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. $\frac{1}{\left|\omega\right|^{2}}$
• D. $\frac{1}{\left|\omega\right|}$
• 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$