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Thread: Residues, describing contours: Find res[exp(1/z)/z^2-16,4], res[exp(1/z)/z^2-16,-4],

  1. #1

    Residues, describing contours: Find res[exp(1/z)/z^2-16,4], res[exp(1/z)/z^2-16,-4],

    Any help would be great.

    Find;

    i) res[exp(1/z)/z^2-16,4]

    ii) res[exp(1/z)/z^2-16,-4]

    iii) res[exp(1/z)/z^2-16,0]

    And then describe all contours of the integral

    iv) ~ exp(1/z)/z^2+16 dz = 0

  2. #2
    Elite Member
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    Well, there is one thing that can help. If you're in Complex Analysis, you should know the difference between

    1/z^2 - 16 = [tex]\dfrac{1}{z^{2}}-16[/tex]

    and

    1/(z^2 - 16) = [tex]\dfrac{1}{z^{2}-16}[/tex]

    After that, have you considered identifying the Order of the Pole and simply computing the appropriate limit?

    Please follow the forum guidelines - this includes showing YOUR work.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Apologies, it was an error on my part, I forgot to include the brackets.

    So far I have exp(1/z)/(z+4)(z-4) so z = +/- 4

    For z = 4 ; exp(1/4)/(4+4)/(4-4) = 0.161

    For z = -4 ; exp(1/-4)/(-4-4)/(-4+4) = -0.097

    For z = 0; exp(1)/-1 = -2.72

    For the second part, the contour exp(1/z)/(z+4i)(z-4i);

    z = +/- 4i

    2(pi)i * [exp(1/4i)/8i + exp(1/-4i)/-8i]

    Not sure if I am correct so any bit of help would be great. Thanks

  4. #4
    Elite Member
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    Let's get a little better with notation, please. Good notation will save you. It matters.

    How can you write "/(4-4)" with a straight face? That's no good.

    Are you sure about z = 0?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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