Question

If $w=\alpha +i\beta$, where $\beta \ne 0$ 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. $\{z:|z|=2\}$
• B. $\{z:z=\overline{z}\}$
• C. $\{z:z\ne 1\}$
• D. $\{z:|z|=1,z\ne 1\}$

Answer 1

The correct option is D $\{z:|z|=1,z\ne 1\}$

Given, $\frac{w-\overline{w}z}{1-z}$ is purely real.

If z is any complex number then

z is purely real if $z=\overline{z}$

So here

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

$\frac{w-\overline{w}z}{1-z}=\frac{\overline{w}-w\overline{z}}{1-\overline{z}}$

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

$w+\overline{w}|z{|}^2=\overline{w}+w|z{|}^2$

$w-\overline{w}+(\overline{w}-w)|z{|}^2=0$

$(w-\overline{w})(1-|z{|}^2)=0\Rightarrow w=\overline{w}$

$|z|=1$

Since $w=\alpha +i\beta$ with $\beta \ne 0$

So $w\ne \overline{w}$

So we are left with

$\{z:|z|=1,z\ne 1\}$

Option D.