# De Moivre's theorem

> IB Mathematics: Analysis and Approaches HL · Number & Algebra
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u1-de-moivre-s-theorem/

This module covers De Moivre's theorem for integer powers of complex numbers, methods to find nth roots of any complex number, and techniques to derive multiple-angle trigonometric identities, a frequent HL exam topic.

**Prerequisites:** [Polar form of complex numbers](https://www.owlsprep.com/study/ib-math-aa-hl-u1-polar-form-complex-numbers/); [Exponential form of complex numbers](https://www.owlsprep.com/study/ib-math-aa-hl-u1-exponential-form-complex-numbers/); [Binomial theorem](https://www.owlsprep.com/study/ib-math-aa-hl-u1-binomial-theorem/)

## Learning objectives

- Apply De Moivre's theorem to raise complex numbers to integer powers
- Use De Moivre's theorem to find all nth roots of a complex number
- Derive multiple-angle trigonometric identities using De Moivre's theorem
- Prove properties of roots of unity using De Moivre's theorem

## De Moivre's Theorem for Integer Powers

**De Moivre's Theorem (Integer Powers)** — For any integer $n$, the nth power of $z$ is given by $z^n = r^n(\cos n\theta + i\sin n\theta) = r^n e^{in\theta}$. This holds for both positive and negative integers $n$.

*Notation:* $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$

*Example:* Works for $n=10, n=-3$, etc.

**Worked example:** Evaluate $(1 + i\sqrt{3})^6$ using De Moivre's theorem.

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

> **tip**
>
> De Moivre's theorem is far faster than binomial expansion for large powers, always use it for integer powers of complex numbers.

## Finding nth Roots of Complex Numbers

**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 $\frac{2\pi}{n}$ between consecutive roots.

**Worked example:** Find all distinct cube roots of $8i$.

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

> **exam_tip**
>
> Always add the $2k\pi$ term to the original argument to get all distinct roots. This is the most frequently missed step in exams.

## Deriving Trigonometric Identities

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

**Worked example:** Derive an expression for $\cos 3\theta$ in terms of $\cos\theta$ using De Moivre's theorem.

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

**Check your understanding**

Test your understanding:

1. What is the correct identity for $\sin 3\theta$, derived using the same method?

   - $3\sin\theta - 4\sin^3\theta$
   - $4\sin^3\theta - 3\sin\theta$
   - $3\cos\theta - 4\cos^3\theta$

   *Answer:* $3\sin\theta - 4\sin^3\theta$

   *Why:* Correct! You get this by equating the imaginary parts of the expansion, then simplifying with $\cos^2\theta = 1-\sin^2\theta$.

## Common pitfalls

- **Wrong:** Forgetting to add $2k\pi$ to the argument when finding nth roots.
  - Why it fails: You only get one root instead of all n distinct roots, resulting in lost marks.
  - Correct: Always write the original argument as $\theta + 2k\pi$ before dividing by n, then take $k=0,1,...,n-1$.
- **Wrong:** Confusing modulus of $z^n$ as $n|z|$ instead of $|z|^n$.
  - Why it fails: The modulus property of complex numbers is often misremembered as additive instead of multiplicative.
  - Correct: Always recall that $|z^n| = |z|^n$, never $n|z|$.
- **Wrong:** Sign errors when expanding binomials for identity derivation.
  - Why it fails: Powers of $i$ often produce negative terms that are missed during expansion.
  - Correct: Simplify powers of $i$ step-by-step: $i^1=i, i^2=-1, i^3=-i, i^4=1$ before grouping terms.
- **Wrong:** Applying De Moivre's theorem directly to complex numbers in rectangular form.
  - Why it fails: The theorem only works for polar/exponential form, you will get an incorrect result.
  - Correct: Always convert to polar form first before applying De Moivre's theorem.

## Cheatsheet

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

## 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](https://www.owlsprep.com/study/ib-math-aa-hl-u1-systems-of-linear-equations/)
- [Functions](https://www.owlsprep.com/study/ib-math-aa-hl-u2-overview/)

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