De Moivre's Theorem: Powers & Roots of ℂ

Introduction to De Moivre's Theorem

What is (1 + i)⁸?

Multiplying out eight brackets of

(1 + i)

is long and easy to get wrong. Watching the length and the angle of each power tells a much simpler story.

Each power multiplies the length by √2 and turns the angle by another 45°.

power	multiplied out	length (modulus)	angle (argument)
z	1 + i	√2	45°
z²	2i	2	90°
z³	−2 + 2i	2√2	135°
z⁴	−4	4	180°
z⁸	16	16	360°

The lengths go

\sqrt{2}, 2, 2\sqrt{2}, 4, \ldots

— powers of

\sqrt{2}

— and the angles go

45°, 90°, 135°, 180°, \ldots

— multiples of

45°

. Eight steps reach length

16

at angle

360°

, the number

16

.

In algebraic form, powers drown in computation; in trigonometric form, they follow one simple rule.

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}

or

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.

The definition

De Moivre's theorem says that for a complex number in trigonometric form and any integer

n

,

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

: raise the modulus to the power, multiply the angle by

n

.

The formula is set out in full, with its special cases, in De Moivre's formula.

So

(1 + i)^8 = (\sqrt{2})^8\,\text{cis}(8 \cdot 45°) = 16\,\text{cis}\,360° = 16

.

The theorem turns a power into one power of a length and one multiple of an angle.

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.

The identities come from computing the same power in two ways. Expanding

(\cos\theta + i\sin\theta)^3

term by term gives the real part

\cos^3\theta - 3\cos\theta\sin^2\theta

, while De Moivre's theorem says the real part is

\cos 3\theta

. Replacing

\sin^2\theta

by

1 - \cos^2\theta

turns this into

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

, and the same comparison for any

n

writes

\cos n\theta

and

\sin n\theta

as polynomials in

\cos\theta

and

\sin\theta

.

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.

Fact: A degree-45 equation solved by angles

In 1593 Adriaan van Roomen challenged the world's mathematicians with

x^{45} - 45x^{43} + 945x^{41} - \cdots + 45x = A

. Viète saw that the left side is

2\sin 45\theta

written in terms of

x = 2\sin\theta

, the identity this theorem gives for

n = 45

, and listed 23 positive solutions.

The number must first be in trigonometric form, and the angle is multiplied by n, not raised to the power n.

(\text{cis}\,45°)^3

is

\text{cis}\,135°

, not an angle of

45^3

degrees. Turning

a + bi

into

r\,\text{cis}\,\theta

takes the modulus and the argument, as described in trigonometric form.

The rest of the page covers De Moivre's formula, its proof, applying it to powers and roots, and the trigonometric identities it produces.

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)

Learn more about this formula: De Moivre's Theorem →

Browse all complex numbers formulas →

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:

De Moivre's Formula

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

Pitfall: Forgetting to raise the modulus

Writing

(2\text{cis}(45°))^4

as

2\text{cis}(180°) = -2

multiplies the angle but leaves the modulus at

2

. The modulus is raised to the fourth power too:

2^4\text{cis}(180°) = 16\text{cis}(180°) = -16

.

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

:

Proof of De Moivre's Theorem

(\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

:

Proof of De Moivre's Theorem

(\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:

Proof of De Moivre's Theorem

= \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

:

Proof of De Moivre's Theorem

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.

Applying De Moivre's Theorem to Powers

(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°

:

Applying De Moivre's Theorem to Powers

= 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.

Pitfall: Taking the argument from the tangent alone

Taking

\theta = \arctan\frac{1}{-1} = -45°

for

-1 + i

gives

(-1 + i)^3 = (\sqrt{2})^3\text{cis}(-135°) = -2 - 2i

. The point lies in the second quadrant, so

\theta = 135°

and

(-1 + i)^3 = 2\sqrt{2}\text{cis}(405°) = 2 + 2i

.

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

A tenth power in a few lines

Multiplying

1 + i

by itself ten times is tedious and error-prone. In trigonometric form the power is one multiplication of the angle.

01

Algebraic formBefore

(1 + i)^{10}

↓Convert to trigonometric form— modulus √2, argument 45° (first quadrant)

02

Trigonometric form

\left(\sqrt{2}\,\text{cis}\,45°\right)^{10}

↓Apply De Moivre's theorem— modulus to the 10th power, angle times 10: (√2)¹⁰ = 32, 10 × 45° = 450°

03

Power in trigonometric form

32\,\text{cis}\,450°

↓Reduce the angle and convert back— 450° = 360° + 90°, and cis 90° = i

04

Algebraic formAfter

32i

(1 + i)¹⁰ by De Moivre's theorem·/complex-numbers/demoivre-theoremLearn Math Class

The power itself is a single move in trigonometric form; the conversions at either end carry the work. The section's second example,

(\sqrt{3} - i)^6 = -64

, follows the same three moves.

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°)

or

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

to

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 three cube roots of 8

Write 8 = 8 cis 0°. Every cube root has modulus 8^(1/3) = 2, so all three sit on the circle of radius 2. Their arguments are (0° + 360°k)/3 for k = 0, 1, 2: 0°, 120° and 240°. The real root 2 is the one real-number experience finds; the other two, −1 ± √3 i, are a conjugate pair, and each cubes to 8 because tripling 120° or 240° lands on a multiple of 360°.

Take k = 3 and the angle is 360°, the same point as k = 0: the list of roots closes after n steps.

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

Learn more about this formula: Nth Roots Formula →

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In degree measure:

The Formula for

n

-th Roots

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.

Pitfall: Adding the full turn after dividing

For

z^3 = 8i

, taking the angles

\frac{90°}{3} + 360°k = 30° + 360°k

gives the single root

2\text{cis}(30°)

three times. Adding

360°k

before dividing gives

\frac{90° + 360°k}{3} = 30°, 150°, 270°

: the roots

\sqrt{3} + i

,

-\sqrt{3} + i

and

-2i

.

The n-th roots of w = R cis φ

one modulus for all n roots; the arguments start at φ/n and step evenly around the circle

zk=R1/ncis(φ+2πkn)

1The k-th root

2Same modulus

3Direction

4Original argument

5Full turns

6Even spacing

1The k-th rootzₖ

Each value of k from 0 to n − 1 gives one root; together they are all solutions of zⁿ = w. Example: the cube roots of 8 are z₀ = 2, z₁ = −1 + √3 i and z₂ = −1 − √3 i. Each cubes to 8.

2Same modulusR¹ᐟⁿ

The positive real n-th root of the original modulus, shared by all n roots; the roots differ only in their arguments. Example: 8 = 8 cis 0°, so R = 8 and every cube root has modulus 8¹ᐟ³ = 2.

3Directioncis

It turns the angle in brackets into a point on the unit circle; the modulus in front scales it out to radius R¹ᐟⁿ. Example: 2 cis 120° = 2(−1/2 + (√3/2) i) = −1 + √3 i.

4Original argumentφ

For k = 0 the argument is φ/n: the original angle divided by n. That first root is where the evenly spaced set starts. Example: φ = 0° for w = 8, so z₀ = 2 cis 0° = 2.

5Full turnsk = 0 … n−1

Adding 2πk (360°k) leaves w unchanged, but after dividing by n each k lands somewhere new. Adding the turns after dividing gives the same root n times. Example: for z³ = 8i, (90° + 360°k)/3 gives 30°, 150°, 270°.

6Even spacingn

Dividing by n spreads the n roots evenly around the circle of radius R¹ᐟⁿ. Example: for the cube roots of 8 the step is 360°/3 = 120°, giving the angles 0°, 120° and 240°.

The n-th roots formula, part by part·/complex-numbers/demoivre-theoremLearn Math Class

The part most often misread is the order of the full turns and the division: add

2\pi k

first, then divide by

n

. The roots of unity are the case

w = 1

.

Geometric Interpretation of Roots

The

n

-th roots of any nonzero 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 nonzero number sit

120°

apart. Fourth roots sit

90°

apart. Fifth roots sit

72°

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

Fact: A polygon's chords multiply to

n

In a regular

n

-gon drawn in a circle of radius

1

, the

n - 1

segments from one corner to all the others multiply to exactly

n

: for a square,

\sqrt{2} \cdot 2 \cdot \sqrt{2} = 4

. The corners are the roots of

z^n - 1

; dividing out

z - 1

leaves

1 + z + \cdots + z^{n-1}

, the product of

z

minus each other corner, and at

z = 1

it equals

n

.

For

n \geq 3

, 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 (for

n \geq 3

), 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. In that case

\phi = 0°

or

180°

, 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 R^1/n
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 (n ≥ 3)

Whatever the nonzero number, its

n

-th roots cut one circle into

n

equal arcs.

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

Learn more about this formula: Roots of Unity →

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RULE

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

an

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.

A primitive

n

-th root of unity, usually written

\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

One number,

\omega

, generates every

n

-th root of unity: its powers walk the unit circle in equal steps.

Notation for $n$ -th Roots

Notation

Notation for $n$ -th Roots

Root-finding invents three marks of its own: the running label that names each answer, the Greek letter that generates them all, and a fractional exponent that means one thing on positive reals and

n

things on complex numbers.

z

and

\mathbb{C}

come from complex numbers basics,

\operatorname{cis}

from the trigonometric form,

e^{i\theta}

and the periodic exponent from the exponential form, the conjugate bar from its page.

z_k

z sub k — the k-th root

The enumeration subscript with a job description:

k

runs

0, 1, \ldots, n-1

, one label per root — see The Formula for n-th Roots above. Beyond

k = n-1

the labels wrap (

z_n

would repeat

z_0

), which is exactly why the list stops.

Cases

k = 0

marks the principal root, argument

\phi/n

; the target number wears capitals to stay distinct —

w = R\operatorname{cis}\phi

in,

z_k

out. The same enumeration job as quadratic roots, here with a running range.

Do not confuseA power.

z_k

is the

k

-th label,

z^k

the

k

-th power — subscript versus superscript, and roots of unity put both in one equation:

z_k = \omega^k

.

\omega

omega — the primitive root of unity

One letter generates them all:

\omega = \operatorname{cis}(2\pi/n)

, and the full set of

n

-th roots of unity is

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

— Roots of Unity above. The convention travels intact into group theory, number theory and Fourier analysis.

CasesThe bare letter assumes context fixes

n

; texts juggling several orders subscript it —

\omega_n = e^{2\pi i/n}

, the spelling signal processing uses for the DFT's twiddle factors.

Do not confuseThe letter

w

. In handwriting

\omega

and

w

collapse into each other — and this very page uses

w

for the number whose roots are sought; keeping the Greek curl visible is not pedantry here.

Same glyph elsewherePhysics' angular frequency — the

\omega

of

Ae^{i\omega t}

in the exponential form's engineering costume — an unrelated job for the same letter.

R^{1/n}

·

w^{1/n}

R to the one over n

Two bases, two behaviors. On the positive real

R

, the fractional exponent means the single positive root — what makes

R^{1/n}

safe in the roots formula. On a complex

w

, the same exponent is

n

-valued:

w^{1/n}

without a chosen branch names all

n

roots at once.

CasesThe formula splits the jobs deliberately: single-valued

R^{1/n}

for the modulus, the

+2\pi k

in the angle for the multiplicity — multivaluedness lives in the periodic exponent, never in the real root.

Do not confuseThe radical habit.

\sqrt[3]{8} = 2

on the real line, but "the" cube root of

8i

does not exist — three do; importing the single-answer instinct from real radicals is how the other

n - 1

roots vanish from solutions.

All complex number symbols →↑ Complex Numbers

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Notation · Cheat sheet

Complex Number Notation at a Glance

The marks from the notation cards across the complex-number pages: how each is read and what it means. Every group links to the full card on its page.

Basics

z, \; w

z, w: complex numbers

The usual letters for complex numbers, as

x

and

y

are for reals.

a + bi

a plus b i

Real part

a

, imaginary part

b

; both are real numbers.

\mathbb{C}

C, the complex numbers

The set of all

a + bi

; it contains

\mathbb{R}

.

Imaginary numbers

i

i, the imaginary unit

The number with

i^2 = -1

.

\sqrt{-1}

the square root of minus one

Another name for

i

; root rules for reals fail with it.

bi

b i

A pure imaginary number: real part

0

.

Algebraic form

\operatorname{Re}(z), \; \operatorname{Im}(z)

real part, imaginary part

For

z = a + bi

:

a

and

b

.

\operatorname{Im}(z)

is

b

, not

bi

.

a + (-b)i

a plus negative b, i

Written

a - bi

; the imaginary part is

-b

.

z = x + yi

z equals x plus y i

Same form with plane letters: the point

(x, y)

.

Complex conjugate

\bar{z}

z bar, the conjugate

For

z = a + bi

,

\bar{z} = a - bi

: the mirror image across the real axis.

\overline{z_1 + z_2}

the conjugate of the sum

The bar covers the whole expression; it equals

\bar{z}_1 + \bar{z}_2

.

z^*

z star

The same conjugate, as written in physics and engineering.

Absolute value

|z|

the modulus of z

Distance from

0

:

\sqrt{a^2 + b^2}

.

|z|^2 = z\bar{z}

mod z squared

Equals

a^2 + b^2

, with no root to take.

|z - c| = R

mod z minus c equals R

The circle of radius

R

around

c

.

Additive inverse

-z

negative z

Both parts change sign:

-a - bi

.

-(a + bi)

minus, quantity a plus b i

The brackets carry the minus to both parts.

z - w

z minus w

Adding the inverse:

z + (-w)

.

Multiplicative inverse

z^{-1}, \; \frac{1}{z}

z inverse; one over z

The number with

z \cdot z^{-1} = 1

; it equals

\frac{\bar{z}}{|z|^2}

.

(3 + 2i)^{-1}

the inverse of three plus two i

Found by multiplying top and bottom by the conjugate:

\frac{3 - 2i}{13}

.

Trigonometric form

r, \; \theta

r and theta, the polar pair

Distance from

0

and angle from the positive real axis.

\arg(z)

the argument of z

Any angle of

z

; there are infinitely many,

2\pi

apart.

\operatorname{Arg}(z)

the principal argument

One chosen angle, usually in

(-\pi, \pi]

; some texts use

[0, 2\pi)

.

\operatorname{cis}\theta

sis theta

Short for

\cos\theta + i\sin\theta

.

Exponential form

e^{i\theta}

e to the i theta

Equals

\operatorname{cis}\theta

(Euler's formula): a point on the unit circle.

re^{i\theta}

r e to the i theta

The whole number

z

: length

r

, angle

\theta

.

e^{i(\theta + 2\pi k)}

theta plus two pi k

The same number for every integer

k

.

De Moivre's theoremThis page

z_k

z sub k, the k-th root

The roots of

z^n = w

, numbered

k = 0, \ldots, n - 1

.

\omega

omega

A primitive root of unity: its powers give all

n

roots of

1

.

R^{1/n}

R to the one over n

The real root of the length; the complex root

w^{1/n}

has

n

values.

Same marks, different job

|z|

The modulus here; for a real number, the absolute value; for a matrix, the determinant.

\bar{z}

The conjugate here; in statistics a bar means a mean, in sets a complement.

z^{-1}

vs

z^*

The reciprocal, against the conjugate: they agree only on the unit circle.

w^{1/n}

n

values for a complex

w

; a single real root for a positive real number.

Complex number symbols table Complex numbers

Complex Number Notation at a Glance·/complex-numbers/basicsLearn Math Class

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)

Learn more about this formula: Sum of Roots of Unity →

Browse all complex numbers formulas →

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:

Properties of Roots of Unity

\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.

DownloadPNG2×PDF

Odd or even n

The

n

-th roots of unity always sum to zero. Their product and which of them are real depend on whether

n

is odd or even.

Roots of unity

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

split by the parity of $n$

1

Case 1

n

odd

Product is 1

product of all roots

only

z = 1

sits on the real axis

−1

Case 2

n

even

Product is −1

product of all roots

z = 1

and

z = -1

sit on the real axis

Why

The product is the constant term of

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

, which gives

(-1)^{n+1}

. Every other root has its conjugate among the roots.

Roots of unity: odd or even n·/complex-numbers/demoivre-theoremLearn Math Class

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 the imaginary unit: 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 negative eight: 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 thirty-two: 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 = R^1/n cis((φ + 360°k) ⁄ n), k = 0…n−1
Modulus operation	r → rⁿ (raise to nth power)	R → R^1/n (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 (n ≥ 3) on the circle of radius R^1/n
Worked example	(1 + i)¹⁰ = 32i (one answer)	z⁴ = −16: four roots √2(±1 ± i) (a square)

Common Pitfalls

Four mistakes account for many of the wrong answers in work with De Moivre's theorem. Three are marked where they happen in the sections above; keeping only one root is the one the introduction to

n

-th roots already points out. Here each is worked through in full, with the wrong and the right working side by side.

Forgetting to raise the modulus

The angle is multiplied by

n

, but the modulus is left as it is, as if every number had modulus

1

.

Wrong working:

(2\text{cis}(45°))^4 = 2\text{cis}(4 \times 45°) = 2\text{cis}(180°) = -2

.

Right working:

(2\text{cis}(45°))^4 = 2^4\text{cis}(180°) = 16\text{cis}(180°) = -16

. In algebraic form

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

, its square is

2 + 4i + 2i^2 = 4i

, and

(4i)^2 = 16i^2 = -16

.

Why it happens: the theorem is first stated for

\cos\theta + i\sin\theta

, whose modulus is

1

, and it is remembered as "multiply the angle".

How to check: the modulus of

z^n

is

|z|^n

. Here

|z^4| = 2^4 = 16

, so

-2

cannot be right.

Taking the argument from the tangent alone

The argument is found as

\arctan\frac{b}{a}

without looking at the quadrant of the point, and De Moivre's theorem then multiplies the wrong angle.

Wrong working: for

z = -1 + i

,

\theta = \arctan\frac{1}{-1} = -45°

, so

(-1 + i)^3 = (\sqrt{2})^3\text{cis}(-135°) = 2\sqrt{2}\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) = -2 - 2i

.

Right working: the point

(-1, 1)

lies in the second quadrant, so

\theta = 180° - 45° = 135°

. Then

(-1 + i)^3 = 2\sqrt{2}\text{cis}(405°) = 2\sqrt{2}\text{cis}(45°) = 2 + 2i

. Directly,

(-1 + i)^2 = 1 - 2i + i^2 = -2i

and

-2i(-1 + i) = 2i - 2i^2 = 2 + 2i

.

Why it happens:

\arctan

only returns angles between

-90°

and

90°

, so it never gives the direction of a point with

a < 0

.

How to check: the signs of

\cos\theta

and

\sin\theta

must match the signs of

a

and

b

. Here

\cos(-45°) > 0

while

a = -1 < 0

.

Keeping only one root

An equation

z^n = w

is solved with the single root that real-number experience suggests, and the other

n - 1

roots are never looked for.

Wrong working:

z^4 = 16

, so

z = 2

, or at most

z = \pm 2

.

Right working:

16 = 16\text{cis}(0°)

, so

z_k = 16^{1/4}\text{cis}\left(\frac{360°k}{4}\right) = 2\text{cis}(90°k)

for

k = 0, 1, 2, 3

. The four roots are

2

,

2i

,

-2

and

-2i

; for example

(2i)^4 = 16i^4 = 16

.

Why it happens: the radical sign and the exponent

\frac{1}{n}

name a single value on the positive reals, and that habit is carried over to

\mathbb{C}

.

How to check: count the roots. A nonzero number has exactly

n

distinct

n

-th roots, evenly spaced

\frac{360°}{n}

apart on one circle.

Adding the full turn after dividing

In the roots formula, the angle

\phi

is divided by

n

first and the multiples of

360°

are added afterwards. Every value of

k

then gives the same direction.

Wrong working: for

z^3 = 8i = 8\text{cis}(90°)

, the angles

\frac{90°}{3} + 360°k = 30°, 390°, 750°

all point the same way, so the only root found is

2\text{cis}(30°) = \sqrt{3} + i

.

Right working: the full turn is added before dividing,

\alpha = \frac{90° + 360°k}{3} = 30°, 150°, 270°

. The roots are

2\text{cis}(30°) = \sqrt{3} + i

,

2\text{cis}(150°) = -\sqrt{3} + i

and

2\text{cis}(270°) = -2i

. For example

(-2i)^3 = -8i^3 = 8i

.

Why it happens: the identity

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

is applied to the root's angle instead of to the angle of

w

, which is the one that can be written in many ways.

How to check: consecutive roots must differ by

\frac{360°}{n}

, here

120°

. If two of the listed roots coincide, the turns were added in the wrong place.

De Moivre's Theorem FAQ

How many nth roots does a complex number have?

+

Exactly n of them, all distinct, provided the number is not zero. The reason is that the argument is only determined up to full turns, so dividing it by n produces n genuinely different angles before the values start repeating. Real numbers are no exception: a nonzero real number has at most two real nth roots, and the rest are not real.Read more →

Why do the nth roots form a regular polygon?

+

Because all n share the same modulus, placing them on one circle, and their arguments differ by equal steps of a full turn divided by n. Equal radii and equal angular spacing is precisely the definition of a regular polygon inscribed in that circle.Read more →

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

+

Zero, for every n greater than one. Geometrically the roots are symmetrically placed around the circle, so as vectors they cancel exactly. Algebraically it follows from the polynomial they satisfy, whose second coefficient is zero and therefore forces the sum of the roots to vanish.Read more →

How do you get the triple-angle formula from De Moivre's theorem?

+

Compute

(\cos\theta + i\sin\theta)^3

in two ways. Expanding the cube gives the real part

\cos^3\theta - 3\cos\theta\sin^2\theta

, while De Moivre's theorem says the real part is

\cos 3\theta

. Replacing

\sin^2\theta

with

1 - \cos^2\theta

gives

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

, and the imaginary parts give

\sin 3\theta = 3\sin\theta - 4\sin^3\theta

the same way. Any

n

works, writing

\cos n\theta

and

\sin n\theta

as polynomials in

\cos\theta

and

\sin\theta

.Read more →

What is the product of all the nth roots of unity?

+

(-1)^{n+1}

: it is

1

when

n

is odd and

-1

when

n

is even. The roots are exactly the roots of

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

, and comparing constant terms gives

(-1)^n z_0z_1\cdots z_{n-1} = -1

. For

n = 4

,

1 \cdot i \cdot (-1) \cdot (-i) = -1

; for

n = 3

, the two non-real roots are conjugates whose product is

1

.Read more →

What do the distances from one corner of a regular polygon to all the others multiply to?

+

Exactly

n

, for a regular

n

-gon drawn in a circle of radius

1

. Place the corners at the

n

th roots of unity, with the chosen corner at

1

. Dividing

z^n - 1

by

z - 1

leaves

1 + z + \cdots + z^{n-1}

, which is the product of

z

minus each other corner, and at

z = 1

it equals

n

. For a square the three distances are

\sqrt{2}

,

2

and

\sqrt{2}

, with product

4

.Read more →

Does De Moivre's theorem work for fractional powers?

+

Not as a single formula. It holds for integer powers, where

z^n

has one value. For the power

\frac{1}{n}

, dividing the angle gives only one of the

n

roots, because

\theta

can be replaced by

\theta + 2\pi k

before dividing. With

-1 = \text{cis}\,\pi

and

n = 2

, the angle

\pi

gives

\text{cis}\,\frac{\pi}{2} = i

, while the same number written as

\text{cis}\,3\pi

gives

\text{cis}\,\frac{3\pi}{2} = -i

; both square to

-1

.Read more →

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

A tenth power in a few lines

Introduction to $n$ -th Roots

The Formula for $n$ -th Roots

The n-th roots of w = R cis φ

Geometric Interpretation of Roots

Roots of Unity

Notation for $n$ -th Roots

Notation for $n$ -th Roots

Properties of Roots of Unity

Worked Examples

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

Common Pitfalls

De Moivre's Theorem FAQ

How many nth roots does a complex number have?

Why do the nth roots form a regular polygon?

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

How do you get the triple-angle formula from De Moivre's theorem?

What is the product of all the nth roots of unity?

What do the distances from one corner of a regular polygon to all the others multiply to?

Does De Moivre's theorem work for fractional powers?

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

A tenth power in a few lines

Introduction to nnn-th Roots

The Formula for nnn-th Roots

The n-th roots of w = R cis φ

Geometric Interpretation of Roots

Roots of Unity

Notation for nnn-th Roots

Notation for nnn-th Roots

Properties of Roots of Unity

Worked Examples

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

Common Pitfalls

De Moivre's Theorem FAQ

How many nth roots does a complex number have?

Why do the nth roots form a regular polygon?

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

How do you get the triple-angle formula from De Moivre's theorem?

What is the product of all the nth roots of unity?

What do the distances from one corner of a regular polygon to all the others multiply to?

Does De Moivre's theorem work for fractional powers?

Introduction to $n$ -th Roots

The Formula for $n$ -th Roots

Notation for $n$ -th Roots

Notation for $n$ -th Roots