Let integers a, b ∈[-3,3] be such that a+ b ≠ 0. Then the number of all possible ordered pairs (a, b), for which | z- a z+ b |=1 and | … |=1, z ∈ C,…

$\begin{aligned} & a, b \in I,-3 \leq a, b \leq 3, a+b \neq 0 \\ & |z-a|=|z+b| \\ & \left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1 \\ & \Rightarrow\left|\begin{array}{ccc}z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1\end{aligned}$​a,b∈I,−3≤a,b≤3,a+b=0∣z−a∣=∣z+b∣​z+1ωω2​ωz+ω21​ω21z+ω​​=1⇒​zωω2​zz+ω21​z1z+ω​​=1​ $\begin{aligned} & \Rightarrow z\left|\begin{array}{ccc}1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1 \\ & \Rightarrow z\left|\begin{array}{ccc}1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2\end{array}\right|=1 \\ & \Rightarrow z^3=1 \\ & \Rightarrow z=\omega, \omega^2, 1\end{aligned}$​⇒z​1ωω2​1z+ω21​11z+ω​​=1⇒z​1ωω2​0z+ω2−ω1−ω2​01−ωz+ω−ω2​​=1⇒z3=1⇒z=ω,ω2,1​ Now $\begin{aligned}
& |1-\mathrm{a}|=|1+\mathrm{b}| \
& \Rightarrow 10 \text { pairs }
\end{aligned}$