De Moivre's theorem and roots of unity

Statement and Proof of De Moivre's Theorem

De Moivre's theorem states that for any complex number in polar form and any integer n:

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

More generally, if z = r(cos θ + i sin θ) then zⁿ = rⁿ(cos nθ + i sin nθ). The theorem is fundamental to Unit 3 Further Complex Numbers in QCAA Specialist Mathematics and must be proved rigorously.

Proof for positive integers by mathematical induction:

Base case (n = 1): (cos θ + i sin θ)¹ = cos θ + i sin θ = cos(1·θ) + i sin(1·θ). True.

Inductive step: Assume the result holds for some positive integer k, i.e., (cos θ + i sin θ)^k = cos(kθ) + i sin(kθ). Consider n = k + 1:

• (cos θ + i sin θ)^k+1 = (cos θ + i sin θ)^k · (cos θ + i sin θ)
• = [cos(kθ) + i sin(kθ)] · (cos θ + i sin θ)    (inductive hypothesis)
• = cos(kθ)cos θ − sin(kθ)sin θ + i[cos(kθ)sin θ + sin(kθ)cos θ]
• = cos(kθ + θ) + i sin(kθ + θ)    (compound angle identities)
• = cos((k+1)θ) + i sin((k+1)θ)

The result holds for n = k + 1. By the principle of mathematical induction, the theorem holds for all positive integers.

Extension to n = 0: (cos θ + i sin θ)⁰ = 1 = cos 0 + i sin 0. Consistent.

Extension to negative integers: For n = −m, where m is a positive integer:

• (cos θ + i sin θ)^−m = 1/[cos(mθ) + i sin(mθ)]    (using positive integer case)
• Multiply numerator and denominator by the conjugate: = [cos(mθ) − i sin(mθ)] / [cos²(mθ) + sin²(mθ)]
• = cos(mθ) − i sin(mθ) = cos(−mθ) + i sin(−mθ) = cos(nθ) + i sin(nθ) ✓

The theorem therefore holds for all n ∈ ℤ. Note carefully: for rational n = p/q, De Moivre's theorem yields one valid value, not the complete set of roots. This distinction is essential when finding nth roots.

Multiplying and Dividing in Polar Form: The Geometric Basis

The inductive proof above rests on the product rule for complex numbers in polar form. Understanding this geometrically clarifies why De Moivre's theorem takes the form it does.

If z₁ = r₁(cos α + i sin α) and z₂ = r₂(cos β + i sin β), then:

• z₁z₂ = r₁r₂[cos(α + β) + i sin(α + β)]

That is: moduli multiply, arguments add. Repeated multiplication by the same number z multiplies the modulus by r each time and adds θ to the argument each time. After n multiplications: modulus becomes rⁿ, argument becomes nθ. This is De Moivre's theorem, expressed geometrically as repeated rotation and scaling on the Argand diagram.

Worked example — Simplification using De Moivre's theorem:

Evaluate (1 + i)⁸.

• Convert to polar form: |1 + i| = √2, arg(1 + i) = π/4.
• So 1 + i = √2(cos π/4 + i sin π/4).
• By De Moivre: (√2)⁸(cos(8 · π/4) + i sin(8 · π/4)) = 16(cos 2π + i sin 2π) = 16(1 + 0i) = 16.

Worked example — Division:

Simplify (cos(π/5) + i sin(π/5))⁷ / (cos(π/5) + i sin(π/5))³.

• = (cos(π/5) + i sin(π/5))⁴ = cos(4π/5) + i sin(4π/5).

In Cartesian form, cos(4π/5) = −cos(π/5) and sin(4π/5) = sin(π/5), using supplementary angle identities. These values are not standard unit-circle values, so the polar form is the preferred final answer unless a decimal approximation is requested. Always reduce the power using index laws before applying De Moivre's theorem.