De Moivre's Theorem: Powers and Roots

De Moivre’s theorem says that to raise a complex number in trigonometric form to a power, raise the modulus to the power and multiply the angle by it: $[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$ . A related formula finds all $n$ of the $n$ th roots of a complex number.

Before this lesson: You should know the Trigonometric Form of Complex Numbers, Coterminal Angles and nth Roots.

Why De Moivre’s Theorem Works

To multiply complex numbers in trigonometric form, you multiply the moduli and add the angles. A power is just the same number multiplied by itself again and again.

So $z^2$ has modulus $r \cdot r = r^2$ and angle $\theta + \theta = 2\theta$ . Then $z^3$ has modulus $r^3$ and angle $3\theta$ . The pattern continues for every whole number $n$ . Without it, $(1 + i)^8$ would take seven long multiplications.

The Formulas

Powers

[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)

Roots

Every nonzero complex number has exactly $n$ different $n$ th roots. They are:

r^{1/n}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i \sin\left(\frac{\theta + 2\pi k}{n}\right)\right], \quad k = 0, 1, \ldots, n - 1

All the roots sit on a circle of radius $r^{1/n}$ , spaced evenly, $\frac{2\pi}{n}$ apart. In degrees, replace $2\pi k$ with $360^\circ k$ .

Complex plane with a dashed circle of radius 2 centered at 0. Three green points on the circle, the cube roots of 8i, are at square root 3 + i, negative square root 3 + i, and −2i. They are 120 degrees apart and form an equilateral triangle shaded in blue. — The three cube roots of 8i lie on a circle of radius 2, spaced 120° apart.

Worked Examples

Example 1: A power that lands on the real axis

Find $(1 + i)^8$ .

Write $1 + i$ in trigonometric form: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $\theta = \frac{\pi}{4}$

(1 + i)^8 = (\sqrt{2})^8 (\cos \frac{8\pi}{4} + i \sin \frac{8\pi}{4})

Apply De Moivre’s theorem.

(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16

Find the power (see exponent rules)

\frac{8\pi}{4} = 2\pi

Simplify the angle.

16(1 + 0i) = 16

Use $\cos 2\pi = 1$ and $\sin 2\pi = 0$ .

Answer: $(1 + i)^8 = 16$

Example 2: A power with a special angle

Find $(1 + i\sqrt{3})^5$ in the form $a + bi$ .

Trigonometric form: $r = \sqrt{1 + 3} = 2$ and $\theta = \frac{\pi}{3}$

2^5 (\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})

Apply the theorem.

2^5 = 32

Find the power.

Use the unit circle: $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$

32(\frac{1}{2} - \frac{\sqrt{3}}{2} i)

Substitute.

16 - 16\sqrt{3} i

Distribute.

Answer: $16 - 16\sqrt{3} i$ , about $16 - 27.71i$

Example 3: Cube roots

Find the three cube roots of $8i$ .

r = 8,\ \theta = \frac{\pi}{2}

Write $8i$ in trigonometric form.

8^{1/3} = 2

Find the modulus of each root.

\frac{\frac{\pi}{2} + 0}{3} = \frac{\pi}{6}

For $k = 0$ , find the angle.

2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2} i)

Write the root and use the unit circle values.

2(\frac{\sqrt{3}}{2} + \frac{1}{2} i) = \sqrt{3} + i

Distribute the 2.

\frac{\frac{\pi}{2} + 2\pi}{3} = \frac{\frac{5\pi}{2}}{3}

For $k = 1$ , add 2π to the angle.

\frac{\frac{5\pi}{2}}{3} = \frac{5\pi}{6}

Divide by 3.

2(-\frac{\sqrt{3}}{2} + \frac{1}{2} i) = -\sqrt{3} + i

Write the root and distribute the 2.

\frac{\frac{\pi}{2} + 4\pi}{3} = \frac{\frac{9\pi}{2}}{3}

For $k = 2$ , add 4π to the angle.

\frac{\frac{9\pi}{2}}{3} = \frac{3\pi}{2}

Divide by 3.

2(0 - 1i) = -2i

Write the root and distribute the 2.

Answer: $\sqrt{3} + i$ , $-\sqrt{3} + i$ and $-2i$ . Check one: $(-2i)^3 = -8i^3 = -8(-i) = 8i$ .

Example 4: Solve an equation with roots

Solve $z^4 = 16$ over the complex numbers.

The solutions are the fourth roots of 16.

r = 16,\ \theta = 0

Write 16 in trigonometric form.

16^{1/4} = 2

Find the modulus of each root.

\frac{0 + 2\pi k}{4} = \frac{\pi k}{2}

Find the angles for $k = 0, 1, 2, 3$ .

0,\ \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2}

List the four angles.

2(\cos 0 + i \sin 0) = 2

Convert the first root.

2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 2i

Convert the second root.

2(\cos \pi + i \sin \pi) = -2

Convert the third root.

2(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = -2i

Convert the fourth root.

Answer: $z = 2, 2i, -2, -2i$ . A degree-4 equation has 4 solutions, as the fundamental theorem of algebra says.

Example 5: Word problem with a bolt pattern

An engineer places 5 bolts evenly around a circle of radius 3 cm, with one bolt straight up from the center. The bolt centers are the fifth roots of $243(\cos 90^\circ + i \sin 90^\circ)$ . Find where each bolt goes.

The modulus of each root is $243^{1/5} = 3$ , since $3^5 = 243$

The angles are $\frac{90^\circ + 360^\circ k}{5} = 18^\circ + 72^\circ k$

For $k = 0, 1, 2, 3, 4$ , the angles are $18^\circ, 90^\circ, 162^\circ, 234^\circ, 306^\circ$

3(\cos 18^\circ + i \sin 18^\circ) \approx 2.853 + 0.927i

Convert the first.

Convert the rest the same way: $3i$ , $-2.853 + 0.927i$ , $-1.763 - 2.427i$ , $1.763 - 2.427i$

Answer: The bolts go at about $(2.85, 0.93)$ , $(0, 3)$ , $(-2.85, 0.93)$ , $(-1.76, -2.43)$ and $(1.76, -2.43)$ cm, $72^\circ$ apart.

Common Mistakes

Mistake	Example of the error	Fix
Multiplying the modulus by n	$(1 + i)^8$ with modulus $8\sqrt{2}$	Raise the modulus to the power: $(\sqrt{2})^8 = 16$ .
Raising the angle to the power	Angle $(\frac{\pi}{4})^8$	Multiply the angle by $n$ : $8 \cdot \frac{\pi}{4} = 2\pi$ .
Finding only one root	Giving only $\sqrt{3} + i$ as a cube root of $8i$	There are $n$ roots. Use $k = 0, 1, \ldots, n - 1$ .
Adding 2πk after dividing	Angles $\frac{\pi}{6} + 2\pi k$	Add $2\pi k$ first, then divide by $n$ : the roots are $\frac{2\pi}{n}$ apart.
Using it in standard form	Raising $1$ and $i$ to the 8th separately	Convert to trigonometric form first.

Related Methods

Roots of unity

The $n$ th roots of 1 are called the $n$ th roots of unity. They are $\cos \frac{2\pi k}{n} + i \sin \frac{2\pi k}{n}$ and make a regular polygon with one corner at 1.

Binomial expansion

You could expand $(1 + i)^8$ with the binomial theorem, but it takes nine terms. De Moivre’s theorem is much faster for high powers.

Trig identities

Expanding $(\cos \theta + i \sin \theta)^2$ and matching it to $\cos 2\theta + i \sin 2\theta$ gives the double-angle formulas.

Practice Problems

Give answers in the form $a + bi$ . Round to the nearest hundredth when needed.

Find $(1 - i)^4$ .
Find $[2(\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})]^6$ .
Find $(\sqrt{3} + i)^4$ .
Find $(-1 + i)^5$ .
Find $[3(\cos 40^\circ + i \sin 40^\circ)]^3$ .
Find the two square roots of $9i$ .
Find the three cube roots of 1.
Solve $z^3 = -8$ over the complex numbers.
Find the four fourth roots of $-16$ .
A designer places the corners of a regular hexagon at the sixth roots of 64, with one corner at 2. Find all six corners.

Show answers

$1 - i$ : $r = \sqrt{2}$ , $\theta = \frac{7\pi}{4}$
Power: $(\sqrt{2})^4 = 4$ ; angle: $4 \cdot \frac{7\pi}{4} = 7\pi$ , which is coterminal with $\pi$

4(\cos \pi + i \sin \pi) = 4(-1 + 0i)

Use the unit circle values.

Answer: $-4$

Power: $2^6 = 64$ ; angle: $6 \cdot \frac{\pi}{12} = \frac{\pi}{2}$

64(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 64(0 + 1i)

Use the unit circle values.

Answer: $64i$

$\sqrt{3} + i$ : $r = \sqrt{3 + 1} = 2$ , $\theta = \frac{\pi}{6}$
Power: $2^4 = 16$ ; angle: $4 \cdot \frac{\pi}{6} = \frac{2\pi}{3}$

16(-\frac{1}{2} + \frac{\sqrt{3}}{2} i)

Use the unit circle values.

Answer: $-8 + 8\sqrt{3} i \approx -8 + 13.86i$

$-1 + i$ : $r = \sqrt{2}$ , $\theta = \frac{3\pi}{4}$
Power: $(\sqrt{2})^5 = 4\sqrt{2}$ ; angle: $5 \cdot \frac{3\pi}{4} = \frac{15\pi}{4}$ , coterminal with $\frac{15\pi}{4} - 2\pi = \frac{7\pi}{4}$

4\sqrt{2}(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)

Use the unit circle values.

4 - 4i

Distribute, using $\sqrt{2} \cdot \sqrt{2} = 2$ .

Answer: $4 - 4i$

Power: $3^3 = 27$ ; angle: $3 \cdot 40^\circ = 120^\circ$

27(-\frac{1}{2} + \frac{\sqrt{3}}{2} i)

Use the unit circle values.

Answer: $-13.5 + 13.5\sqrt{3} i \approx -13.5 + 23.38i$

$9i$ : $r = 9$ , $\theta = \frac{\pi}{2}$ ; root modulus $9^{1/2} = 3$
$k = 0$ : angle $\frac{\pi}{4}$ , root $3(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i) = \frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} i$
$k = 1$ : angle $\frac{\pi}{4} + \pi = \frac{5\pi}{4}$ , root $-\frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} i$
Answer: $\pm(\frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} i) \approx \pm(2.12 + 2.12i)$

$1$ : $r = 1$ , $\theta = 0$ ; angles $\frac{2\pi k}{3}$ : $0, \frac{2\pi}{3}, \frac{4\pi}{3}$
Roots: $1$ , $\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$ , $\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i$
Answer: $1, -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

$-8$ : $r = 8$ , $\theta = \pi$ ; root modulus $8^{1/3} = 2$
Angles: $\frac{\pi + 2\pi k}{3}$ gives $\frac{\pi}{3}, \pi, \frac{5\pi}{3}$
Roots: $2(\frac{1}{2} + \frac{\sqrt{3}}{2} i) = 1 + \sqrt{3} i$ , $2(-1) = -2$ , $2(\frac{1}{2} - \frac{\sqrt{3}}{2} i) = 1 - \sqrt{3} i$
Answer: $z = -2, 1 \pm \sqrt{3} i$

$-16$ : $r = 16$ , $\theta = \pi$ ; root modulus $16^{1/4} = 2$
Angles: $\frac{\pi + 2\pi k}{4}$ gives $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

2(\pm\frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2} i) = \pm\sqrt{2} \pm \sqrt{2} i

Write each root and distribute.

Answer: $\sqrt{2} + \sqrt{2} i$ , $-\sqrt{2} + \sqrt{2} i$ , $-\sqrt{2} - \sqrt{2} i$ , $\sqrt{2} - \sqrt{2} i$

10.

$64$ : $r = 64$ , $\theta = 0$ ; root modulus $64^{1/6} = 2$
Angles: $\frac{2\pi k}{6} = \frac{\pi k}{3}$ : $0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}$
Convert: $2$ , $1 + \sqrt{3} i$ , $-1 + \sqrt{3} i$ , $-2$ , $-1 - \sqrt{3} i$ , $1 - \sqrt{3} i$
Answer: those six corners

FAQ

What is De Moivre’s theorem used for?

It raises complex numbers to powers quickly and finds all of their roots. It also helps derive trig identities like the double-angle formulas.

How many nth roots does a complex number have?

Every nonzero complex number has exactly $n$ different $n$ th roots. They are evenly spaced around a circle of radius $r^{1/n}$ .

Does De Moivre’s theorem work for negative powers?

Yes. For a negative whole number $n$ , the same formula works, with $r^n$ meaning $\frac{1}{r^{\lvert n \rvert}}$ .

Why are the roots evenly spaced?

Each root’s angle is $\frac{\theta + 2\pi k}{n}$ . Raising $k$ by 1 adds $\frac{2\pi}{n}$ to the angle, so the roots are equal steps apart.

Summary

Powers: $[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$ .
Raise the modulus to the power; multiply the angle by the power.
Roots: modulus $r^{1/n}$ and angles $\frac{\theta + 2\pi k}{n}$ for $k = 0$ to $n - 1$ .
The $n$ roots are evenly spaced on a circle, $\frac{2\pi}{n}$ apart.
Convert to trigonometric form first.

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