-3)^2(z+3)^2}

    As you can see, only one of the poles (z=-3) lies within the given circle, so we let f(z)=\frac{e^z}{(z-3)^2} and use Cauchy's integral formula for f(-3):

    f'(-3)=\frac{1!}{2\pi i}\int_{|z+2|=2}\frac{f(z)}{(z+3)^2}dz=\frac{1!}{2  \pi i}\int_{|z+2|=2}g(z)dz
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  6. #6
    MHF Contributor Amer's Avatar
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    we have two singular points -3,3 but 3 is outside the circle
    but -3 is inside so the singular point for the function is -3


    the function is


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