JEE MAIN - Mathematics (2015 (Offline) - No. 19)
A complex number z is said to be unimodular if $$\,\left| z \right| = 1$$. Suppose $${z_1}$$ and $${z_2}$$ are complex numbers such that $${{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}$$ is unimodular and $${z_2}$$ is not unimodular. Then the point $${z_1}$$ lies on a :
circle of radius 2.
circle of radius $${\sqrt 2 }$$.
straight line parallel to x-axis
straight line parallel to y-axis.
Explanation
$$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$$
$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$$
$$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$$
$$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$$
$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$$
$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$$
$$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$$
As $$\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$$ $$\therefore$$ $$\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$$
$$ \Rightarrow \,$$ Point $$\,{z_1}\,$$ lies on circle of radius $$2.$$
$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$$
$$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$$
$$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$$
$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$$
$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$$
$$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$$
As $$\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$$ $$\therefore$$ $$\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$$
$$ \Rightarrow \,$$ Point $$\,{z_1}\,$$ lies on circle of radius $$2.$$
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