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 } \( a, b \in [-3, 3] \), \( a + b \neq 0 \). We are 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{Using the fact that } \( \omega \) \text{ and } \(\omega^2\) \text{ are the roots of } \(x^2 + x + 1\) = 0, we can proceed as follows: \[ \left| \frac{z-a}{z+b} \right| = |z - a| = |z + b| \] \text{From this, we know that } \( |z - a| = |z + b| \). \text{Next, solve for } z: \[ z^2 = 1 \quad \Rightarrow \quad z = \omega, \omega^2, 1 \] \text{Now, compute the possible values for } a \text{ and } b: \[ | - a | = | + b | \] \text{Thus, we get 10 possible ordered pairs for } (a, b).