Let integers a, b ∈ [-3 , 3] be such that a+b ≠ 0. Then the number of all possible ordered pairs (a,b) for which frac z - a z + b = 1\;and\; begin array 20 c z + 1 &ω & ω 2 \\ω & z + ω 2 &1\\ ω 2 &1& z + ω end array =1, Z∈ C, where ω \;and\; ω 2 are the roots x 2 + x + 1 = 0,\;is\;equal\;to\;

Mathematics · Complex Numbers

JEE Main 2025 — 29 January, Evening Shift — Question 65

Let integers a, b ∈ [-3 , 3] be such that a+b ≠ 0.

Then the number of all possible ordered pairs (a,b) for which

$\left| {\frac{{z - a}}{{z + b}}} \right| = 1\;and\;\left| {\begin{array}{*{20}{c}}{z + 1}&\omega &{{\omega ^2}}\\\omega &{z + {\omega ^2}}&1\\{{\omega ^2}}&1&{z + \omega }\end{array}} \right|$ =1, Z∈ C, where $\omega \;and\;{\omega ^2}$ are the roots ${x^2} + x + 1 = 0,\;is\;equal\;to\;$

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

a, b \in I, \quad -3 \leq a, b \leq 3, \quad a + b \neq 0 \\ |z - a| = |z + b| \\ \begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 1 \\ \Rightarrow \begin{vmatrix} z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 1 \\ \Rightarrow z \begin{vmatrix} 1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 1 \\ \Rightarrow z \begin{vmatrix} 1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2 \end{vmatrix} = 1 \\ \Rightarrow z^3 = 1 \\ \Rightarrow z = \omega, \, \omega^2, \, 1 \\ \text{Now} \\ |1 - a| = |1 + b| \\ \Rightarrow 10 \text{ pairs} \end{gathered}$$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers