α=ω, β=ω2 (cube roots of unity).
Let z1=7−7ω+9ω2, z2=9+7ω−7ω2, z3=−7+9ω+7ω2.
Calculating: z1=6−8i3, z2=9+7i3, z3=−15+i3.
All have ∣zk∣=228.
Verify: z2=z1ω, z3=z1ω2.
z120+z220+z320=z120(1+ω20+ω40)=z120(1+ω2+ω)=0.
Fourth term: 14+7ω+7ω2=14+7(−1)=7. So 720=4910.
Total =0+4910=m10⇒m=49.