If $\mathrm{z}$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\frac{\pi}{2}$, then $\bar{z} \omega$ is equal to

Question

Aicharya · Accepted Answer

$|z \omega|=1$ $ 	ext { As, } \operatorname{Arg}\left(\frac{z}{\omega}ight)=\frac{\pi}{2} 	ext { therefore } \frac{z}{\omega}=i $ $	herefore\left|\frac{z}{\omega}ight|=1$ From (1) \& (2), $|z|=|\omega|=1$ and $\frac{z}{\omega}+\frac{\bar{z}}{\omega}=0 ; z \bar{\omega}+\bar{z} \omega=0$ $\bar{z} \omega=-z \bar{\omega}=\frac{-z}{\omega} \cdot \bar{\omega} \cdot \omega ; \bar{z} \omega=-i|\omega|^2=-\mathrm{i}$

If $\mathrm{z}$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and…

Solution