A complex number z is said to be unimodular if |z|=1. Suppose z1 and z2 are complex numbers such that z1−2z22−z1z¯2 is unimodular and z2 is not unimodular. Then the point z1 lies on a

z1−2z22−z1z¯2=1z1−2z2z¯1−2z¯2=2−z1z¯22−z¯1z2⇒z12−2z2z¯1−2z¯2z1+4z22=4−2z1z¯2−2z¯1z2+z12z22⇒z12+4z22−4-z12z22=0z121−z22−41−z22=0⇒z1=2 as z2≠1

A complex number z is said to be unimodular if |z|=1. Suppose z1 and z2 are complex numbers such that z1−2z22−z1z¯2 is unimodular and z2 is not unimodular. Then the point z1 lies on a

z1−2z22−z1z¯2=1z1−2z2z¯1−2z¯2=2−z1z¯22−z¯1z2⇒z12−2z2z¯1−2z¯2z1+4z22=4−2z1z¯2−2z¯1z2+z12z22⇒z12+4z22−4-z12z22=0z121−z22−41−z22=0⇒z1=2 as z2≠1

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A complex number z is said to be unimodular if $| z | = 1$ . Suppose $z_{1} and z_{2}$ are complex numbers such that $\frac{z_{1} - 2 z_{2}}{2 - z_{1} {\bar{z}}_{2}}$ is unimodular and z₂ is not unimodular. Then the point z₁ lies on a

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Straight line parallel to x–axis

Circle of radius $\sqrt{2}$

Straight line parallel to y–axis

Circle of radius 2

answer is C.

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Detailed Solution

$\begin{array}{l} |\frac{z_{1} - 2 z_{2}}{2 - z_{1} {\bar{z}}_{2}}| = 1 \\ (z_{1} - 2 z_{2}) ({\bar{z}}_{1} - 2 {\bar{z}}_{2}) = (2 - z_{1} {\bar{z}}_{2}) (2 - {\bar{z}}_{1} z_{2}) \\ \Rightarrow {|z_{1}|}^{2} - 2 z_{2} {\bar{z}}_{1} - 2 {\bar{z}}_{2} z_{1} + 4 {|z_{2}|}^{2} \\ = 4 - 2 z_{1} {\bar{z}}_{2} - 2 {\bar{z}}_{1} z_{2} + {|z_{1}|}^{2} {|z_{2}|}^{2} \\ \Rightarrow {|z_{1}|}^{2} + 4 {|z_{2}|}^{2} - 4 - {|z_{1}|}^{2} {|z_{2}|}^{2} = 0 \\ {|z_{1}|}^{2} (1 - {|z_{2}|}^{2}) - 4 (1 - {|z_{2}|}^{2}) = 0 \\ \Rightarrow |z_{1}| = 2 (as |z_{2}| \neq 1) \end{array}$