a,b∈I,−3≤a,b≤3,a+b=0∣z−a∣=∣z+b∣z+1ωω2ωz+ω21ω21z+ω=1⇒zωω2zz+ω21z1z+ω=1 ⇒z1ωω21z+ω2111z+ω=1⇒z1ωω20z+ω2−ω1−ω201−ωz+ω−ω2=1⇒z3=1⇒z=ω,ω2,1 Now $\begin{aligned}
& |1-\mathrm{a}|=|1+\mathrm{b}| \
& \Rightarrow 10 \text { pairs }
\end{aligned}$