solving logarithmic complex variable equation

**flametag3** · November 13th 2012, 04:28 PM

Can someone please verify if I have solved this equation properly

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

taking exp on both sides

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

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

**Prove It** · November 13th 2012, 08:00 PM

Originally Posted by flametag3

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

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