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Math Help - Find all solutions of the equation, please help, complex analysis?

  1. #1
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    Wah Cantt
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    Find all solutions of the equation, please help, complex analysis?

    cos z = 2 i

    where z is a complex number, find z?
    and i is (iota)
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  2. #2
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    Crna Gora
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    Re: Find all solutions of the equation, please help, complex analysis?

    Hint :

    \cos z=\frac{e^{iz}+e^{-iz}}{2}
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  3. #3
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    Re: Find all solutions of the equation, please help, complex analysis?

    I assume you mean that i is the imaginary unit \displaystyle \sqrt{-1}. Anyway

    \displaystyle \begin{align*} \cos{(z)} &= 2i \\ \cos{( x + i\, y)} &= 2i \\ \cos{(x)}\cos{(i\,y)} - \sin{(x)}\sin{(i\,y)} &= 2i \\ \cos{(x)}\cosh{(y)} - i\sin{(x)}\sinh{(y)} &= 0 + 2i \\ \cos{(x)}\cosh{(y)} = 0 \textrm{ and } -\sin{(x)}\sinh{(y)} &= 2   \end{align*}

    From the first equation, we have \displaystyle \cos{(x)} = 0 \implies x = \frac{(2n+1)\pi}{2} \textrm{ where } n \in \mathbf{Z} or  \displaystyle \cosh{(y)} = 0 which does not have any real solutions. Substituting x we find

    \displaystyle \begin{align*} -\sin{\left[ \frac{(2n+1)\pi}{2} \right] } \sinh{(y)} &= 2 \\ (-1)^{n+1} \sinh{(y)} &= 2 \\ \sinh{(y)} &= \frac{2}{(-1)^{n+1}} \\ y &= \sinh^{-1}{\left[ \frac{2}{(-1)^{n+1}} \right]} \end{align*}

    So there you have it, the solutions are \displaystyle z = \frac{(2n+1)\pi}{2} + i\sinh^{-1}{\left[ \frac{2}{(-1)^{n+1}} \right]} \textrm{ where } n \in \mathbf{Z}.
    Thanks from Shakarri
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  4. #4
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    Re: Find all solutions of the equation, please help, complex analysis?

    the answer is (2n + 1)(pi/2) - i ( (-1)^n ) 1.443
    where as I got (2n+1)(pi/2) + 1.443
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