Problem type

The n-th roots of unity

What it solves: Solve z to the power n equals 1. Find all n solutions, see why they are the corners of a regular polygon, why each is a power of the first, and prove that they add up to zero.
Subject: Complex Numbers
Topic: Length and angle
Also called: roots of unity, nth roots of 1, cyclotomic
Free example: Read the complete worked example

Femtobytes

The 5 roots of unity: solve z^5 = 1, step by step

femtobytes.com/solve/roots-of-unity/5

The problem

Find every complex number $z$ that satisfies this equation.

z^{5} = 1

The result

\omega_k = \cos \frac{2k\pi}{5} + i \sin \frac{2k\pi}{5}, \quad k = 0, 1, \ldots, 4

There are exactly $5$ solutions, equally spaced round the unit circle, and their sum is $0$ .

checked 10 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: Say what is being asked, and what we already expect

The number 1, where every root of unity must land after being multiplied by itself n times. Each root is a turn that, done n times, comes back exactly to here.

Say what is being asked, and what we already expect

Why this step

Among real numbers, $z^{5} = 1$ has one solution, $1$ . Yet an equation of degree $5$ is entitled to $5$ solutions. The missing ones are not on the number line at all. They are in the plane, and the only way to find them is to stop thinking of multiplication as repeated adding and start thinking of it as turning.

The number $1$ , as a point of the plane, has length $1$ and points along the positive real axis, so its angle is $0$ , or equally $2\pi$ , $4\pi$ , and so on: $1 = \cos(2k\pi) + i \sin(2k\pi)$ for every whole number $k$ . Here $i$ is the imaginary unit with $i^2 = -1$ , and angles are in radians, where a full turn is $2\pi$ .

Write an unknown solution as $z = \rho \left( \cos \varphi + i \sin \varphi \right)$ with length $\rho$ (rho) and angle $\varphi$ (phi). De Moivre's theorem says raising to the power $5$ raises the length to that power and multiplies the angle by $5$ : $z^{5} = \rho^{5} \left( \cos 5\varphi + i \sin 5\varphi \right)$ .

\rho^{5} = 1 \qquad \text{and} \qquad 5\varphi = 2k\pi

Equal complex numbers have equal lengths, and angles that agree up to whole turns.

The number 1, where every root of unity must land after being multiplied by itself n times. Each root is a turn that, done n times, comes back exactly to here.

Solve for the length and the angles

Why this step

The length equation has no freedom in it: a length is a positive real number and the only positive real number whose power is $1$ is $1$ itself. So every solution lies on the unit circle. The angle equation is where the $5$ different solutions come from: a full turn, shared out $5$ ways, gives $5$ different directions before it starts repeating.

\rho = 1, \qquad \varphi_k = \frac{2k\pi}{5}, \quad k = 0, 1, \ldots, 4

Dividing

2k\pi

5

. Taking

k = 5

would give

2\pi

, a full turn, which is the same direction as

k = 0

; so

k

stops at

4

\varphi_{0} = 0

With

k = 0

. checked

\varphi_{1} = \frac{2 \pi}{5}

With

k = 1

\frac{2 \pi}{5}

in lowest terms, which is

72^\circ

. checked

\varphi_{2} = \frac{4 \pi}{5}

With

k = 2

\frac{4 \pi}{5}

in lowest terms, which is

144^\circ

. checked

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

With

k = 3

\frac{6 \pi}{5}

in lowest terms, which is

216^\circ

. checked

\varphi_{4} = \frac{8 \pi}{5}

With

k = 4

\frac{8 \pi}{5}

in lowest terms, which is

288^\circ

. checked

All the roots have length 1, so they sit on the unit circle, and their angles are the equal shares of a full turn.

The n-th roots of unity

Say what is being asked, and what we already expect

Solve for the length and the angles

Write out the roots

See that every root is a power of the first one

Prove that the roots add up to zero