Question
If $$\frac{{w - \overline w z}}{{1 - z}}$$ is purely real where $$w = \alpha + i\beta ,\beta \ne 0\,{\mkern 1mu} {\mkern 1mu} {\text{and}}\,\,z \ne 1,$$ then the set of the values of $$z$$ is
A.
$$\left\{ {z:\left| z \right| = 1} \right\}$$
B.
$$\left\{ {z:z = \bar z} \right\}$$
C.
$$\left\{ {z:z \ne 1} \right\}$$
D.
$$\left\{ {z:\left| z \right| = 1,z \ne 1} \right\}$$
Answer :
$$\left\{ {z:\left| z \right| = 1,z \ne 1} \right\}$$
Solution :
$$\because \frac{{w - wz}}{{1 - z}}$$ is purely real
$$\eqalign{
& \therefore \,\,\overline {\left( {\frac{{w - \overline w z}}{{1 - z}}} \right)} = \left( {\frac{{w - \overline w z}}{{1 - z}}} \right) \cr
& \Rightarrow \,\,\frac{{\overline w - w\overline z }}{{1 - \overline z }} = \frac{{w - \overline w z}}{{1 - z}} \cr
& \Rightarrow \,\,\overline w - \overline w z - w\overline z + wz\overline z = w - w\overline z - \overline w z + \overline w z\overline z \cr
& \Rightarrow \,\,w - \overline w = \left( {w - \overline w } \right){\left| z \right|^2} \cr
& \Rightarrow \,\,{\left| z \right|^2} = 1\,\,\,\,\,\,\left( {\because \,\,w = \alpha + i\beta \,\,{\text{and }}\beta \ne {\text{0}}} \right) \cr
& \Rightarrow \,\,\left| z \right| = 1\,\,{\text{also given }}z \ne 1 \cr} $$
∴ The required set is $$\left\{ {z:\left| z \right| = 1,z \ne 1} \right\} = 3\omega \left( {\omega - 1} \right)$$