Question

A complex number z is said to be unimodular if $\left| z \right|eq1$. If ${z_1}$ and ${z_2}$ are complex numbers such that $\left| {\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}} \right|$ is unimodular and ${z_2}$ is not unimodular. Then, the point ${z_1}$ lies on a
A. straight line parallel to X-axis.
B. straight line parallel to Y-axis.
C. circle of radius 2.
D. circle of radius $\sqrt 2$.

Answer 1

Hint: To solve this question, we will use the concept of complex numbers and quadratic equations. We will use the method of finding the modulus and conjugate of a complex number. The conjugate of a complex number $z = a + ib$is denoted by $\overline z$, where $\overline z = a - ib$ and modulus of z is $r = \sqrt {{a^2} + {b^2}}$.

Complete step-by-step answer:
A number of the form $a + ib$, where a and b are real numbers, is called a complex number.
We know that a complex number z is said to be unimodular if $\left| z \right| = 1$.
Here, $\left| {\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}} \right|$ is unimodular and ${z_2}$ is not unimodular.
So,
$\Rightarrow \left| {\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}} \right| = 1$.
Thus, we can say that
$\Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}\overline {{z_2}} } \right|^2}$. ……… (i)
We know that, ${\left| a \right|^2} = a \times \overline a$.
Therefore equation (i) will become,
$\Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1}} - 2\overline {{z_2}} } \right) = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - \overline {{z_1}} {z_2}} \right)$.
Solving this, we will get

$$\Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 2{z_1}\overline {{z_2}} - 2\overline {{z_1}} {z_2} = 4 - 2\overline {{z_1}} {z_2} - 2{z_1}\overline {{z_2}} + {\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 \\\$$

Taking common,

$$\Rightarrow {\left| {{z_1}} \right|^2}\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) - 4\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0 \\\ \Rightarrow \left( {1 - {{\left| {{z_2}} \right|}^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} - 4} \right) = 0 \\\$$

here two cases are possible,

$$\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0 \\\ {\left| {{z_2}} \right|^2} = 1 \\\$$

This cannot be possible because as according to the question ${z_2}$ is not unimodular, i.e. ${\left| {{z_2}} \right|^2} \ne 1$.
Therefore,

$$\Rightarrow \left( {{{\left| {{z_1}} \right|}^2} - 4} \right) = 0 \\\ \Rightarrow {\left| {{z_1}} \right|^2} = 4 \\\ \Rightarrow \left| {{z_1}} \right| = 2 \\\$$

Thus, we can clearly see that this is the locus of circle of radius 2.
Hence, the correct answer is option (C).

Note: Whenever we ask such types of questions, we have to remember some basic points related to complex numbers like modulus, conjugate etc. First, we will use the statements given in the question to make a equation and then by using some basic points we will solve that equation step by step and generate a new solution. After that by equating that solution to zero, we will get the required answer.