1998

JEE Main 1998 Mathematics question on Complex Numbers and Quadratic Equations

If ω\omega is an imaginary cube root of unity, then (1+ω−ω2)7(1 + \omega - \omega^2)^7 is equal to
  1. A.
    128ω128\omega
  2. B.
    −128ω-128\omega
  3. C.
    128ω2128\omega^2
  4. D.
    −128ω2-128\omega^2
Show answer and solution

Correct option: D

Solution

Since ω is a primitive cube root of unity, 1 + ω + ω² = 0, so 1 + ω = -ω².

Therefore (1 + ω - ω²) = -ω² - ω² = -2ω².

Raising to the 7th power: (-2ω²)⁷ = -128 · ω¹⁴ = -128 · ω² (since ω³ = 1, so ω¹⁴ = ω²).

The answer is -128ω².

[1998]

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