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Math Help - Inverse cosine of i

  1. #1
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    Inverse cosine of i

    How do you find: cos^-1(i)
    THANK YOU
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  2. #2
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    Re: Inverse cosine of i

    The first thing I can think of is to use the power series of arccos.
    Thanks from gokujosh
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  3. #3
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    Re: Inverse cosine of i

    Quote Originally Posted by gokujosh View Post
    How do you find: cos^-1(i)
    THANK YOU
    Let \displaystyle z = \cos^{-1}{(w)}, then

     \displaystyle \begin{align*} \cos{(z)} &= w \\ \frac{e^{i\,z} + e^{-i\,z}}{2} &= w \\ e^{i\,z} + e^{-i\,z} &= 2w \\ \left( e^{i\,z} \right)^2 + 1 &= 2w\,e^{i\,z} \\ \left( e^{i\,z} \right) ^2 - 2w\,e^{i\,z} + 1 &= 0 \\ \left( e^{i\,z} \right) ^2 - 2w\,e^{i\,z} + \left( -w \right) ^2 - \left( -w \right) ^2 + 1 &= 0 \\ \left( e^{i\,z} - w \right) ^2 - w^2 + 1 &= 0 \\ \left( e^{i\,z} - w \right) ^2 &=  w^2 - 1 \\ e^{i\,z} - w &= \pm \sqrt{ w^2 - 1 } \\ e^{i\,z} &= w \pm \sqrt{ w^2 - 1 } \\ i\,z &= \log{ \left( w \pm \sqrt{ w^2 - 1 } \right) } \\ z &= -i\log{ \left( w \pm \sqrt{ w^2 - 1 } \right) }  \end{align*}

    So in your case, where \displaystyle z = \cos^{-1}{(i)}, you would have

    \displaystyle \begin{align*} z &= \cos^{-1}{(i)} \\ &= -i \log{ \left( i \pm \sqrt{ i^2 - 1 } \right) } \\ &= - i \log{ \left( i \pm \sqrt{ -2} \right) } \\ &= - i\log{ \left( i \pm i\,\sqrt{2} \right) } \\ &= -i \log{ \left[ i\left( 1 \pm \sqrt{2} \right) \right] } \end{align*}

    Case 1:

     \displaystyle \begin{align*} z &= -i \log{ \left[ i \left( 1 - \sqrt{2} \right) \right] } \\ &= -i \left\{ \ln{ \left| i\left( 1 - \sqrt{2} \right) \right| } + i\arg{ \left[ i \left( 1 - \sqrt{2} \right) \right] } + 2\pi i m \right\} \textrm{ where } m \in \mathbf{Z} \\ &= -i \left[ \ln{ \left( 1 - \sqrt{2} \right) } - \frac{\pi}{2}i + 2\pi i m \right] \\ &= 2\pi m -\frac{\pi}{2} - i\ln{ \left( 1 - \sqrt{2} \right) }  \end{align*}

    Case 2:

    \displaystyle \begin{align*} z &= -i \log{ \left[ i \left( 1 + \sqrt{2} \right) \right] } \\ &= -i \left\{ \ln{ \left| i \left( 1 + \sqrt{2} \right) \right| } + i\arg{ \left[ i \left( 1 + \sqrt{2} \right) \right] } + 2\pi i n  \right\} \textrm{ where } n \in \mathbf{Z} \\ &= -i \left[ \ln{ \left( 1 + \sqrt{2} \right) } + \frac{\pi}{2}i + 2\pi i n \right] \\ &= 2\pi n + \frac{\pi}{2} - i \ln{ \left( 1 + \sqrt{2} \right) } \end{align*}

    Well there you go, there are the infinitely many values for \displaystyle \cos^{-1}{(i)}
    Thanks from gokujosh and Gusbob
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  4. #4
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    Re: Inverse cosine of i

    Thank you, this was so confusing, but that makes sense now .
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