# solving logarithmic complex variable equation

• Nov 13th 2012, 04:28 PM
flametag3
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

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

$\displaystyle z^2 - 1 = -1$
$\displaystyle z^2 = 2$
$\displaystyle z$ = $\displaystyle \sqrt{2}$ + $\displaystyle 2k\pi i$
• Nov 13th 2012, 08:00 PM
Prove It
Re: solving logarithmic complex variable equation
Quote:

Originally Posted by flametag3
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*}