Complex Numbers and Quadratic Equations

Complex number z = a + bi can be written in polar form:

z = r(cos θ + i sin θ) = r·e^(iθ), where r = |z|, θ = arg(z).

De Moivre's theorem:

(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)

Equivalently, (r·e^(iθ))^n = rⁿ · e^(inθ).

Why this matters: raising complex numbers to powers becomes trivial. (1 + i)¹⁰⁰ would be brutal directly; in polar form:
|1 + i| = √2, arg = π/4. So (1+i)¹⁰⁰ = (√2)¹⁰⁰ · (cos(25π) + i sin(25π)) = 2⁵⁰ · (cos π + i sin π) = −2⁵⁰.

n-th roots of complex number z = r·e^(iθ):

z^(1/n) = r^(1/n) · e^(i(θ + 2kπ)/n), k = 0, 1, ..., n−1.

There are exactly n distinct n-th roots, equally spaced on a circle of radius r^(1/n) centered at origin.

n-th roots of unity (z = 1, so r = 1, θ = 0):

ω_k = e^(2πki/n) = cos(2πk/n) + i sin(2πk/n), k = 0, 1, ..., n−1.

Properties:

• All n-th roots of unity are equally spaced on the unit circle.
• They form a regular n-gon inscribed in unit circle.
• Sum: 1 + ω + ω² + ... + ω^(n−1) = 0 (for n ≥ 2).
• Product: 1 · ω · ω² · ... · ω^(n−1) = (−1)^(n−1).
• If ω is any primitive n-th root, then ω^n = 1 and {1, ω, ω², ..., ω^(n−1)} = all n-th roots.

Cube roots of unity: 1, ω, ω² where ω = e^(2πi/3) = −½ + (√3/2)i, ω² = −½ − (√3/2)i.

• ω³ = 1, 1 + ω + ω² = 0, ω² = ω̄ (complex conjugate).
• Identities: x³ − 1 = (x−1)(x−ω)(x−ω²) = (x−1)(x² + x + 1).

Worked example. Find all cube roots of 8.

8 = 8·e^(i·0). Cube roots: 8^(1/3) · e^(i·2πk/3) = 2·e^(i·2πk/3), k = 0, 1, 2.
k=0: 2 (real). k=1: 2·(cos 120° + i sin 120°) = −1 + i√3. k=2: −1 − i√3.

So cube roots of 8 are 2, −1 + i√3, −1 − i√3.