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Math Help - Compex integral evaluation

  1. #1
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    Compex integral evaluation

    Evaluate the integral around C of [(e^3z)/(z+pi/2)] dz if C is the circle |z| = 5.





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  2. #2
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    Quote Originally Posted by kiwijoey View Post
    Evaluate the integral around C of [(e^3z)/(z+pi/2)] dz if C is the circle |z| = 5.




    f(z) has a simple pole at -\frac{pi}{2}

    so the residue is \lim_{z}^{-\frac{\pi}{2}}\left( z+\frac{\pi}{2}\right)f(z)=e^{-\frac{3\pi}{2}}

    so

    \oint_{|z|=5} \frac{e^{3z}}{z+\frac{\pi}{2}}dz=2\pi i e^{-\frac{3\pi}{2}}=2\pi i (\cos\left(\frac{3\pi}{2} \right)+i\sin\left(\frac{3\pi}{2} \right))=2\pi
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  3. #3
    MHF Contributor chiph588@'s Avatar
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     e^{-\frac{3\pi}{2}} \neq \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post

    \oint_{|z|=5} \frac{e^{3z}}{z+\frac{\pi}{2}}dz=2\pi i e^{-\frac{3\pi}{2}}=2\pi i (\cos\left(\frac{3\pi}{2} \right)+i\sin\left(\frac{3\pi}{2} \right))=2\pi
    You can also use Cauchy's theorem:
    Let f(z) = e^{3z} then \frac{1}{2\pi i}\oint_{|z|=5} \frac{f(z)}{z+\frac{\pi}{2}} dz = f ' \left( -\frac{\pi}{2} \right)
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  5. #5
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    Is this Cauchy's int.theorem - where alpha is -pi/2 - oh Sorry new to this and as I type this I see you already have an answer! Glad to see I was on the right track. Good luck! Ruth
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