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Math Help - solving logarithmic complex variable equation

  1. #1
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    solving logarithmic complex variable equation

    Can someone please verify if I have solved this equation properly

    Log (z^{2} -1) = \frac{i \pi}{2}

    taking exp on both sides

    e^{Log (z^{2} -1)} = e^{\frac{i \pi}{2}}

    z^2 - 1 = -1
    z^2 = 2
    z = \sqrt{2} + 2k\pi i
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  2. #2
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    Re: solving logarithmic complex variable equation

    Quote Originally Posted by flametag3 View Post
    Can someone please verify if I have solved this equation properly

    Log (z^{2} -1) = \frac{i \pi}{2}

    taking exp on both sides

    e^{Log (z^{2} -1)} = e^{\frac{i \pi}{2}}

    z^2 - 1 = -1
    z^2 = 2
    z = \sqrt{2} + 2k\pi i
    It's \displaystyle \begin{align*} e^{i\pi} \end{align*} which equals \displaystyle \begin{align*} -1 \end{align*}, not \displaystyle \begin{align*} e^{\frac{i\pi}{2}} \end{align*}.

    Rather \displaystyle \begin{align*} e^{\frac{i\pi}{2}} = \cos{\left( \frac{\pi}{2} \right)} + i\sin{\left( \frac{\pi}{2} \right)} = 0 + 1i = i \end{align*}
    Thanks from flametag3
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