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Math Help - Proof required for expressing ARCSIN in complex exponential function.

  1. #1
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    Question Proof required for expressing ARCSIN in complex exponential function.

    The question goes this way:

    Given the complex inverse sine function w(z)=sin-(z) is defined such that z=sin(w).

    Writing sin(w) in terms of complex exponential function, show that:

    arcsin(z) = - i log[ i z √(1-z)]

    Thanks in advance.
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  2. #2
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    Start with

    e^{iz} = \cos z + i \sin z \ \ \ \  (1)
    e^{-iz} = \cos z - i \sin z \ \ \ (2)

    (1) - (2)

     e^{iz} - e^{-iz} = 2 i \sin z

    giving \sin z = \frac{ e^{iz} - e^{-iz}}{2i}

    rewrite this as a quadratic in  e^{iz}

    you should get

     (e^{iz})^2  - 2 i \sin z e^{iz} - 1

    use the quadratic formula.

    to get e^{iz} = i \sin z \pm \sqrt{1 - \sin^2 z}

    then take logs z = -i \ \ln (\  i \sin z \pm \sqrt{1 - \sin^2 z} \ )

    remember i^2 = -1 and \frac{1}{i} = -i

    Bobak
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  3. #3
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    thanks

    Thanks a lot mate. :-)
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