Roots of Unity

[MathDefinition title="nth Roots of Unity"] The nth roots of unity are the solutions to the equation zⁿ = 1. There are exactly n distinct roots, given by: z_k = cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, 2, ..., n−1 [/MathDefinition]

[MathDefinition title="Properties of nth Roots of Unity"]

1. The nth roots of unity form a regular n-sided polygon inscribed in the unit circle
2. The sum of all nth roots of unity is 0 (for n > 1)
3. The product of all nth roots of unity is (−1)ⁿ⁻¹
4. They are equally spaced on the unit circle at angles 2kπ/n [/MathDefinition]

[MathKeyFormula title="Key Formulas for Roots of Unity"]

1. nth roots of unity: z_k = e^(2kπi/n) = cos(2kπ/n) + i sin(2kπ/n)
2. ω = e^(2πi/n) is a primitive nth root of unity
3. 1 + ω + ω² + ... + ωⁿ⁻¹ = 0
4. ωⁿ = 1
5. For cube roots: 1, ω, ω² where ω = e^(2πi/3) = −1/2 + i√3/2
6. 1 + ω + ω² = 0 and ω³ = 1 [/MathKeyFormula]

[MathNote title="Understanding Roots of Unity"] The nth roots of unity divide the unit circle into n equal parts. The root corresponding to k = 0 is always 1. The other roots are complex numbers equally spaced around the circle. [/MathNote]

[MathExample title="Example 1: Cube Roots of Unity"] Problem: Find the cube roots of unity and verify their sum is 0.

Solution: The cube roots of unity satisfy z³ = 1.

z = 1^(1/3) = e^(2kπi/3) for k = 0, 1, 2

For k = 0: z₀ = e⁰ = 1 For k = 1: z₁ = e^(2πi/3) = cos(2π/3) + i sin(2π/3) = −1/2 + i√3/2 For k = 2: z₂ = e^(4πi/3) = cos(4π/3) + i sin(4π/3) = −1/2 − i√3/2

Sum: z₀ + z₁ + z₂ = 1 + (−1/2 + i√3/2) + (−1/2 − i√3/2) = 1 − 1/2 − 1/2 + i(√3/2 − √3/2) = 0

The cube roots of unity are: 1, −1/2 + i√3/2, −1/2 − i√3/2 Their sum is 0. [/MathExample]

[MathExample title="Example 2: Fourth Roots of Unity"] Problem: Find the fourth roots of unity and represent them on the complex plane.

Solution: The fourth roots of unity satisfy z⁴ = 1.

z = e^(2kπi/4) = e^(kπi/2) for k = 0, 1, 2, 3

For k = 0: z₀ = e⁰ = 1 For k = 1: z₁ = e^(πi/2) = cos(π/2) + i sin(π/2) = i For k = 2: z₂ = e^(πi) = cos π + i sin π = −1 For k = 3: z₃ = e^(3πi/2) = cos(3π/2) + i sin(3π/2) = −i

The fourth roots of unity are: 1, i, −1, −i

On the complex plane, these form a square with vertices at (1,0), (0,1), (−1,0), (0,−1). [/MathExample]

[MathExample title="Example 3: Sum of Squares"] Problem: Find the sum of squares of the cube roots of unity.

Solution: The cube roots of unity are: 1, ω, ω² where ω = −1/2 + i√3/2

Sum of squares = 1² + ω² + (ω²)² = 1 + ω² + ω⁴ = 1 + ω² + ω³·ω = 1 + ω² + 1·ω [since ω³ = 1] = 1 + ω + ω² = 0 [property of cube roots of unity]

Therefore, the sum of squares of the cube roots of unity is 0. [/MathExample]

[MathExample title="Example 4: Product of Roots"] Problem: Find the product of the nth roots of unity.

Solution: The nth roots of unity are: z_k = e^(2kπi/n) for k = 0, 1, ..., n−1

Product = z₀ × z₁ × ... × z_{n−1} = e⁰ × e^(2πi/n) × e^(4πi/n) × ... × e^(2(n−1)πi/n) = e^(0 + 2πi/n + 4πi/n + ... + 2(n−1)πi/n) = e^(2πi/n × (0 + 1 + 2 + ... + (n−1))) = e^(2πi/n × n(n−1)/2) = e^(πi(n−1)) = (e^(πi))^(n−1) = (−1)^(n−1)

Therefore, the product of the nth roots of unity is (−1)^(n−1). [/MathExample]