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Thread: 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

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

    $\displaystyle (1) - (2)$

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

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

    rewrite this as a quadratic in $\displaystyle e^{iz}$

    you should get

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

    use the quadratic formula.

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

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

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

    Bobak
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  3. #3
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    thanks

    Thanks a lot mate. :-)
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