Home › JEE Main › Mathematics › Complex Numbers › Cube roots of unity are 1, ω, ω². Find 1 + ω + ω²:
Cube roots of unity are 1, ω, ω². Find 1 + ω + ω²:
Aω³
B0
C3
D1
Answer & Solution
Correct answer: B. 0
Roots of z³ - 1 = 0 are 1, ω, ω². Sum of roots = -coefficient of z² / coefficient of z³ = 0. So 1 + ω + ω² = 0. Useful identity for many problems.
Related questions
The number of distinct $n$th roots of a non-zero complex number is:Using De Moivre's theorem, $(\cos\theta + i in\theta)^4$ equals:The modulus of $z = 5 + 12i$ is:The conjugate of $z = 3 - 4i$ is:Find (3 + 4i)/(1 + 2i):If z1, z2, z3 are vertices of an equilateral triangle, then:Solve: z² + z + 1 = 0:For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) =