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1 solution
Moderator note:
Simple standard approach.
An alternative approach would be to use the identity z z ˉ = ∣ z ∣ 2 and to recall that the absolute value of all roots of unity is 1 .
Hence, a general result would be, ∀ z ∈ C , ∣ z ∣ = 1 ⟺ z ˉ = z 1
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And by the same token the complex conjugate of ω 2 1 = ω is ω 2 . :)
I suppose that we could also just note that, since the cube roots of unity are 1 , ω , ω 2 , and since complex roots occur in conjugate pairs, ω and ω 2 must be each other's conjugate.
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Yes , that was a shortcut :P
Why do complex roots appear in conjugate pairs?
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This is known as the complex conjugate root theorem . A proof of the theorem is provided in the link.
First note that ω 1 = ω 3 ω 2 = ω 2 = − 2 1 − 2 i 3
Thus complex conjugate of ω 2 = − 2 1 + 2 i 3 = ω