Question:
If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{\pi}{2}$, then:
Correct Option: , 3
Solution:
Given $|z \omega|=1$.............(1)
and $\arg \left(\frac{z}{\omega}\right)=\frac{\pi}{2}$.............(2)
$\therefore \frac{z}{\omega}+\frac{\bar{z}}{\omega}=0$ $\left[\because \operatorname{Re}\left(\frac{z}{\omega}\right)=0\right]$
$\Rightarrow z \bar{\omega}=-\bar{z} \bar{\omega}$
from equation (i), $z \bar{z} \omega \bar{\omega}=1$ [using $z \bar{z}=|z|^{2}$ ]
$(\bar{z} \omega)^{2}=-1 \Rightarrow \bar{z} \omega=\pm i$
from equation (ii), $-\arg (\bar{z})-\arg \omega=\frac{\pi}{2}-\arg (\bar{z} \omega)=\frac{-\pi}{2}$
Hence, $\bar{z} w=-i$