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Thread: Another Cauchy integral

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    Another Cauchy integral

    Another question to be done using Cauchy's integral formula:

    $\displaystyle \oint\frac{e^z}{z(z^2-9)} \, {\color{red}dz}$ with $\displaystyle |z-3|=1 $

    Start by letting $\displaystyle f(z) = \frac {e^z}{z}$ since z has no zeros in $\displaystyle |z-3|=1$

    then $\displaystyle \oint\frac{f(z)}{(z^2-9) - z_0}$ with $\displaystyle z_0 = +-3$

    So $\displaystyle 2\pi i * f(z_0) = 2\pi i*\frac{e^3}{3}$ and $\displaystyle 2\pi i*\frac{e^{-3}}{-3}$

    Did i go wrong anywhere?
    Last edited by mr fantastic; Jul 12th 2009 at 06:56 PM. Reason: Fixed some latex and added the red dz. Edited the first sentance and changed post title.
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  2. #2
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    Quote Originally Posted by Richmond View Post
    Another similiar question,

    $\displaystyle \oint\frac{e^z}{z(z^2-9)} \, {\color{red}dz}$ with $\displaystyle |z-3|=1 $

    Start by letting $\displaystyle f(z) = \frac {e^z}{z}$ since z has no zeros in $\displaystyle |z-3|=1$

    then $\displaystyle \oint\frac{f(z)}{(z^2-9) - z_0}$ with $\displaystyle z_0 = +-3$

    So $\displaystyle 2\pi i * f(z_0) = 2\pi i*\frac{e^3}{3}$ and $\displaystyle 2\pi i*\frac{e^{-3}}{-3}$

    Did i go wrong anywhere?
    The contour encloses only one singularity: A simple pole at z = 3. So take $\displaystyle f(z) = \frac{e^z}{z(z + 3)}$.
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