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Math Help - Residues

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    Residues

    Calculate \int_{|z|=3}\frac{e^{iz}}{z^2(z-2)(z+5i)}dz So the point Z= -5i is not enclosed by the contour.

    So am i right that we only calculate the residues at z=0 and z=2 and sum them then multiply by 2i*pi to evaluate the integral?

    Residue at z=0 = \frac{-3}{25} +\frac{i}{20}

    Residue at z=2 = \frac{e^{2i}}{58} - \frac{5ie^{2i}}{116}

    These just seem like strange answers, have i made a mistake somewhere?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Residues

    Quote Originally Posted by linalg123 View Post
    So am i right that we only calculate the residues at z=0 and z=2 and sum them then multiply by 2i*pi to evaluate the integral?
    Right.


    These just seem like strange answers, have i made a mistake somewhere?
    Why do you think those are strange answers?
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    Re: Residues

    Quote Originally Posted by FernandoRevilla View Post
    Right.




    Why do you think those are strange answers?
    In the ones we have done previously, usually the residues have had similar denominators or something so it could be further simplified. So are these correct?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Residues

    Quote Originally Posted by linalg123 View Post
    In the ones we have done previously, usually the residues have had similar denominators or something so it could be further simplified. So are these correct?
    No problem, show the way you have calculated the residues and we can review it.
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    Re: Residues

    res(f,2)= lim z-->2 \frac{e^{iz}}{z^2(z+5i)} = \frac{e^{2i}}{4(2+5i)} = \frac{e^{2i}}{8+20i} \frac{8-20i}{8-20i} = \frac{8e^{2i}-20ie^{2i}}{464} = (\frac{1}{58} - \frac{5i}{116})e^{2i}

    res (f,0) = lim z-->0 \frac{d}{dz}{\frac{e^{iz}}{(z-2)(z+5i)}} = lim z-->0 \frac{ie^{iz}(z^2-(2-7i)z-(5+12i)}{(z-2)^2(z+5i)^2}= \frac{i(-5-12i)}{-100}

     =\frac{-12+5i}{100} = \frac{-3}{25} + \frac{i}{20}
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Residues

    Quote Originally Posted by linalg123 View Post
    res(f,2)= lim z-->2 \frac{e^{iz}}{z^2(z+5i)} = \frac{e^{2i}}{4(2+5i)} = \frac{e^{2i}}{8+20i} \frac{8-20i}{8-20i} = \frac{8e^{2i}-20ie^{2i}}{464} = (\frac{1}{58} - \frac{5i}{116})e^{2i}
    res (f,0) = lim z-->0 \frac{d}{dz}{\frac{e^{iz}}{(z-2)(z+5i)}} = lim z-->0 \frac{ie^{iz}(z^2-(2-7i)z-(5+12i)}{(z-2)^2(z+5i)^2}= \frac{i(-5-12i)}{-100}
     =\frac{-12+5i}{100} = \frac{-3}{25} + \frac{i}{20}
    Right.

    P.S. With \lim_{z\to 0}f(z), you'll obtain \lim_{z\to 0}f(z) .
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