Question

If $z$ and $\omega$ are two non-zero complex numbers such that $|z\omega| = 1$ and Arg $(z) - $ Arg $(\omega) = \frac{\pi}{2} $, and $\bar{z} \omega$ is equal to $i^k$, where $k$ is smallest natural number then $k$ is

Answer 1

$|\bar{z} \omega| = |\bar{z}||\omega| = |z||\omega| = |z \omega |= 1$
Arg $(\bar{z} \omega) =$ arg $(\bar{z}) +$ arg$(\omega)$
$ = -$ arg $(z) + $ arg $\omega = -\frac{\pi}{2}$
$\therefore \bar{z} \omega = -1$