Question

If $w=\alpha +i\beta ,where\beta \ne 0andz\ne 1,$ satisfies the condition that $\left(\frac{w-\overline{w}z}{1-z}\right)$ is purely real, then the set of values of z is

• A. $|z|=1,z\ne 2$
• B. $|z|=1,andz\ne 1$
• C. $z=\overline{z}$
• D. None of these

Answer 1

The correct option is B $|z|=1,andz\ne 1$

$Let{z}_1=\frac{w-\overline{w}z}{1-z}bepurelyreal\Rightarrow z_1={\overline{z}}_1\therefore \frac{w-\overline{w}z}{1-z}=\frac{\overline{w}-w\overline{z}}{1-\overline{z}}\Rightarrow w-w\overline{z}-\overline{w}z+\overline{w}z.\overline{z}=\overline{w}-z\overline{w}-w\overline{z}+wz.\overline{z}\Rightarrow (w-\overline{w})+(\overline{w}-w)|z{|}^2=0\Rightarrow (w-\overline{w})(1-|z{|}^2)=0\Rightarrow |z{|}^2=1[asw-\overline{w}\ne 0,since\beta \ne 0]\Rightarrow |z|=1andz\ne 1$