Let integers be such that . Then the number of all possible ordered pairs (a, b) for which and , where and are the roots of , is equal to __________. [2025]

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Solution

Q.
Let integers $a, b \in [- 3, 3]$ be such that $a + b \neq 0$ . Then the number of all possible ordered pairs (a, b) for which $| \frac{z - a}{z + b} |$ $= 1$ and $| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} |$ $= 1, z \in C$ , where $ω$ and $ω^{2}$ are the roots of $x^{2} + x + 1 = 0$ , is equal to __________. [2025]

Ans.
(10)

We have, $a, b \in z, - 3 \leq a, b \leq 3$ and $a + b \neq 0$ .

Also, $| \frac{z - a}{z + b} |$ $= 1$ and $| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} |$ $= 1$

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$

$z$ $| \begin{matrix} 1 & ω & ω^{2} \\ 1 & z + ω^{2} & 1 \\ 1 & 1 & z + ω \end{matrix} |$ $= 1$ [ $∵ 1 + ω + ω^{2} = 0$ ]

On expanding, we get

$z [(z^{2} + (ω^{2} + ω) z + ω^{3} - 1) - ω z - ω^{2} - ω^{2} + ω - ω^{2} z - ω] = 1$

$\Rightarrow z (z^{2}) = 1 \Rightarrow z^{3} = 1 \Rightarrow z = 1, ω, ω^{2}$

Case 1 : $z = ω$ , then $a + b \neq 0$ and a – b = –1

a = –3, b = –2; a = –2; b = –1;

a = –1, b = 0; a = 0, b = 1

a = 1, b = 2; a = 2, b = 3

Case 2 : z = 1; then a – b = 2 and $a + b \neq 0$

a = –1, b = –3; a = 0, b = –2; a = 2, b = 0; a = 3, b = 1

Total pairs = 10.

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