NDA 2024 · PYQ · Complex Numbers / Cube Roots of Unity · medium
If ω ≠ 1 is a cube root of unity, then what are the solutions of (z − 100)³ + 1000 = 0?
- A.10(1 − ω), 10(10 − ω²), 100
- B.10(10 − ω), 10(10 − ω²), 90✓ Correct
- C.10(1 − ω), 10(10 − ω²), 1000
- D.(1 + ω), (10 + ω²), −1
(z−100)³ = −1000, so z−100 = −10·(cube roots of unity) = −10, −10ω, −10ω². Thus z = 100−10 = 90, z = 100−10ω = 10(10−ω), z = 100−10ω² = 10(10−ω²). The solutions are 10(10−ω), 10(10−ω²), 90.
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