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Question:

Let z0 be a root of the quadratic equation x² + x + 1 = 0. If z = 3 + 6i(z0)^810 - i(z0)^930, then arg(z) is equal to: π/4, π/3, π/6, 0

π4

0

π3

π6

Solution:

We know that z0 = ω or ω² (where ω is a non-real cube root of unity)
z = 3 + 6i(ω)^810 - i(ω)^930
z = 3 + 3i [∵ ω³ = 1]
z = 3 + 3i
comparing with z = x + iy ⇒ arg(z) = tan⁻¹(y/x) = π/4