Let integers $a, b \in [-3,3]$ be such that $a+b\neq0$. Then the number of all possible ordered pairs $(a,b)$ for which $\left|\dfrac{z-a}{z+b}\right|=1$ and $\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix}=1$, $z\in\mathbb{C}$, where $\omega$ and $\omega^2$ are the roots of $x^2+x+1=0$, is equal to
Step-by-Step Solution
Key Concept: Evaluate the determinant (it equals $z^3$) to get $z^3=1$, giving $z=1,\omega,\omega^2$; then the perpendicular-bisector condition $|z-a|=|z+b|$ constrains $\text{Re}(z)=(a-b)/2$, yielding two families of integer pairs.
**Step 1 — Evaluate the determinant.**
Adding all three columns: each row sums to $z+(1+\omega+\omega^2)=z+0=z$. Factor out $z$ from $C_1\to C_1+C_2+C_3$, then subtract Row 1 from Rows 2 and 3. After simplification using $\omega-\omega^2=i\sqrt{3}$ and $1+\omega+\omega^2=0$:
$$D = z\bigl[(z+\omega^2-\omega)(z+\omega-\omega^2)-(1-\omega^2)(1-\omega)\bigr] = z(z^2+3-3)=z^3.$$
So $z^3=1\Rightarrow z\in\{1,\omega,\omega^2\}$.
**Step 2 — Perpendicular bisector condition.**
$|z-a|=|z+b|$ is the perpendicular bisector of $a$ and $-b$ on the real axis: $\text{Re}(z)=\dfrac{a-b}{2}$.
$z=1$: $\text{Re}=1\Rightarrow a-b=2$.
$z=\omega$ or $\omega^2$: $\text{Re}=-\tfrac{1}{2}\Rightarrow a-b=-1$.
**Step 3 — Count integer pairs in $[-3,3]$ with $a+b\neq0$.**
$a-b=2$: pairs $(b+2,b)$ with $b\in\{-3,-2,-1,0,1\}$; exclude $b=-1$ (gives $a+b=0$). Valid: 4.
$a-b=-1$: pairs $(b-1,b)$ with $b\in\{-2,-1,0,1,2,3\}$; none have $a+b=0$ (since $a+b=2b-1$ is odd and non-zero for all integer $b$). Valid: 6.
Total: $4+6=\mathbf{10}$.
Correct Answer: 10