Question

Let $z\ne 1$ be a complex number and let $\omega =x+iy$, $y\ne 0$. If $\frac{\omega -\bar{\omega }z}{1-z}$ is purely real, then $|z|$ is equal to

• A. $\left|\omega \right|$
• B. ${\left|\omega \right|}^2$
• C. $\frac{1}{{|\omega |}^2}$
• D. $\frac{1}{|\omega |}$
• E. 1

Answer 1

Answer: (E)

Since, $\frac{\omega -\bar{\omega }z}{1-z}$ is purely real.
$\therefore \left(\frac{\omega -\bar{\omega }z}{1-z}\right)=\left(\frac{\omega -\bar{\omega }z}{1-z}\right)$
$\Rightarrow \frac{\omega -\bar{\omega }z}{1-z}=\frac{\bar{\omega }-\omega \bar{z}}{1-\bar{z}}$
$\Rightarrow \omega -\omega \bar{z}-\bar{\omega }z+\omega z\bar{z}=\bar{\omega }-\omega \bar{z}-z\bar{\omega }+\omega z\bar{z}$
$\Rightarrow \omega -\bar{\omega }=(\omega -\bar{\omega })|z{|}^2$
$\Rightarrow |z{|}^2=1\Rightarrow |z|=1$