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Math Help - Equation of type sin(z) = w

  1. #1
    Junior Member
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    Equation of type sin(z) = w

    COMPLEX ANALYSIS:
    Can somebody please thoroughly explain to me why the solution to

    sin(z) = w

    is

    z = -i*log (i*w + (1-w^2)^(1/2))

    and not

    z = -i*log (i*w sqrt(1-w^2))
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  2. #2
    MHF Contributor
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    Re: Equation of type sin(z) = w

    If sin(z) = w then cos(z) = \sqrt{1-w^2} not \pm \sqrt{1-w^2} by the Pythagorean theorem.

    So, you have:
    e^{iz} = cos(z) + i w
    e^{iz} = iw + \sqrt{1-w^2}
    iz = \log{(iw + \sqrt{1-w^2})}
    z = -i \log{(iw + \sqrt{1-w^2})}
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  3. #3
    Junior Member
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    Re: Equation of type sin(z) = w

    Thanks
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