Let integers $a,b\in[-3,3]$ be such that $a+b\ne 0$. Then the number of all possible ordered pairs $(a,b)$, for which $\left|\dfrac{z-a}{z+b}\right|=1$ and $\left|\,\begin{matrix} z+1 & \omega & \omega^{2}\\ \omega & z+\omega^{2} & 1\\ \omega^{2} & 1 & z+\omega\end{matrix}\,\right|=1$, where $\omega$ and $\omega^{2}$ are the complex cube roots of unity, is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: The determinant identity for $\omega$: row-summing gives a factor of $z$, then column reduction yields $z^{3}=1$, so $z\in\{1,\omega,\omega^{2}\}$. The other condition $|z-a|=|z+b|$ places $z$ on the perpendicular bisector of $a$ and $-b$ — a vertical line $\operatorname{Re}(z)=\dfrac{a-b}{2}$.
\textbf{Determinant condition.} Add all three columns to column 1; the new column 1 becomes $(z+1+\omega+\omega^{2},\ z+1+\omega+\omega^{2},\ z+1+\omega+\omega^{2})^{T}=(z,z,z)^{T}$ (using $1+\omega+\omega^{2}=0$). Pull out $z$, perform row reductions, and the determinant simplifies to $z^{3}=1$, giving $z\in\{1,\omega,\omega^{2}\}$.
\textbf{Modulus condition.} $|z-a|=|z+b|\Longleftrightarrow z$ lies on the perpendicular bisector of $a$ and $-b$, i.e.\ $\operatorname{Re}(z)=\dfrac{a-b}{2}$ (since $a,b\in\mathbb{Z}$ are real).
\textbf{Case } $z=1$: $\operatorname{Re}(1)=1\Rightarrow a-b=2$. Pairs $(a,b)$ with $a,b\in[-3,3]$: $(-1,-3),(0,-2),(1,-1),(2,0),(3,1)$. Exclude $(1,-1)$ since $a+b=0$. $\Rightarrow 4$ pairs.
\textbf{Case } $z=\omega$ (or $\omega^{2}$, same equation): $\operatorname{Re}(\omega)=-\tfrac{1}{2}\Rightarrow a-b=-1$. Pairs: $(-3,-2),(-2,-1),(-1,0),(0,1),(1,2),(2,3)$. None of these satisfy $a+b=0$. $\Rightarrow 6$ pairs.
\textbf{Total.} $4+6 = 10.$
Correct Answer: 10