Question

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

• A. 1
• B. – 1
• C. I
• D. – i

Answer 1

Answer: (D)

– i

Answer 2

Sol.$|z|\omega| = 1$ .....(i) and arg$\left( \frac{z}{\omega} \right) = \frac{\pi}{2}$

⇒ $\frac{z}{\omega} = i$ ⇒ $\left| \frac{z}{\omega} \right| = 1$ .....(ii)

From equation (i) and (ii),

$|z| = |\omega| = 1$and $\frac{z}{\omega} + \frac{\bar{z}}{\bar{\omega}} = 0;z\bar{\omega} + \bar{z}\omega = 0$

⇒ $\bar{z}\omega = - z\bar{\omega} = \frac{- z}{\omega}\bar{\omega}\omega$ ⇒ $\bar{z}\omega = - i|\omega|^{2} = - i.$