Question

If \omega = $\alpha$ + i$\beta$ where $\alpha$, $\beta$ are real, $\beta$ $\ne$ 0 and z $\ne$ 1 satisfies the condition that $\frac{\omega -\overline{\omega }z}{1-z}$ is purely real then the set of values of z is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Hint: If Z is purely real then z = $\overline{z}$

i.e., $\frac{\omega -\overline{\omega }z}{1-z}$ = $\overline{\left(\frac{\omega -\overline{\omega }z}{1-z}\right)}$ then $\frac{\omega -\overline{\omega }z}{1-z}$ = $\frac{\overline{\omega }-\omega \overline{z}}{1-\overline{z}}$

so that $\omega +\overline{\omega }z\overline{z}$ = $\overline{\omega }+\omega z\overline{z}$ then $(\omega -\overline{\omega })(1-z\overline{z})$ = 0

which means 1 = z $\overline{z}$ ⇒ $|z{|}^2$ = 1 ⇒ |z| = 1 but given z $\ne$ 1