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

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

$\displaystyle z^2 - 1 = -1$

$\displaystyle z^2 = 2$

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