JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1
If $z$ and $\omega$ are two complex numbers such that $z\omega=1$ and $\arg z-\arg\omega=\frac\pi2$, then :
Complex Numbers
Solution
$|z||\omega|=1$ and $\arg z+\arg\omega=0$, with $\arg z-\arg\omega=\frac\pi2$: $\arg z=\frac\pi4$, $\arg\omega=-\frac\pi4$. Then $z\bar\omega=e^{i\pi/2}=i$ and $\bar z\omega=-i$.