De Moivre's Theorem

Core Concepts

De Moivre's Theorem — Statement

FUNDAMENTAL

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) where n is any integer (extends to rational n for roots)

Works for all integer n (positive, negative, zero)
For rational n = p/q, gives multiple values (q roots exist)
Also written as: (cis θ)ⁿ = cis(nθ), where cis θ = cos θ + i sin θ
Polar form: r(cos θ + i sin θ) — always convert before applying

Polar Form & Modulus-Argument

CONVERSION

z = r(cos θ + i sin θ) = r·cis θ r = |z| = √(a² + b²), θ = arg(z) = arctan(b/a) [with quadrant check]

When multiplying: multiply moduli, add arguments
When dividing: divide moduli, subtract arguments
Argument range: -π < θ ≤ π (principal argument)

nth Roots of a Complex Number

ROOTS

The n roots of z = r·cis α are: z_k = r^(1/n) · cis((α + 2πk)/n), k = 0, 1, 2, ..., n-1

There are always exactly n distinct roots
Roots are equally spaced at angles of 2π/n apart on a circle of radius r^(1/n)
Roots of unity: roots of zⁿ = 1, all lie on the unit circle
If ω is a primitive nth root of unity: 1 + ω + ω² + ... + ω^(n-1) = 0

Trig Identities via De Moivre

IDENTITY

cos(nθ) + i sin(nθ) = (cos θ + i sin θ)ⁿ Expand RHS using binomial theorem, then equate real/imaginary parts

cos(2θ) = cos²θ − sin²θ, sin(2θ) = 2 sin θ cos θ
cos(3θ) = 4cos³θ − 3cos θ, sin(3θ) = 3sin θ − 4sin³θ
Key tool: z + z⁻¹ = 2 cos θ and z − z⁻¹ = 2i sin θ

Worked Examples

Example 1 — Powers

Compute (1 + i)⁸

Convert: |1+i| = √2, arg = π/4, so 1+i = √2·cis(π/4)
(1+i)⁸ = (√2)⁸ · cis(8·π/4) = 16 · cis(2π) = 16(cos 2π + i sin 2π)

= 16

Example 2 — Cube Roots of Unity

Find the cube roots of 1 and verify their sum is 0.

z³ = 1 = cis(0). Roots: z_k = cis(2πk/3), k = 0,1,2
z₀ = 1, z₁ = cis(2π/3) = −½ + i(√3/2), z₂ = cis(4π/3) = −½ − i(√3/2)
Sum = 1 + (−½ + i√3/2) + (−½ − i√3/2)

= 1 − 1 = 0 ✓

Example 3 — Trig Identity

Express cos(3θ) in terms of cos θ using De Moivre's theorem.

(cos θ + i sin θ)³ = cos(3θ) + i sin(3θ)
Expand LHS: cos³θ + 3cos²θ(i sin θ) + 3cos θ(i sin θ)² + (i sin θ)³
= cos³θ − 3cos θ sin²θ + i(3cos²θ sin θ − sin³θ)
Real part: cos(3θ) = cos³θ − 3cos θ(1−cos²θ)

= 4cos³θ − 3cos θ