De Moivre's Theorem

Expanding $(1 + \sqrt{3}\,i)^7$ by brackets would take a while. In polar form it takes one line: raise the modulus to the power and multiply the argument. That’s De Moivre’s theorem. Run it backwards and it finds roots of complex numbers, which turn out to sit at the corners of regular polygons. It even turns into a machine for trig identities. This page is part of AA HL (subtopic AHL 1.14). All angles are in radians.

Key ideas

The theorem

For any integer $n$,

$$\big(r(\cos\theta + i\sin\theta)\big)^n = r^n(\cos n\theta + i\sin n\theta)$$

In cis and Euler notation: $(r\,\text{cis}\,\theta)^n = r^n\,\text{cis}\,n\theta$, and $(re^{i\theta})^n = r^n e^{in\theta}$. The Euler version looks just like an exponent law, which is a good way to remember it.

To find a power: write the number in polar form, raise the modulus to the power $n$, multiply the argument by $n$, then adjust the argument to $-\pi \lt \theta \le \pi$ if needed.

Proof by induction for positive integers

The guide expects you to be able to prove the theorem by induction for $n \in \mathbb{Z}^+$. Since $r^n$ just multiplies through, it’s enough to prove

$$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$$

Base case, $n = 1$: both sides are $\cos\theta + i\sin\theta$. True.

Inductive step. Assume the result is true for $n = k$, that is, $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$. Then for $n = k + 1$:

$$\begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta) \\ &= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) && \text{by the assumption} \\ &= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta) && i^2 = -1 \\ &= \cos(k\theta + \theta) + i\sin(k\theta + \theta) && \text{compound angle formulas} \\ &= \cos(k + 1)\theta + i\sin(k + 1)\theta \end{aligned}$$

So if the result is true for $n = k$, it’s true for $n = k + 1$. Since it’s true for $n = 1$, it’s true for all $n \in \mathbb{Z}^+$ by mathematical induction. (The step uses the compound angle formulas.)

For $n = 0$ both sides are $1$, and for negative integers it follows from $\dfrac{1}{\text{cis}\,\theta} = \text{cis}(-\theta)$.

Roots of a complex number

To solve $z^n = w$, write $w = R\,\text{cis}\,\varphi$. Because adding $2\pi$ to an argument doesn’t change the number, $w$ is also $R\,\text{cis}(\varphi + 2k\pi)$ for any integer $k$. Taking the $n$th root of each of these (De Moivre with exponent $\dfrac{1}{n}$):

$$z = R^{1/n}\,\text{cis}\left(\frac{\varphi + 2k\pi}{n}\right), \qquad k = 0, 1, 2, \ldots, n - 1$$

• There are exactly $n$ different $n$th roots. Other values of $k$ just repeat them.
• They all have the same modulus $R^{1/n}$, so they lie on a circle centred at the origin.
• Their arguments are spaced $\dfrac{2\pi}{n}$ apart, so they are the vertices of a regular $n$-sided polygon.

This is the extension of De Moivre’s theorem to rational exponents: a power like $w^{1/n}$ (or $w^{p/q}$) has several values, and adding multiples of $2\pi$ to the argument before you multiply it by the exponent is how you find them all.

Roots of unity

The solutions of $z^n = 1$ are the $n$th roots of unity:

$$z = \text{cis}\,\frac{2k\pi}{n}, \qquad k = 0, 1, \ldots, n - 1$$

If $\omega = \text{cis}\,\dfrac{2\pi}{n}$, the roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$, and for $n \ge 2$,

$$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$$

One reason: the equation $z^n - 1 = 0$ has no $z^{n-1}$ term, so the sum of its roots is $0$. (Or use the geometric series formula: the sum is $\dfrac{\omega^n - 1}{\omega - 1} = 0$.)

Multiple-angle identities

Expand $(\cos\theta + i\sin\theta)^n$ two ways, by De Moivre and by the binomial theorem, then equate real parts and imaginary parts. This gives formulas for $\cos n\theta$ and $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$ (Example 4).

Worked examples

Example 1: Powers

Use De Moivre’s theorem to find, in the form $a + bi$:

• (a) $(1 + \sqrt{3}\,i)^7$
• (b) $(1 - i)^{-6}$

Solution.

(a) $|1 + \sqrt{3}\,i| = \sqrt{1 + 3} = 2$ and the argument is $\dfrac{\pi}{3}$ (first quadrant). So

$$\begin{aligned} (1 + \sqrt{3}\,i)^7 &= 2^7\,\text{cis}\,\frac{7\pi}{3} \\ &= 128\,\text{cis}\,\frac{\pi}{3} && \tfrac{7\pi}{3} - 2\pi = \tfrac{\pi}{3} \\ &= 128\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) \\ &= 64 + 64\sqrt{3}\,i \end{aligned}$$

(b) $1 - i = \sqrt{2}\,\text{cis}\left(-\dfrac{\pi}{4}\right)$. With $n = -6$:

$$(1 - i)^{-6} = (\sqrt{2})^{-6}\,\text{cis}\,\frac{6\pi}{4} = \frac{1}{8}\,\text{cis}\,\frac{3\pi}{2} = \frac{1}{8}(0 - i) = -\frac{1}{8}i$$

Check (b): $(1 - i)^2 = -2i$, so $(1 - i)^6 = (-2i)^3 = -8i^3 = 8i$, and $\dfrac{1}{8i} = \dfrac{-i}{8}$. ✓

Example 2: Cube roots

Solve $z^3 = 8i$, giving the solutions in Cartesian form, and show them on an Argand diagram.

Solution. In polar form, $8i = 8\,\text{cis}\,\dfrac{\pi}{2}$. The cube roots have modulus $8^{1/3} = 2$ and arguments

$$\frac{\frac{\pi}{2} + 2k\pi}{3} = \frac{\pi}{6} + \frac{2k\pi}{3}, \qquad k = 0, 1, 2$$

• $k = 0$: $z = 2\,\text{cis}\,\dfrac{\pi}{6} = 2\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) = \sqrt{3} + i$
• $k = 1$: $z = 2\,\text{cis}\,\dfrac{5\pi}{6} = 2\left(-\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) = -\sqrt{3} + i$
• $k = 2$: $z = 2\,\text{cis}\,\dfrac{3\pi}{2} = 2\,\text{cis}\left(-\dfrac{\pi}{2}\right) = -2i$

Check one: $(-2i)^3 = -8i^3 = 8i$. ✓

The cube roots of $8i$ lie on the circle $|z| = 2$, spaced $\dfrac{2\pi}{3}$ apart: an equilateral triangle.

Example 3: Fifth roots of unity

Find the solutions of $z^5 = 1$, show they form a regular pentagon, and show that they add up to $0$.

Solution. $1 = \text{cis}(0 + 2k\pi)$, so the roots are $z = \text{cis}\,\dfrac{2k\pi}{5}$ for $k = 0, 1, 2, 3, 4$. With principal arguments:

$$1, \quad \text{cis}\,\frac{2\pi}{5}, \quad \text{cis}\,\frac{4\pi}{5}, \quad \text{cis}\left(-\frac{4\pi}{5}\right), \quad \text{cis}\left(-\frac{2\pi}{5}\right)$$

(For $k = 3$ and $k = 4$, $\dfrac{6\pi}{5} - 2\pi = -\dfrac{4\pi}{5}$ and $\dfrac{8\pi}{5} - 2\pi = -\dfrac{2\pi}{5}$.) To 3 s.f., $\text{cis}\,\dfrac{2\pi}{5} \approx 0.309 + 0.951i$ and $\text{cis}\,\dfrac{4\pi}{5} \approx -0.809 + 0.588i$; the other two are their conjugates.

All five have modulus $1$, and consecutive arguments differ by $\dfrac{2\pi}{5}$. Points equally spaced around a circle are the vertices of a regular polygon, here a pentagon.

The fifth roots of unity $1, \omega, \omega^2, \omega^3, \omega^4$, where $\omega = \text{cis}\,\dfrac{2\pi}{5}$, form a regular pentagon.

Sum. The equation $z^5 + 0z^4 + 0z^3 + 0z^2 + 0z - 1 = 0$ has coefficient $a_4 = 0$, so the sum of its roots is $-\dfrac{0}{1} = 0$. You can also see it numerically: the real parts give $1 + 2(0.309) + 2(-0.809) = 0.000$ to 3 d.p., and the imaginary parts cancel in conjugate pairs.

Example 4: A triple-angle formula

Use De Moivre’s theorem to show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$, and find a similar expression for $\sin 3\theta$.

Solution. Write $c = \cos\theta$ and $s = \sin\theta$. By De Moivre,

$$(c + is)^3 = \cos 3\theta + i\sin 3\theta$$

By the binomial theorem (with $i^2 = -1$ and $i^3 = -i$),

$$(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = (c^3 - 3cs^2) + i(3c^2 s - s^3)$$

Real parts: $\cos 3\theta = c^3 - 3cs^2$. Replace $s^2$ with $1 - c^2$:

$$\cos 3\theta = c^3 - 3c(1 - c^2) = 4c^3 - 3c = 4\cos^3\theta - 3\cos\theta$$

Imaginary parts: $\sin 3\theta = 3c^2 s - s^3$. Replace $c^2$ with $1 - s^2$:

$$\sin 3\theta = 3(1 - s^2)s - s^3 = 3\sin\theta - 4\sin^3\theta$$

Check with $\theta = \dfrac{\pi}{3}$: $\cos\pi = -1$, and $4\left(\dfrac{1}{2}\right)^3 - 3\left(\dfrac{1}{2}\right) = \dfrac{1}{2} - \dfrac{3}{2} = -1$. ✓

Common mistakes

Raising the parts of a + bi to the power. $(1 + \sqrt{3}\,i)^7$ is not $1^7 + (\sqrt{3}\,i)^7$. De Moivre’s theorem only works in polar or Euler form, so convert first.

Forgetting to raise the modulus to the power. $(2\,\text{cis}\,\theta)^5 = 32\,\text{cis}\,5\theta$, not $2\,\text{cis}\,5\theta$. Both parts change: the modulus is raised to the power, the argument is multiplied.

Finding only one root. $z^3 = 8i$ has three solutions, not just $\sqrt{3} + i$. Add $2k\pi$ to the argument before dividing by $n$, and use $k = 0, 1, \ldots, n - 1$.

Dividing the modulus instead of taking a root. The cube roots of $8\,\text{cis}\,\dfrac{\pi}{2}$ have modulus $\sqrt[3]{8} = 2$, not $\dfrac{8}{3}$.

Leaving arguments outside the principal range. $\text{cis}\,\dfrac{3\pi}{2}$ is a correct root, but a question asking for $-\pi \lt \theta \le \pi$ wants $\text{cis}\left(-\dfrac{\pi}{2}\right)$.

Mixing up which parts to equate. In multiple-angle questions, $\cos n\theta$ comes from the real part and $\sin n\theta$ from the imaginary part. Keep track of the powers of $i$: $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

Practice

1. (Warm-up) Simplify, giving the answer in the form $a + bi$.

• (a) $\left(\text{cis}\,\dfrac{\pi}{8}\right)^4$
• (b) $\left(2e^{i\pi/6}\right)^3$

Solution

(a) $\text{cis}\,\dfrac{4\pi}{8} = \text{cis}\,\dfrac{\pi}{2} = i$.

(b) $2^3 e^{3i\pi/6} = 8e^{i\pi/2} = 8i$.

2. (Warm-up) Use De Moivre’s theorem to find $(1 + i)^{10}$.

Solution

$1 + i = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}$, so

$$(1 + i)^{10} = (\sqrt{2})^{10}\,\text{cis}\,\frac{10\pi}{4} = 32\,\text{cis}\,\frac{5\pi}{2} = 32\,\text{cis}\,\frac{\pi}{2} = 32i$$

Check: $(1 + i)^2 = 2i$, so $(1 + i)^{10} = (2i)^5 = 32i^5 = 32i$. ✓

3. (Core) Find each power in the form $a + bi$.

• (a) $(-\sqrt{3} + i)^5$
• (b) $(\sqrt{3} + i)^{-3}$

Solution

(a) $|-\sqrt{3} + i| = 2$; the number is in the second quadrant with reference angle $\dfrac{\pi}{6}$, so its argument is $\dfrac{5\pi}{6}$.

$$(-\sqrt{3} + i)^5 = 32\,\text{cis}\,\frac{25\pi}{6} = 32\,\text{cis}\,\frac{\pi}{6} = 32\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = 16\sqrt{3} + 16i$$

(using $\dfrac{25\pi}{6} - 4\pi = \dfrac{\pi}{6}$).

(b) $\sqrt{3} + i = 2\,\text{cis}\,\dfrac{\pi}{6}$, so

$$(\sqrt{3} + i)^{-3} = 2^{-3}\,\text{cis}\left(-\frac{\pi}{2}\right) = \frac{1}{8}(0 - i) = -\frac{1}{8}i$$

4. (Core) Solve $z^4 = -16$, giving the solutions in Cartesian form.

Solution

$-16 = 16\,\text{cis}\,\pi$. The roots have modulus $16^{1/4} = 2$ and arguments $\dfrac{\pi + 2k\pi}{4}$ for $k = 0, 1, 2, 3$: that is, $\dfrac{\pi}{4}$, $\dfrac{3\pi}{4}$, $\dfrac{5\pi}{4}$ and $\dfrac{7\pi}{4}$ (or $-\dfrac{3\pi}{4}$ and $-\dfrac{\pi}{4}$ as principal arguments).

$$z = \sqrt{2} + \sqrt{2}\,i, \quad -\sqrt{2} + \sqrt{2}\,i, \quad -\sqrt{2} - \sqrt{2}\,i, \quad \sqrt{2} - \sqrt{2}\,i$$

They form a square on the circle $|z| = 2$.

5. (Core) Find the three cube roots of $-8 + 8\sqrt{3}\,i$, in the form $r\,\text{cis}\,\theta$ with $-\pi \lt \theta \le \pi$.

Solution

$|-8 + 8\sqrt{3}\,i| = \sqrt{64 + 192} = \sqrt{256} = 16$. The number is in the second quadrant with reference angle $\arctan\sqrt{3} = \dfrac{\pi}{3}$, so its argument is $\dfrac{2\pi}{3}$.

The cube roots have modulus $16^{1/3} = 2\sqrt[3]{2}$ and arguments

$$\frac{\frac{2\pi}{3} + 2k\pi}{3} = \frac{2\pi}{9} + \frac{2k\pi}{3}, \qquad k = 0, 1, 2$$

That’s $\dfrac{2\pi}{9}$, $\dfrac{8\pi}{9}$ and $\dfrac{14\pi}{9}$. The last one is outside the principal range: $\dfrac{14\pi}{9} - 2\pi = -\dfrac{4\pi}{9}$.

$$z = 2\sqrt[3]{2}\,\text{cis}\,\frac{2\pi}{9}, \quad 2\sqrt[3]{2}\,\text{cis}\,\frac{8\pi}{9}, \quad 2\sqrt[3]{2}\,\text{cis}\left(-\frac{4\pi}{9}\right)$$

6. (Core) Let $\omega$ be the root of $z^3 = 1$ with positive imaginary part.

• (a) Find all three cube roots of unity in Cartesian form, and identify $\omega$.
• (b) Show that $\omega^2 = \omega^*$ and that $1 + \omega + \omega^2 = 0$.
• (c) Hence find $(1 + \omega)^3$.

Solution

(a) $z = \text{cis}\,\dfrac{2k\pi}{3}$ for $k = 0, 1, 2$, which gives $1$, $\text{cis}\,\dfrac{2\pi}{3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ and $\text{cis}\,\dfrac{4\pi}{3} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$. So $\omega = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$.

(b) $\omega^2 = \text{cis}\,\dfrac{4\pi}{3} = \text{cis}\left(-\dfrac{2\pi}{3}\right)$, which has the same modulus as $\omega$ and the opposite argument, so $\omega^2 = \omega^*$. Then

$$1 + \omega + \omega^2 = 1 + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 0$$

(c) From (b), $1 + \omega = -\omega^2$. So $(1 + \omega)^3 = (-\omega^2)^3 = -\omega^6 = -(\omega^3)^2 = -1$.

7. (Core) Let $z = 1 + \sqrt{3}\,i$.

• (a) Find the smallest positive integer $n$ for which $z^n$ is real, and find $z^n$ for that $n$.
• (b) Find the smallest positive integer $n$ for which $z^n$ is a positive real number.

Solution

$z = 2\,\text{cis}\,\dfrac{\pi}{3}$, so $z^n = 2^n\,\text{cis}\,\dfrac{n\pi}{3}$.

(a) $z^n$ is real when $\sin\dfrac{n\pi}{3} = 0$, that is, when $\dfrac{n\pi}{3}$ is a multiple of $\pi$. The smallest is $n = 3$: $z^3 = 8\,\text{cis}\,\pi = -8$.

(b) For a positive real number, $\dfrac{n\pi}{3}$ must be a multiple of $2\pi$, so $n = 6$: $z^6 = 64\,\text{cis}\,2\pi = 64$.

8. (Challenge) Use De Moivre’s theorem to show that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$.

Solution

Let $c = \cos\theta$, $s = \sin\theta$. By De Moivre, $\cos 4\theta$ is the real part of $(c + is)^4$. By the binomial theorem,

$$(c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4$$

The real terms are those with even powers of $i$: $c^4 + 6c^2 i^2 s^2 + i^4 s^4 = c^4 - 6c^2 s^2 + s^4$. Replace $s^2$ with $1 - c^2$:

$$\begin{aligned} \cos 4\theta &= c^4 - 6c^2(1 - c^2) + (1 - c^2)^2 \\ &= c^4 - 6c^2 + 6c^4 + 1 - 2c^2 + c^4 \\ &= 8c^4 - 8c^2 + 1 \\ &= 8\cos^4\theta - 8\cos^2\theta + 1 \end{aligned}$$

Check with $\theta = 0$: $\cos 0 = 1$ and $8 - 8 + 1 = 1$. ✓

9. (Challenge) The solutions of $z^6 = 64$ are plotted on an Argand diagram.

• (a) Find the solutions in Cartesian form.
• (b) Explain why they form a regular hexagon, and find its exact area.

Solution

(a) $64 = 64\,\text{cis}(0 + 2k\pi)$, so $z = 2\,\text{cis}\,\dfrac{k\pi}{3}$ for $k = 0, 1, \ldots, 5$:

$$2, \quad 1 + \sqrt{3}\,i, \quad -1 + \sqrt{3}\,i, \quad -2, \quad -1 - \sqrt{3}\,i, \quad 1 - \sqrt{3}\,i$$

(b) All six have modulus $2$, and their arguments are spaced $\dfrac{\pi}{3}$ apart, so they are equally spaced around the circle $|z| = 2$: a regular hexagon. Joining each vertex to the origin splits it into six triangles, each with two sides of length $2$ and an angle of $\dfrac{\pi}{3}$ between them, so each is equilateral. Using $\text{area} = \dfrac{1}{2}ab\sin C$:

$$\text{area} = 6 \times \frac{1}{2}(2)(2)\sin\frac{\pi}{3} = 6 \times \sqrt{3} = 6\sqrt{3}$$