Question

If $(w-\overline{w}z)/(1-z)$ is purely real where $w=\alpha +i\beta ,\beta \ne 0$ and $z\ne 1$, then set of the values of $z$ is

• A. $z:\left|z\right|=1$
• B. $z:z=\overline{z}$
• C. $z:z\ne 1$
• D. $z:\left|z\right|=1,z\ne 1$

Answer 1

The correct option is D $z:\left|z\right|=1,z\ne 1$

$\frac{(\omega -\overline{\omega }z)}{1-z}$ is purely real and $z\ne 1$

$\frac{(\omega -\overline{\omega }z)}{1-z}=\frac{\overline{\omega -\overline{\omega }z}}{\left(\overline{1-z}\right)}$

$\frac{(\omega -\overline{\omega }z)}{1-z}=\frac{\overline{\omega }-\omega \overline{z}}{1-\overline{z}}$

$(\omega -\overline{\omega }z)(1-\overline{z})=(1-z)(\overline{\omega }-\omega \overline{z})$

$\omega -\omega \overline{z}-\overline{\omega }z+\overline{\omega }|z{|}^2=\overline{\omega }+\omega |z{|}^2-\overline{\omega }z-\omega \overline{z}$

$\omega -\overline{\omega }=|z{|}^2(\omega -\overline{\omega })$

$\therefore |z{|}^2=1$ (as $Im\left(\omega \right)=\beta \ne 0$)

$\Rightarrow |z|=1$ But $z\ne 1$

$\therefore$ Set of values of $z$ is $z:|z|=1,z\ne 1$