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Math Help - singularity

  1. #1
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    singularity

    f(z)=\frac{1}{sinh(z)}
    It's ok that z=0 is a singularity, but why
    2ik\pi, where k is a integer, are singularities too? sinh(z)=0 only when z=0

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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    \sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    \sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0
    sinh(ki\pi)=0

    so, why the argument of the sinh must be an even multiple of i\pi ?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hurz View Post
    sinh(ki\pi)=0 so, why the argument of the sinh must be an even multiple of i\pi ?

    You asked if 2k\pi i were singularities of f(z) and the answer is yes. If you want to find all of them:


    \sinh z=\dfrac{e^z-e^{-z}}{2}=0\Leftrightarrow\ldots\Leftrightarrow e^{2z}=1\Leftrightarrow\ldots\Leftrightarrow z=k\pi i\quad (k\in\mathbb{Z})
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