Question
A complex number $$z$$ is said to be unimodular if $$\left| z \right| = 1.$$ Suppose $${z_1}\,{\text{and }}{z_2}$$ are complex numbers such that $$\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}}$$ is unimodular and $${{z_2}}$$ is not unimodular. Then the point $${{z_2}}$$ lies on a:
A.
circle of radius 2.
B.
circle of radius $$\sqrt 2. $$
C.
straight line parallel to $$x$$ - axis.
D.
straight line parallel to $$y$$ - axis.
Answer :
circle of radius 2.
Solution :
$$\eqalign{
& \left| {\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}}} \right| = 1 \cr
& \Rightarrow \,{\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2} \cr
& \Rightarrow \,\left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right) = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right) \cr
& \Rightarrow \,\left( {{z_1} - 2{z_2}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right) = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {2 - {{\overline z }_1}{z_2}} \right) \cr
& \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2} = \,4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2} \cr
& \Rightarrow \,{\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} \cr
& \Rightarrow \,{\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0 \cr
& \left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0 \cr
& \because \,\left| {{z_2}} \right| \ne 1 \cr
& \therefore \,{\left| {{z_1}} \right|^2} = 4 \cr
& \Rightarrow \,\left| {{z_1}} \right| = 2 \cr} $$
⇒ Point $${{z_1}}$$ lies on circle of radius 2.