What is De Moivre's theorem?

De Moivre's theorem states that (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ) for any integer n. In compact notation, (cis θ)^n = cis(nθ). The angle simply multiplies by the exponent, making complex powers trivial to compute in trigonometric form.

How do you apply De Moivre's theorem to complex powers?

Convert the complex number to trigonometric form z = r·cis θ, then apply the formula: z^n = r^n·cis(nθ). The modulus raises to the nth power and the argument multiplies by n. Finally, convert back to algebraic form if needed.

Does De Moivre's theorem work for negative exponents?

Yes. For negative integers, z^(-n) = r^(-n)·cis(-nθ). The modulus inverts and the argument multiplies by the negative exponent. The same structural simplicity applies to all integer powers, positive or negative.

How do you prove De Moivre's theorem?

The proof uses mathematical induction. The base case n = 1 is trivial. For the inductive step, multiply (cis θ)^k by cis θ, which adds arguments to give cis((k+1)θ). Extension to negative integers follows from the reciprocal relationship z^(-n) = 1/z^n.

How many nth roots does a complex number have?

Every nonzero complex number has exactly n distinct nth roots. They all have the same modulus R^(1/n) and are evenly spaced around a circle, with arguments differing by 360°/n. The roots form vertices of a regular n-gon.

What is the formula for nth roots of a complex number?

For w = R·cis φ, the n roots are z_k = R^(1/n)·cis((φ + 360°k)/n) for k = 0, 1, ..., n-1. Each value of k gives a distinct root. Beyond k = n-1, the angles repeat and no new roots appear.

Why do nth roots form a regular polygon?

All n roots share the same modulus R^(1/n), placing them on a circle. Consecutive roots differ in argument by exactly 360°/n, spacing them evenly. This uniform distribution around the circle creates vertices of a regular n-gon.

What are the roots of unity?

The nth roots of unity are the n solutions to z^n = 1. They are z_k = cis(360°k/n) for k = 0, 1, ..., n-1. All lie on the unit circle and form a regular n-gon. The root ω = cis(360°/n) generates all others as powers: 1, ω, ω², ..., ω^(n-1).

What is a primitive root of unity?

The primitive nth root of unity is ω = cis(2π/n) = cis(360°/n). It generates all other nth roots through its powers: the complete set is {1, ω, ω², ..., ω^(n-1)}. This generator connects roots of unity to group theory.

What is the product of all nth roots of unity?

The product of all nth roots of unity is (-1)^(n+1). For odd n, the product is 1. For even n, the product is -1. This relates to the constant term of the polynomial z^n - 1.

What are the fourth roots of unity?

The fourth roots of unity are 1, i, -1, and -i. They sit at 0°, 90°, 180°, and 270° on the unit circle, forming a square. Each satisfies z⁴ = 1.

How do you find the cube roots of a negative number?

Write the negative number in trigonometric form with argument 180°. For -8 = 8·cis(180°), the cube roots are 2·cis((180° + 360°k)/3) for k = 0, 1, 2, giving 2·cis(60°), 2·cis(180°) = -2, and 2·cis(300°). One root is real (-2), and two are complex conjugates.

What are the square roots of i?

Write i = 1·cis(90°). The square roots are cis(45°) and cis(225°), which equal (√2/2)(1 + i) and (√2/2)(-1 - i). They are opposite each other on the unit circle, separated by 180°.

Demoivre Theorem

Q: What is the sum of all nth roots of unity?

The sum of all nth roots of unity equals zero: 1 + ω + ω² + ... + ω^(n-1) = 0. Geometrically, the roots form a regular polygon centered at the origin, so their vector sum returns to zero. This is a geometric series with ratio ω ≠ 1.

Key Terms

Introduction to De Moivre's Theorem

De Moivre's Formula

Proof of De Moivre's Theorem

Applying De Moivre's Theorem to Powers

Introduction to

n

-th Roots

The Formula for

n

-th Roots

Geometric Interpretation of Roots

Roots of Unity

Properties of Roots of Unity

Worked Examples

Summary: Forward and Reverse Directions of De Moivre's Theorem

Powers and Roots Made Simple

Computing

(1 + i)^{100}

through repeated multiplication would fill pages with tedious algebra. Yet the answer emerges in seconds using De Moivre's theorem — a formula that transforms exponentiation into elementary arithmetic on angles and lengths. This same principle runs in reverse to extract roots, revealing that every nonzero complex number has exactly

n

distinct

n

-th roots, arranged in perfect geometric symmetry around the origin.

Key Terms

Trigonometric Form— the representation De Moivre's theorem operates on

Modulus— raised to the

n

th power

Argument— multiplied by

n

Roots of Unity— found using the theorem's inverse

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Introduction to De Moivre's Theorem

Raising complex numbers to powers in algebraic form quickly becomes unmanageable. Squaring

z = a + bi

requires expanding

(a + bi)^2 = a^2 + 2abi + b^2i^2 = (a^2 - b^2) + 2abi

. Cubing demands multiplying this result by

a + bi

again. By the time we reach

z^{10}

z^{100}

, the algebraic approach drowns in computation.

The trigonometric form offers rescue. When

z = r\text{cis}\theta

, multiplication becomes elegant: multiply moduli, add arguments. Squaring gives

z^2 = r^2\text{cis}(2\theta)

. Cubing gives

z^3 = r^3\text{cis}(3\theta)

. The pattern suggests a general rule — and De Moivre's theorem confirms it.

Abraham de Moivre, a French mathematician living in England, discovered this relationship in the early eighteenth century. Though he never stated the theorem in its modern form, his work on probability and trigonometric identities laid the foundation. The theorem bearing his name connects complex exponentiation to the simple operation of multiplying an angle by an integer.

De Moivre's theorem stands among the most useful results in complex analysis. It reduces high powers to trivial calculations, enables extraction of roots, derives trigonometric identities, and appears throughout physics and engineering wherever oscillations and rotations arise.

De Moivre's Formula

The theorem states a precise relationship between powers and angles. For any integer

n

De Moivre's Theorem

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

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Raise

\text{cis}\theta

to the

n

-th power, and the result is

\text{cis}(n\theta)

. The angle simply multiplies by

n

. No expansion, no collecting terms, no tracking powers of

i

— just multiply the argument.

In the condensed notation:

(\text{cis}\theta)^n = \text{cis}(n\theta)

.

The formula extends naturally to complex numbers with arbitrary modulus. If

z = r\text{cis}\theta

, then:

z^n = (r\text{cis}\theta)^n = r^n\text{cis}(n\theta)

The modulus raises to the

n

-th power while the argument multiplies by

n

. Both operations are elementary. The number

2\text{cis}(30°)

raised to the sixth power becomes

2^6\text{cis}(180°) = 64\text{cis}(180°) = -64

. What would require extensive algebra in rectangular form reduces to mental arithmetic in trigonometric form.

The formula works for negative integers as well. Since

z^{-n} = 1/z^n

, we have:

z^{-n} = r^{-n}\text{cis}(-n\theta)

Negative exponents invert the modulus and negate the angle multiplier, maintaining the same structural simplicity.

Proof of De Moivre's Theorem

Mathematical induction establishes De Moivre's theorem for positive integers. The proof builds from the multiplication rule for trigonometric form: when two complex numbers multiply, their moduli multiply and their arguments add.

The base case

n = 1

holds trivially:

(\cos\theta + i\sin\theta)^1 = \cos(1 \cdot \theta) + i\sin(1 \cdot \theta)

.

For the inductive step, assume the formula holds for

n = k

(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)

We must prove it holds for

n = k + 1

. Multiply both sides by

\cos\theta + i\sin\theta

(\cos\theta + i\sin\theta)^{k+1} = (\cos(k\theta) + i\sin(k\theta))(\cos\theta + i\sin\theta)

The right side is a product of two numbers in trigonometric form. By the multiplication rule, we add arguments:

= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos((k+1)\theta) + i\sin((k+1)\theta)

This completes the induction. The theorem holds for all positive integers.

Extension to negative integers follows from the reciprocal. For

n > 0

z^{-n} = \frac{1}{z^n} = \frac{1}{r^n\text{cis}(n\theta)} = \frac{1}{r^n} \cdot \text{cis}(-n\theta) = r^{-n}\text{cis}(-n\theta)

The formula

z^n = r^n\text{cis}(n\theta)

thus holds for all integers

n

, positive, negative, or zero.

Applying De Moivre's Theorem to Powers

Computing high powers follows a three-step procedure. First convert to trigonometric form, then apply De Moivre's formula, finally convert back to algebraic form if required.

Consider

(1 + i)^{10}

.

Step 1: Convert

1 + i

to trigonometric form. The modulus is

|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}

. The argument satisfies

\tan\theta = 1/1 = 1

with the point in the first quadrant, giving

\theta = 45° = \pi/4

. Thus

1 + i = \sqrt{2}\text{cis}(45°)

.

Step 2: Apply De Moivre's theorem.

(1 + i)^{10} = (\sqrt{2})^{10}\text{cis}(10 \times 45°) = 2^5\text{cis}(450°) = 32\text{cis}(450°)

Since

450° = 360° + 90°

, the angle reduces to

90°

= 32\text{cis}(90°) = 32(\cos 90° + i\sin 90°) = 32(0 + i) = 32i

Step 3: The answer

32i

is already in algebraic form.

Verification through direct calculation would require multiplying

1 + i

by itself ten times — tedious and error-prone. De Moivre's theorem completes the same computation in a few lines.

Another example: find

(\sqrt{3} - i)^6

. The modulus is

\sqrt{3 + 1} = 2

. The argument, with

a = \sqrt{3} > 0

and

b = -1 < 0

, lies in the fourth quadrant:

\theta = -30°

. So

\sqrt{3} - i = 2\text{cis}(-30°)

, and:

(\sqrt{3} - i)^6 = 2^6\text{cis}(-180°) = 64\text{cis}(-180°) = 64(-1) = -64

z	n	r, θ	rⁿ cis(nθ)	Result
1 + i	10	√2, 45°	(√2)¹⁰ cis(450°) = 32 cis(90°)	32i
√3 − i	6	2, −30°	2⁶ cis(−180°) = 64 · (−1)	−64
2 cis(30°)	6	2, 30°	2⁶ cis(180°)	−64
3 cis(60°)	−2	3, 60°	3⁻² cis(−120°) = (1⁄9) cis(−120°)	−1⁄18 − (√3⁄18) i

Introduction to $n$ -th Roots

De Moivre's theorem runs in reverse to solve equations of the form

z^n = w

. Given a complex number

w

, what values of

z

satisfy this equation? The answer reveals surprising multiplicity: every nonzero complex number has exactly

n

distinct

n

-th roots.

Real number experience suggests uniqueness. The equation

x^3 = 8

has one real solution:

x = 2

. But in

\mathbb{C}

, the equation

z^3 = 8

has three solutions. The cube root of

8

is not just

2

but three different complex numbers, each of which cubes to

8

.

Where do these extra roots come from? The key lies in angle periodicity. Adding

360°

(or

2\pi

radians) to any argument produces the same complex number:

\text{cis}(\theta) = \text{cis}(\theta + 360°)

. When we write

w = R\text{cis}\phi

, we could equally write

w = R\text{cis}(\phi + 360°)

w = R\text{cis}(\phi + 720°)

.

Now suppose

z^n = w

with

z = r\text{cis}\alpha

. By De Moivre's theorem,

r^n\text{cis}(n\alpha) = R\text{cis}\phi

. Matching moduli gives

r = R^{1/n}

. Matching arguments gives

n\alpha = \phi + 360°k

for some integer

k

, so

\alpha = \frac{\phi + 360°k}{n}

.

Different values of

k

yield different angles

\alpha

. As

k

runs from

0

n-1

, we obtain

n

distinct arguments. Beyond

k = n-1

, the angles repeat modulo

360°

. Exactly

n

roots exist, no more, no fewer.

The Formula for $n$ -th Roots

Let

w = R\text{cis}\phi

be any nonzero complex number. The

n

-th roots of

w

— all solutions to

z^n = w

— are given by:

Nth Roots Formula

z_k = R^{1/n} \bigl[\cos\!\left(\tfrac{\phi + 2\pi k}{n}\right) + i\sin\!\left(\tfrac{\phi + 2\pi k}{n}\right)\bigr], \quad k = 0, 1, \ldots, n-1

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In degree measure:

z_k = R^{1/n}\text{cis}\left(\frac{\phi + 360°k}{n}\right) \quad \text{for } k = 0, 1, 2, \ldots, n-1

Each root has the same modulus:

|z_k| = R^{1/n}

, the positive real

n

-th root of the original modulus. The roots differ only in their arguments.

The first root (

k = 0

) has argument

\phi/n

, the original angle divided by

n

. Each subsequent root adds

360°/n

to the argument. The

n

roots space themselves evenly around the circle of radius

R^{1/n}

.

Consider finding the cube roots of

8

. Write

8 = 8\text{cis}(0°)

, so

R = 8

and

\phi = 0°

. The formula gives:

z_k = 8^{1/3}\text{cis}\left(\frac{0° + 360°k}{3}\right) = 2\text{cis}(120°k)

For

k = 0

z_0 = 2\text{cis}(0°) = 2

.
For

k = 1

z_1 = 2\text{cis}(120°) = 2(-\frac{1}{2} + \frac{\sqrt{3}}{2}i) = -1 + \sqrt{3}i

.
For

k = 2

z_2 = 2\text{cis}(240°) = 2(-\frac{1}{2} - \frac{\sqrt{3}}{2}i) = -1 - \sqrt{3}i

.

Three roots:

2

-1 + \sqrt{3}i

, and

-1 - \sqrt{3}i

. Each cubes to

8

Geometric Interpretation of Roots

The

n

-th roots of any complex number arrange themselves with perfect geometric regularity. Plotting them on the complex plane reveals a pattern of elegant symmetry.

All

n

roots share the same modulus

R^{1/n}

, placing them on a circle centered at the origin with that radius. They do not cluster or scatter randomly — they distribute themselves at equal angular intervals around this circle.

The angular separation between consecutive roots is exactly

\frac{360°}{n}

(or

\frac{2\pi}{n}

radians). The cube roots of any number sit

120°

apart. Fourth roots sit

90°

apart. Fifth roots sit

72°

apart. This uniform spacing holds regardless of which number we take roots of — only the radius and starting angle vary.

The roots form vertices of a regular

n

-gon inscribed in the circle. Cube roots give an equilateral triangle. Fourth roots give a square. Fifth roots give a regular pentagon. The polygon may appear rotated depending on the argument of the original number, but its shape is always perfectly regular.

This geometric picture aids computation and verification. If one root is known, the others follow by rotating

360°/n

around the origin. If roots fail to form a regular polygon, an error has occurred. The visual pattern provides both a computational shortcut and a consistency check.

The conjugate relationship appears when the original number

w

is real and positive. In that case

\phi = 0°

, and the roots distribute symmetrically about the real axis. Complex roots come in conjugate pairs, with the real axis serving as a line of mirror symmetry.

n	Angular separation	Polygon shape on the circle of radius RŁⁿ
2	180°	two diametrically opposite points (a line segment through the origin)
3	120°	equilateral triangle
4	90°	square
5	72°	regular pentagon
6	60°	regular hexagon
general n	360°⁄n	regular n-gon centered at origin

Roots of Unity

The

n

-th roots of unity — solutions to

z^n = 1

— form a special case of fundamental importance. Since

1 = 1\text{cis}(0°)

, the formula simplifies:

Roots of Unity

z_k = e^{i \, 2\pi k / n}, \quad k = 0, 1, \ldots, n-1

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All roots of unity have modulus

1

— they lie on the unit circle. Their arguments are multiples of

\frac{360°}{n}

, dividing the circle into

n

equal arcs.

The root

z_0 = \text{cis}(0°) = 1

always appears. Every integer power of

1

equals

1

, making

1

n

-th root of unity for every

n

.

The fourth roots of unity illustrate the pattern:

z_0 = 1

z_1 = i

z_2 = -1

z_3 = -i

. These four points mark the cardinal directions on the unit circle, forming a square. Each satisfies

z^4 = 1

.

The sixth roots of unity are

1

\text{cis}(60°)

\text{cis}(120°)

-1

\text{cis}(240°)

\text{cis}(300°)

. In algebraic form:

1

\frac{1}{2} + \frac{\sqrt{3}}{2}i

-\frac{1}{2} + \frac{\sqrt{3}}{2}i

-1

-\frac{1}{2} - \frac{\sqrt{3}}{2}i

\frac{1}{2} - \frac{\sqrt{3}}{2}i

. They form a regular hexagon on the unit circle.

The primitive

n

-th root of unity, denoted

\omega = \text{cis}\left(\frac{2\pi}{n}\right)

, generates all others through its powers. Since

z_k = \omega^k

, the complete set of

n

-th roots of unity is

\{1, \omega, \omega^2, \ldots, \omega^{n-1}\}

. This generator perspective connects roots of unity to group theory and number theory.

n	n-th roots of unity (as cis)	In algebraic form a + bi	Polygon
2	cis(0°), cis(180°)	1, −1	two points on the real axis
3	cis(0°), cis(120°), cis(240°)	1, −1⁄2 + (√3⁄2)i, −1⁄2 − (√3⁄2)i	equilateral triangle
4	cis(0°), cis(90°), cis(180°), cis(270°)	1, i, −1, −i	square
6	cis(60°k), k = 0…5	1, 1⁄2 + (√3⁄2)i, −1⁄2 + (√3⁄2)i, −1, −1⁄2 − (√3⁄2)i, 1⁄2 − (√3⁄2)i	regular hexagon

Properties of Roots of Unity

The

n

-th roots of unity satisfy algebraic identities that reflect their geometric symmetry. These properties appear throughout mathematics, from polynomial factorization to Fourier analysis.

The sum of all

n

-th roots of unity equals zero:

Sum of Roots of Unity

\sum_{k=0}^{n-1} e^{i \, 2\pi k / n} = 0 \quad (n \geq 2)

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Geometrically, the roots form a regular polygon centered at the origin. Their vector sum — placing all arrows tip to tail — returns to the starting point, yielding zero. Algebraically, this is the sum of a geometric series with ratio

\omega \neq 1

\frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0

.

The product of all

n

-th roots of unity follows a pattern:

\prod_{k=0}^{n-1} z_k = (-1)^{n+1}

For odd

n

, the product is

1

. For even

n

, the product is

-1

. This connects to the constant term in the polynomial

z^n - 1

, which factors as

(z - z_0)(z - z_1)\cdots(z - z_{n-1})

.

When

n \geq 2

, complex roots of unity come in conjugate pairs. If

\omega^k

is a root, so is

\overline{\omega^k} = \omega^{n-k}

. The roots

\omega

and

\omega^{n-1}

are conjugates;

\omega^2

and

\omega^{n-2}

are conjugates; and so on. Only

z = 1

(always) and

z = -1

(when

n

is even) sit on the real axis without conjugate partners among the roots.

These symmetries constrain the algebraic behavior of expressions involving roots of unity and simplify many calculations.

Worked Examples

Concrete examples consolidate the root-finding technique. Each follows the same pattern: express the target in trigonometric form, apply the formula, enumerate all

n

roots.

Square roots of $i$: Solve

z^2 = i

.

Write

i = 1\text{cis}(90°)

. The roots are

z_k = 1^{1/2}\text{cis}\left(\frac{90° + 360°k}{2}\right)

for

k = 0, 1

z_0 = \text{cis}(45°) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i

z_1 = \text{cis}(225°) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i

Two roots, opposite each other on the unit circle.

Cube roots of $-8$: Solve

z^3 = -8

.

Write

-8 = 8\text{cis}(180°)

. The roots are

z_k = 2\text{cis}\left(\frac{180° + 360°k}{3}\right)

for

k = 0, 1, 2

z_0 = 2\text{cis}(60°) = 1 + \sqrt{3}i

z_1 = 2\text{cis}(180°) = -2

z_2 = 2\text{cis}(300°) = 1 - \sqrt{3}i

Three roots forming an equilateral triangle with radius

2

.

Fourth roots of unity: Solve

z^4 = 1

.

Write

1 = 1\text{cis}(0°)

. The roots are

z_k = \text{cis}(90°k)

for

k = 0, 1, 2, 3

z_0 = 1

z_1 = i

z_2 = -1

z_3 = -i

A square on the unit circle at the four cardinal points.

Fifth roots of $32$: Solve

z^5 = 32

.

Write

32 = 32\text{cis}(0°)

. The roots are

z_k = 2\text{cis}(72°k)

for

k = 0, 1, 2, 3, 4

z_0 = 2

z_1 = 2\text{cis}(72°)

z_2 = 2\text{cis}(144°)

z_3 = 2\text{cis}(216°)

z_4 = 2\text{cis}(288°)

A regular pentagon with radius

2

, one vertex on the positive real axis.

Summary: Forward and Reverse Directions of De Moivre's Theorem

De Moivre's theorem operates in two directions. Run forward, it turns a complex number into its

n

th power — a single answer obtained by raising the modulus and multiplying the argument. Run in reverse, it produces every

n

th root of a given number —

n

distinct values arranged as a regular polygon. The table below collects both directions side by side: formula, what happens to modulus and argument, how many results to expect, and the geometric picture each direction draws on the complex plane.

Aspect	Forward direction (powers)	Reverse direction (n-th roots)
Question	given z, what is zⁿ ?	given w, what are the z with zⁿ = w ?
Formula	zⁿ = rⁿ cis(nθ)	zₖ = RŁⁿ cis((φ + 360°k) ⁄ n), k = 0…n−1
Modulus operation	r → rⁿ (raise to nth power)	R → RŁⁿ (take positive nth root)
Argument operation	θ → nθ (multiply by n)	φ → (φ + 360°k) ⁄ n (divide, then shift by k turns)
Number of results	exactly 1 (a single value)	exactly n (distinct roots)
Geometric picture	a single point at distance rⁿ from the origin	vertices of a regular n-gon on the circle of radius RŁⁿ
Worked example	(1 + i)¹⁰ = 32i (one answer)	z⁴ = −16: four roots √2(±1 ± i) (a square)

Demoivre Theorem

Powers and Roots Made Simple

Key Terms

Introduction to De Moivre's Theorem

De Moivre's Formula

Proof of De Moivre's Theorem

Applying De Moivre's Theorem to Powers

Introduction to nnn-th Roots

The Formula for nnn-th Roots

Geometric Interpretation of Roots

Roots of Unity

Properties of Roots of Unity

Worked Examples

Summary: Forward and Reverse Directions of De Moivre's Theorem

Introduction to $n$ -th Roots

The Formula for $n$ -th Roots