De Moivre's Theorem
De Moivre's Theorem
De Moivre's theorem is one of the most powerful tools in A-Level Further Maths, connecting complex number multiplication with trigonometric identities. It states that for any complex number in polar form and any integer n:
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
Polar (Modulus-Argument) Form Recap
Every complex number z = a + bi can be written as:
z = r(cos θ + i sin θ) or equivalently z = re^(iθ)
where:
- r = |z| = √(a² + b²) is the modulus
- θ = arg(z) is the argument (angle from the positive real axis)
Statement of De Moivre's Theorem
For any integer n (positive, negative, or zero):
| Form | Expression |
|---|---|
| Trigonometric | (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ |
| Exponential | (e^(iθ))ⁿ = e^(inθ) |
| With modulus | [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) |
Proof by Induction (Positive Integers)
Base case (n = 1): (cos θ + i sin θ)¹ = cos θ + i sin θ = cos(1·θ) + i sin(1·θ) ✓
Inductive step: Assume true for n = k, so (cos θ + i sin θ)^k = cos kθ + i sin kθ.
For n = k + 1:
- (cos θ + i sin θ)^(k+1) = (cos kθ + i sin kθ)(cos θ + i sin θ)
- Expanding: = cos kθ cos θ − sin kθ sin θ + i(sin kθ cos θ + cos kθ sin θ)
- By addition formulae: = cos(k+1)θ + i sin(k+1)θ ✓
Application 1: Finding Powers of Complex Numbers
To compute (1 + i)^10:
1. Convert to polar: |1 + i| = √2, arg = π/4
2. So 1 + i = √2(cos π/4 + i sin π/4)
3. Apply De Moivre: (√2)^10 (cos 10π/4 + i sin 10π/4)
4. = 32(cos 5π/2 + i sin 5π/2) = 32(0 + i) = 32i
Application 2: Deriving Trigonometric Identities
Express cos 3θ in terms of cos θ:
- (cos θ + i sin θ)³ = cos 3θ + i sin 3θ (by De Moivre)
- Expand the left side using the binomial theorem:
- cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ
- Equating real parts: cos 3θ = cos³θ − 3 cos θ sin²θ = 4cos³θ − 3cos θ
- Equating imaginary parts: sin 3θ = 3cos²θ sin θ − sin³θ = 3sin θ − 4sin³θ
Application 3: Roots of Unity
The nth roots of unity are the solutions to zⁿ = 1. Using De Moivre:
z_k = cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, 2, ..., n−1
Key properties:
- There are exactly n distinct nth roots of unity
- They are equally spaced around the unit circle (separated by 2π/n)
- They sum to zero: z₀ + z₁ + ... + z_(n−1) = 0
- The primitive root ω = e^(2πi/n) generates all others: z_k = ωᵏ
Example: The cube roots of unity are 1, e^(2πi/3), e^(4πi/3), which simplify to:
- z₀ = 1
- z₁ = −1/2 + i√3/2
- z₂ = −1/2 − i√3/2
Application 4: Roots of Any Complex Number
To find the nth roots of w = R(cos φ + i sin φ):
z_k = R^(1/n) [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] for k = 0, 1, ..., n−1
Example: Find the square roots of 2i.
- 2i = 2(cos π/2 + i sin π/2)
- z₀ = √2(cos π/4 + i sin π/4) = 1 + i
- z₁ = √2(cos 5π/4 + i sin 5π/4) = −1 − i
The z + 1/z Technique
If z = cos θ + i sin θ, then:
- z + z⁻¹ = 2 cos θ
- z − z⁻¹ = 2i sin θ
- zⁿ + z⁻ⁿ = 2 cos nθ
- zⁿ − z⁻ⁿ = 2i sin nθ
This lets you express powers of cos and sin as sums of multiple angles.
Example: Express cos⁴θ in terms of multiple angles.
- (2cos θ)⁴ = (z + z⁻¹)⁴ = z⁴ + 4z² + 6 + 4z⁻² + z⁻⁴
- 16cos⁴θ = 2cos 4θ + 8cos 2θ + 6
- cos⁴θ = (1/8)cos 4θ + (1/2)cos 2θ + 3/8
Exam Tips
- Always convert to polar form before applying De Moivre's theorem
- When finding nth roots, check you have found exactly n roots
- For trig identity questions, expand using binomial theorem then equate real and imaginary parts separately
- The z + 1/z technique is essential for integration of powers of trig functions
- Remember that De Moivre works for negative n too: (cos θ + i sin θ)⁻¹ = cos θ − i sin θ