Question

Let integers \( a, b \in [-3, 3] \) be such that \( a + b \neq 0 \). \(\text{Then the number of all possible ordered pairs}\) \( (a, b) \), \(\text{for which}\) \[ \left| \frac{z - a}{z + b} \right| = 1 \quad \text{and} \quad \left| \begin{matrix} z + 1 & \omega & \omega^2 \\ \omega^2 & 1 & z + \omega \\ \omega^2 & 1 & z + \omega \end{matrix} \right| = 1, \] \(\text{is equal to:}\) 

Hint:

For problems involving modulus and complex numbers, simplify using roots of unity and utilize symmetry to count valid pairs.

Answer 1

Correct Answer: 10

\(\text{Let }\) \( a, b \in [-3, 3] \), \( a + b eq 0 \). Given the conditions: 
\[ \left| \frac{z - a}{z + b} \right| = 1 \quad \text{and} \quad \left| \begin{matrix} z + 1 & \omega & \omega^2 \\ \omega^2 & 1 & z + \omega \\ \omega^2 & 1 & z + \omega \end{matrix} \right| = 1 \] \(\text{Since }\) \( \omega \) \(\text{ and }\) \(\omega^2\) \(\text{ are the roots of }\) \(x^2 + x + 1 = 0\), we have: 
\[ \left| \frac{z - a}{z + b} \right| = |z - a| = |z + b| \] \(\text{This implies}\) \( |z - a| = |z + b| \). \(\text{Solving for } z\) yields: 
\[ z^2 = 1 \quad \Rightarrow \quad z = \omega, \omega^2, 1 \] \(\text{The possible values for } a \text{ and } b \text{ are determined by}\) :
\[ | - a | = | + b | \] \(\text{Consequently, there are 10 possible ordered pairs for } (a, b).\)