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Math Help - Simple complex equality driving me nuts

  1. #1
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    Simple complex equality driving me nuts

    Find all solutions to sinz = 2i.

    I've gone down the road of sin (z-z^(-1))/2i and its polar counterpart, and also laurent series, but the solutions I get are very complicated. This is a problem from a competency exam so I'm guessing it should have a clever but simple path and answer. Any guidance?
    Last edited by igopogo; June 2nd 2010 at 11:28 AM. Reason: sin z, not sin x!
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  2. #2
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    Quote Originally Posted by igopogo View Post
    Find all solutions to sinz = 2i.
    \sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)
    So in this solve this system:
    \sin(x)\cosh(y)=0~\&~\cos(x)\sinh(y)=2
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