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Math Help - complex analysis4

  1. #1
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    complex analysis4

    a. Evaluate the integral
    \int e^z/((z^2+pi)^2) where the integral is the circle of radius 4 centered at the origin.

    b. Evaluate the integral
    \int 1/((z^2+1)(z^2-1))
    where the integral is the circle of radius 2 centered at the origin.
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  2. #2
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    Quote Originally Posted by shaheen7 View Post
    a. Evaluate the integral
    \int e^z/((z^2+pi)^2) where the integral is the circle of radius 4 centered at the origin.

    b. Evaluate the integral
    \int 1/((z^2+1)(z^2-1))
    where the integral is the circle of radius 2 centered at the origin.
    Use the residue theorem

    \oint \frac{e^z}{(z^2+\pi^2)^2}dz=\oint \frac{e^z}{(z+i\pi)^2(z-i\pi)^2}dz

    So the function has two poles of order 2.

    So we need to calculate the residues at
    \pm i\pi

    So to calculate at

    i\pi

    we have

    \text{Res}(f,i\pi) = \lim_{z \to i\pi }\frac{d}{dz}(z-i\pi)^2\left(\frac{e^z}{(z+i\pi)^2(z-i\pi)^2} \right)

    \lim_{z \to i\pi} \frac{d}{dz}\frac{e^{z}}{(z+i\pi)^2}= \lim_{z \to i\pi} \frac{e^z(z+i\pi)^2-2e^{z}(z + i \pi)}{(z+ i\pi )^4}

    \frac{e^{i\pi}(2i\pi)(2i\pi-2)}{(2i\pi)^4}=\frac{-1(2 i \pi -2)}{(2i \pi)^3 }=\frac{2-2\pi i}{-8 i\pi^3}=\frac{1}{4}\left(\frac{\pi  +i}{\pi^3} \right)

    Now you compute the other one and remember that the answer is

    2 \pi i \sum \text{Res}(f)

    at all of the poles in the contour.
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