1. Tardigrade
  2. Question
  3. Mathematics
  4. A complex number z is said to be unimodular if |.z|.=1 . Let, z1 and z2 are complex numbers such that ( textz1 - 2z text2/2 - textz1 bar textz2) is unimodular and z2 is not unimodular, then the point z1 lies on a

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A complex number $z$ is said to be unimodular if $\left|\right.z\left|\right.=1$ . Let, $z_{1}$ and $z_{2}$ are complex numbers such that $\frac{\text{z}_{1} - 2z \text{}_{2}}{2 - \text{z}_{1} \bar{\text{z}}_{2}}$ is unimodular and $z_{2}$ is not unimodular, then the point $z_{1}$ lies on a

NTA AbhyasNTA Abhyas 2020Complex Numbers and Quadratic Equations

Solution:

Given, $\frac{\text{z}_{1} - 2z \text{}_{2}}{2 - \text{z}_{1} \bar{\text{z}}_{2}}$ is unimodular
$\Rightarrow \left|\frac{\text{z}_{1} - 2 \text{z}_{2}}{2 - \text{z}_{1} \bar{\text{z}}_{2}}\right| = 1$
$\Rightarrow \left|\text{z}_{1} - 2 \text{z}_{2}\right| = \left|2 - \text{z}_{1} \bar{\text{z}}_{2}\right|$
Squaring both the sides, we get,
$\left|\text{z}_{1} - 2z \text{}_{2}\right|^{2} = \left|2 - \text{z}_{1} \bar{\text{z}}_{2}\right|^{2}$
$\Rightarrow \left(\left(\text{z}\right)_{1} - 2 \left(\text{z}\right)_{2}\right) \left(\left(\bar{\text{z}}\right)_{1} - 2 \left(\bar{\text{z}}\right)_{2}\right) = \left(2 - \left(\text{z}\right)_{1} \left(\bar{\text{z}}\right)_{2}\right) \left(2 - \left(\bar{\text{z}}\right)_{1} \left(\text{z}\right)_{2}\right)$
$\left(\because \left(\left|\text{z}\right|\right)^{2} = \text{z } \bar{\text{z}}\right)$
$\Rightarrow \text{z}_{1} \bar{\text{z}}_{1} - 2 \text{z}_{1} \bar{\text{z}}_{2} - 2 \bar{\text{z}}_{1} \text{z}_{2} + 4 \text{z}_{2} \bar{\text{z}}_{2}$
$= 4 - 2 \bar{\text{z}}_{1} \text{z}_{2} - 2 \text{z}_{1} \bar{\text{z}}_{2} + \text{z}_{1} \bar{\text{z}}_{1} \text{z}_{2} \bar{\text{z}}_{2}$
$\Rightarrow \left|\text{z}_{1}\right|^{2} + 4 \left|\text{z}_{2}\right|^{2} = 4 + \left|\text{z}_{1}\right|^{2} \left|\text{z}_{2}\right|^{2}$
$\Rightarrow \left|\text{z}_{1}\right|^{2} - 4 + 4 \left|\text{z}_{2}\right|^{2} - \left|\text{z}_{1}\right|^{2} \left|\text{z}_{2}\right|^{2} = 0$
$\Rightarrow \left(\left(\left|\left(\text{z}\right)_{1}\right|\right)^{2} - 4\right) \left(1 - \left(\left|\left(\text{z}\right)_{2}\right|\right)^{2}\right) = 0$
$\Rightarrow \left|\text{z}_{1}\right|=2\text{or }\left|\text{z}_{2}\right|=1$
Given, $z_{2}$ is not unimodular
$\therefore \left|\text{z}_{1}\right| = 2$
$\therefore $ Point $z_{1}$ lies on a circle of radius $2$ .