ω=−21+23i (cube root of unity)
So, 1+ω+ω2=0 and ω3=1
Now ∣1111ωω21ω2ω∣=3(ω2−ω)
So k=ω2−ω
⇒k=(−21−23i)−(−21+23i)
⇒k=−3i
⇒k=−z
Let ω be a complex number such that 2ω+1=z where z=−3 . If
∣1111−ω2−1ω21ω2ω7∣=3k,
Then k can be equal to:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
–z
z1
−1
1
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