AMC12 Complex Numbers 2 — Questions and Answers

Correct answer: 0

The cube roots of unity 1, ω, ω² are roots of z³ − 1 = 0, and by Vieta's formulas their sum equals 0.

Correct answer: 1

By De Moivre's theorem: cos(8·π/4) + i·sin(8·π/4) = cos(2π) + i·sin(2π) = 1.

Correct answer: 4

A degree-4 polynomial has exactly 4 complex roots; the four 4th roots of unity are 1, i, −1, −i.

Correct answer: 2

In polar form z = r·(cos θ + i·sin θ), the modulus is r = 2.

Correct answer: 2π/3

The point −1 + i√3 lies in the second quadrant; tan⁻¹(√3/1) = π/3, so the argument is π − π/3 = 2π/3.

Correct answer: 1

Since 100 ≡ 0 (mod 4), i^100 = (i⁴)^25 = 1^25 = 1.

Correct answer: −1 + i

(1+i)² = 2i; then 2i/(1−i) · (1+i)/(1+i) = 2i(1+i)/2 = i + i² = −1 + i.

AMC12 Complex Numbers 2

What is the sum of all cube roots of unity?

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