De Moivre's Theorem and Roots of Complex Numbers

Quick answer

In polar form, multiplying complex numbers multiplies their lengths and adds their angles. Repeating that n times gives De Moivre's theorem: the nth power of r(cos θ + i sin θ) is rⁿ(cos nθ + i sin nθ). Reading it backward gives the nth roots. Every nonzero complex number has exactly n of them, all with length the real nth root of r, and their angles start at θ/n and step by 360°/n, so they sit at the corners of a regular n-gon on a circle.

What you'll learn

Raise a complex number to a power with De Moivre's theorem
Find all n complex nth roots of a number
Place the roots on a circle as a regular polygon
Derive a multiple-angle identity from De Moivre's theorem

Powers multiply the angle

In polar form, a complex number is a length and a direction:

z = r(\cos\theta + i\sin\theta)

Multiplying two such numbers multiplies the lengths and adds the angles. Multiplying $z$ by itself $n$ times therefore raises the length to the $n$ th power and adds the angle to itself $n$ times:

De Moivre’s theorem. For every whole number $n$ , $\bigl(r(\cos\theta + i\sin\theta)\bigr)^n = r^n\bigl(\cos n\theta + i\sin n\theta\bigr)$ .

Why the roots spread evenly

Reverse the theorem. To solve $w^n = z$ , write $w = s(\cos\varphi + i\sin\varphi)$ . Matching lengths gives $s^n = r$ , so $s = \sqrt[n]{r}$ , the ordinary real root of a positive number. Matching directions is the interesting part: two angles name the same direction when they differ by a full turn, so

n\varphi = \theta + 360°k \quad\Longrightarrow\quad \varphi = \frac{\theta}{n} + \frac{360°k}{n}

Taking $k = 0, 1, \ldots, n - 1$ gives $n$ different directions. At $k = n$ the angle has grown by a full $360°$ and the first root comes back. An $n$ th root divides the angle by $n$ and takes the $n$ th root of the length, and the full turn splits into $n$ equally spaced choices. So the roots sit at the corners of a regular $n$ -gon on a circle of radius $\sqrt[n]{r}$ .

The count matches the Fundamental Theorem of Algebra: $w^n - z$ has degree $n$ , so it has exactly $n$ zeros.

The three cube roots of 8

Worked examples

Common mistakes

Practice problems

Write $1 + i$ in polar form.

Answer

$\sqrt{2}\left(\cos 45° + i\sin 45°\right)$

Full solution

$|1 + i| = \sqrt{1 + 1} = \sqrt{2}$ , and the point $(1, 1)$ lies at $45°$ .
Find $\bigl(2(\cos 30° + i\sin 30°)\bigr)^6$ .

Answer

$-64$

Full solution

The length becomes $2^6 = 64$ and the angle $6(30°) = 180°$ , so the answer is $64(\cos 180° + i\sin 180°) = -64$ .
Find $(1 + i)^{10}$ .

Answer

$32i$

Full solution

The length is $\left(\sqrt{2}\right)^{10} = 32$ and the angle is $10(45°) = 450°$ , the same direction as $90°$ . So the answer is $32i$ .
Find the three cube roots of $27i$ .

Answer

$3(\cos 30° + i\sin 30°)$ , $3(\cos 150° + i\sin 150°)$ and $-3i$

Full solution

$27i$ has length $27$ and angle $90°$ . The roots have length $3$ and angles $30°$ , $150°$ and $270°$ . The first is $\tfrac{3\sqrt{3}}{2} + \tfrac{3}{2}i$ , and the last is $-3i$ .
Find the four fourth roots of $1$ .

Answer

$1$ , $i$ , $-1$ and $-i$

Full solution

Length $1$ , angles $0°$ , $90°$ , $180°$ and $270°$ .
How far apart in angle are the sixth roots of $64$ , and how long is each?

Answer

$60°$ apart, each of length $2$

Full solution

A full turn divided by $6$ is $60°$ , and $\sqrt[6]{64} = 2$ .
Add the four fourth roots of $1$ .

Answer

$0$

Full solution

$1 + i - 1 - i = 0$ . The roots pair off into opposites, as the corners of a square centered at the origin always do.
Use De Moivre’s theorem to show $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .

Hint

Match imaginary parts after cubing.

Answer

$\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$ , and $\cos^2\theta = 1 - \sin^2\theta$ .

Full solution

Cubing $\cos\theta + i\sin\theta$ gives an imaginary part of $3\cos^2\theta\sin\theta - \sin^3\theta$ , which must equal $\sin 3\theta$ . Replacing $\cos^2\theta$ gives $3\sin\theta - 3\sin^3\theta - \sin^3\theta$ .
How many solutions does $z^5 = 32$ have, and what is the length of each?

Answer

Five solutions, each of length $2$

Full solution

The equation has degree $5$ , so there are five roots, spaced $72°$ apart, and each has length $\sqrt[5]{32} = 2$ .
A student says the only cube root of $8$ is $2$ . What went wrong?

Hint

How many zeros does $z^3 - 8$ have?

Answer

There are three cube roots: $2$ , $-1 + \sqrt{3}\,i$ and $-1 - \sqrt{3}\,i$ .

Full solution

Among real numbers, $2$ is the only cube root of $8$ . Among complex numbers, $z^3 - 8 = (z - 2)\left(z^2 + 2z + 4\right)$ , and the quadratic factor contributes the conjugate pair $-1 \pm \sqrt{3}\,i$ .

Frequently asked questions

What is De Moivre's theorem?

For z = r(cos θ + i sin θ) and a whole number n, zⁿ = rⁿ(cos nθ + i sin nθ). Powers multiply the angle and raise the length.

How many nth roots does a complex number have?

Exactly n, as long as the number is not zero. They all have the same length and are spaced 360°/n apart in angle.

How do you find the nth roots of a complex number?

Take the real nth root of the modulus, divide the argument by n, then add 360°/n repeatedly until you have n roots.

What are the roots of unity?

The solutions of zⁿ = 1. They lie on the unit circle at the corners of a regular n-gon, starting at 1.

Why does 8 have three cube roots?

z³ = 8 is a polynomial equation of degree 3, so the Fundamental Theorem of Algebra gives three roots: 2, −1 + √3 i and −1 − √3 i.

What to learn next

Proof by Mathematical InductionProve a statement for every positive integer with two steps: check it for n = 1, then show that whenever it holds for k it holds for k + 1.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSN.CN.B.5The Complex Number System(+) Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation.
CCSS.MATH.CONTENT.HSN.CN.C.9The Complex Number System(+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.

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