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AIEEE 2004

JEE Main 2004 Mathematics question on Complex Numbers and Quadratic Equations

If ω(≠1)\omega (\neq 1) be a cube root of unity and (1+ω2)n=(1+ω4)n(1 + \omega^2)^n = (1 + \omega^4)^n, then the least positive value of nn is
  1. A.
    2
  2. B.
    3
  3. C.
    5
  4. D.
    6
Show answer and solution

Correct option: B

Solution

Since ω³ = 1, we have ω⁴ = ω. Also, 1 + ω² = -ω (since 1 + ω + ω² = 0) and 1 + ω⁴ = 1 + ω = -ω². So the equation becomes (-ω)ⁿ = (-ω²)ⁿ, which gives ωⁿ = ω²ⁿ, i.e., ω²ⁿ/ωⁿ = 1, so ωⁿ = 1. The least positive n for which ωⁿ = 1 is n = 3.

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