De Moivre's theorem

IB Mathematics: Analysis and Approaches HL· Unit 1: Number & Algebra, Topic 1.9 Complex numbers· 25 min read

1. De Moivre's Theorem for Integer Powers★★☆☆☆⏱ 10 min

📘 Definition

De Moivre's Theorem (Integer Powers)

$z = r (cos θ + i sin θ) = r e^{i θ}$

For any integer $n$ , the nth power of $z$ is given by $z^{n} = r^{n} (cos n θ + i sin n θ) = r^{n} e^{in θ}$ . This holds for both positive and negative integers $n$ .

Example:

Works for $n = 10, n = - 3$ , etc.

📐 Worked Example

Evaluate $(1 + i 3)^{6}$ using De Moivre's theorem.

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1. Convert $z = 1 + i 3$ to polar form:
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Modulus: $r = 1^{2} + (3)^{2} = 2$ , Argument: $θ = \frac{π}{3}$
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$z = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)$
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1. Apply De Moivre's theorem with $n = 6$ :
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$z^6 = 2^6\left(\cos\left(6 \cdot \frac{\pi}{3}\right) + i\sin\left(6 \cdot \frac{\pi}{3}\right)\right) = 64(\cos 2\pi + i\sin 2\pi)$
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1. Simplify to get the final result:
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$z^6 = 64(1 + 0i) = 64$

2. Finding nth Roots of Complex Numbers★★★☆☆⏱ 10 min

📘 Definition

nth Root of a Complex Number

Any complex number $w$ that satisfies $w^{n} = z$ for a given non-zero $z$ , where $n$ is a positive integer. All non-zero $z$ have exactly $n$ distinct nth roots.

All nth roots of a complex number with modulus $r$ are equally spaced around a circle of radius $r^{1/ n}$ centered at the origin, with an angular separation of $\frac{2 π}{n}$ between consecutive roots.

📐 Worked Example

Find all distinct cube roots of $8 i$ .

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1. Write $8 i$ in general polar form:
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$z = 8\left(\cos\left(\frac{\pi}{2} + 2k\pi\right) + i\sin\left(\frac{\pi}{2} + 2k\pi\right)\right), \quad k \in \mathbb{Z}$
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1. Let $w = ρ (cos ϕ + i sin ϕ)$ be a root, so $w^{3} = z$ by De Moivre:
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$\rho^3 (\cos 3\phi + i\sin 3\phi) = 8\left(\cos\left(\frac{\pi}{2} + 2k\pi\right) + i\sin\left(\frac{\pi}{2} + 2k\pi\right)\right)$
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1. Equate modulus and arguments: $ρ^{3} = 8 ⟹ ρ = 2$ , and $3 ϕ = \frac{π}{2} + 2 k π ⟹ ϕ = \frac{π}{6} + \frac{2 k π}{3}$
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1. Substitute $k = 0, 1, 2$ to get 3 distinct roots:
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- $k = 0 : 2 (cos \frac{π}{6} + i sin \frac{π}{6}) = 3 + i$
- $k = 1 : 2 (cos \frac{5 π}{6} + i sin \frac{5 π}{6}) = - 3 + i$
- $k = 2 : 2 (cos \frac{3 π}{2} + i sin \frac{3 π}{2}) = - 2 i$

3. Deriving Trigonometric Identities★★★★☆⏱ 15 min

A common IB exam question asks to derive identities for $sin n θ$ and $cos n θ$ in terms of powers of $sin θ$ and $cos θ$ . This uses De Moivre's theorem combined with the binomial theorem.

📐 Worked Example

Derive an expression for $cos 3 θ$ in terms of $cos θ$ using De Moivre's theorem.

1
1. By De Moivre's theorem:
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$\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$
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1. Expand the right-hand side with the binomial theorem:
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$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta$
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1. Simplify powers of $i$ and group real/imaginary parts:
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$= \left(\cos^3\theta - 3\cos\theta\sin^2\theta\right) + i\left(3\cos^2\theta\sin\theta - \sin^3\theta\right)$
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1. Equate real parts from both sides, then substitute $sin^{2} θ = 1 - cos^{2} θ$ :
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$\cos 3\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta - 3\cos\theta$

✓ Quick check

Test your understanding:

What is the correct identity for $sin 3 θ$ , derived using the same method?
- $3 sin θ - 4 sin^{3} θ$
- $4 sin^{3} θ - 3 sin θ$
- $3 cos θ - 4 cos^{3} θ$

4. Common Pitfalls

Wrong move:

Forgetting to add $2 k π$ to the argument when finding nth roots.

Why:

You only get one root instead of all n distinct roots, resulting in lost marks.

Correct move:

Always write the original argument as $θ + 2 k π$ before dividing by n, then take $k = 0, 1, ..., n - 1$ .

Wrong move:

Confusing modulus of $z^{n}$ as $n ∣ z ∣$ instead of $∣ z ∣^{n}$ .

Why:

The modulus property of complex numbers is often misremembered as additive instead of multiplicative.

Correct move:

Always recall that $∣ z^{n} ∣ = ∣ z ∣^{n}$ , never $n ∣ z ∣$ .

Wrong move:

Sign errors when expanding binomials for identity derivation.

Why:

Powers of $i$ often produce negative terms that are missed during expansion.

Correct move:

Simplify powers of $i$ step-by-step: $i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1$ before grouping terms.

Wrong move:

Applying De Moivre's theorem directly to complex numbers in rectangular form.

Why:

The theorem only works for polar/exponential form, you will get an incorrect result.

Correct move:

Always convert to polar form first before applying De Moivre's theorem.

5. Quick Reference Cheatsheet

Scenario	Formula / Rule
Integer power of $z = r (cos θ + i sin θ)$	$z^{n} = r^{n} (cos n θ + i sin n θ)$
nth roots of $z = r (cos (θ + 2 k π) + i sin (θ + 2 k π))$	Roots: $r^{1/ n} (cos (\frac{θ + 2 k π}{n}) + i sin (\frac{θ + 2 k π}{n})), k = 0, ..., n - 1$
Derive $cos n θ$	Equate real parts of $(cos θ + i sin θ)^{n}$
Derive $sin n θ$	Equate imaginary parts of $(cos θ + i sin θ)^{n}$
nth roots of unity	Roots = $e^{\frac{2 k π i}{n}}, k = 0, ..., n - 1$ , equally spaced on unit circle

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Find 5th roots of unity
2023 · 2
Derive identity for cos 3θ
2021 · 1
Evaluate (1+i)^10 by theorem

Going deeper

syllabusIB AA HL Official Syllabus Topic 1.9Full official syllabus reference

What's Next

De Moivre's theorem is a foundational tool for all further work with complex numbers in IB AA HL, and it regularly appears in combined questions with other topics like polynomials, trigonometry, and argument properties. Mastery of this topic is required for understanding roots of unity, which have applications to polynomial factorization and geometry of the complex plane, both common extended response questions in HL papers. Building on this topic, you will explore more advanced properties of complex numbers and their applications to calculus.

→Systems of linear equations
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