For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) = 0?
A2
BOnly n = 3
C1
DAny n > 1
Answer & Solution
Correct answer: D. Any n > 1
Sum 1 + ω + ω² + ... + ω^(n-1) = (ω^n - 1)/(ω - 1) = 0 when ω is a primitive n-th root (ω ≠ 1). So for any n ≥ 2 (with non-trivial primitive root), sum is 0.
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:Find principal argument of -1 - i: