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Thread: 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

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

    taking exp on both sides

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

    $\displaystyle z^2 - 1 = -1$
    $\displaystyle z^2 = 2$
    $\displaystyle z$ = $\displaystyle \sqrt{2}$ + $\displaystyle 2k\pi i$
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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

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

    taking exp on both sides

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

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

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