Question ID: #579
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 $\left|\frac{z-a}{z+b}\right|=1$ and $\left|\begin{matrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{matrix}\right| = 1$, $z \in C$ where $\omega$ and $\omega^2$ are the roots of $x^2+x+1=0$, is equal to
Solution:
First, solve the determinant equation. Applying $C_1 \to C_1+C_2+C_3$ and using $1+\omega+\omega^2=0$:
$$ \left|\begin{matrix} z & \omega & \omega^2 \\ z & z+\omega^2 & 1 \\ z & 1 & z+\omega \end{matrix}\right| = 1 \Rightarrow z^3 = 1 \Rightarrow z = 1, \omega, \omega^2\ $$
The second condition is $|z-a|=|z+b|$, meaning $z$ is equidistant from $a$ and $-b$.
Case 1: $z=1$
$$ |1-a| = |1+b| \Rightarrow (1-a)^2 = (1+b)^2 \Rightarrow 1-a = \pm(1+b) $$
Since $a+b \neq 0$, we must have $1-a = -(1+b) \Rightarrow a-b=2$.
Pairs in $[-3, 3]$: $(3, 1), (2, 0), (0, -2), (-1, -3)$. (4 pairs).
Case 2: $z=\omega$
$$ |\omega-a| = |\omega+b| \Rightarrow (\omega-a)(\bar{\omega}-a) = (\omega+b)(\bar{\omega}+b) $$
$$ \omega\bar{\omega} – a(\omega+\bar{\omega}) + a^2 = \omega\bar{\omega} + b(\omega+\bar{\omega}) + b^2 $$
$$ 1 – a(-1) + a^2 = 1 + b(-1) + b^2 \Rightarrow a^2+a = b^2-b $$
$$ a(a+1) = b(b-1) $$
Checking integers in $[-3, 3]$ excluding $a+b=0$:
Pairs: $(1, 2), (2, 3), (0, 1), (-1, 0), (-2, -1), (-3, -2)$. (6 pairs).
Total pairs = $4 + 6 = 10$.
Ans. 10
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