The 6 roots of unity: solve z^6 = 1, step by step

Femtobytes

femtobytes.com/solve/roots-of-unity/6

The problem

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

z^{6} = 1

The result

\omega_k = \cos \frac{2k\pi}{6} + i \sin \frac{2k\pi}{6}, \quad k = 0, 1, \ldots, 5

There are exactly $6$ solutions, equally spaced round the unit circle, and their sum is $0$ .

checked 12 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 5Step 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, $z^{6} = 1$ has two solutions, $1$ and $-1$ . Yet an equation of degree $6$ is entitled to $6$ 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 $6$ raises the length to that power and multiplies the angle by $6$ : $z^{6} = \rho^{6} \left( \cos 6\varphi + i \sin 6\varphi \right)$ .

\rho^{6} = 1 \qquad \text{and} \qquad 6\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 $6$ different solutions come from: a full turn, shared out $6$ ways, gives $6$ different directions before it starts repeating.

\rho = 1, \qquad \varphi_k = \frac{2k\pi}{6}, \quad k = 0, 1, \ldots, 5

Dividing

2k\pi

6

. Taking

k = 6

would give

2\pi

, a full turn, which is the same direction as

k = 0

; so

k

stops at

5

\varphi_{0} = 0

With

k = 0

. checked

\varphi_{1} = \frac{\pi}{3}

With

k = 1

\frac{2 \pi}{6}

in lowest terms, which is

60^\circ

. checked

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

With

k = 2

\frac{4 \pi}{6}

in lowest terms, which is

120^\circ

. checked

\varphi_{3} = \pi

With

k = 3

\frac{6 \pi}{6}

in lowest terms, which is

180^\circ

. checked

\varphi_{4} = \frac{4 \pi}{3}

With

k = 4

\frac{8 \pi}{6}

in lowest terms, which is

240^\circ

. checked

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

With

k = 5

\frac{10 \pi}{6}

in lowest terms, which is

300^\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.

Write out the roots

Why this step

Each root is the point of the unit circle in one of those directions: $\omega_k = \cos \varphi_k + i \sin \varphi_k$ ( $\omega$ is the Greek letter omega, the customary name for a root of unity). For this $n$ every angle is a special one, so every root can be written exactly.

root	angle	in the form $a + bi$
$\omega_{0}$	$0$	$1$
$\omega_{1}$	$\frac{\pi}{3}$	$\frac{1}{2} + \frac{\sqrt{3}}{2} i$
$\omega_{2}$	$\frac{2 \pi}{3}$	$-\frac{1}{2} + \frac{\sqrt{3}}{2} i$
$\omega_{3}$	$\pi$	$-1$
$\omega_{4}$	$\frac{4 \pi}{3}$	$-\frac{1}{2} - \frac{\sqrt{3}}{2} i$
$\omega_{5}$	$\frac{5 \pi}{3}$	$\frac{1}{2} - \frac{\sqrt{3}}{2} i$

The 6 solutions of z to the power 6 equals 1.

The 6 roots of unity. All have length 1, so they lie on the unit circle, and neighbours are 60.00 degrees apart: they are the corners of a regular polygon with 6 sides, one corner at the number 1.

See that every root is a power of the first one

Why this step

The list of roots looks like $6$ unrelated numbers. It is really ONE number, $\omega = \omega_1$ , and its powers. Multiplying by $\omega$ turns the plane by $\frac{2\pi}{6}$ without stretching it, so each multiplication steps to the next corner of the polygon, and after $6$ steps the turn is complete and we are back at $1$ .

\omega^{2} = \omega_{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i

By De Moivre,

\omega^{2}

has angle

2 \cdot \frac{2\pi}{6}

, which is exactly

\varphi_{2}

. checked

\omega^{3} = \omega_{3} = -1

By De Moivre,

\omega^{3}

has angle

3 \cdot \frac{2\pi}{6}

, which is exactly

\varphi_{3}

. checked

\omega^{4} = \omega_{4} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i

By De Moivre,

\omega^{4}

has angle

4 \cdot \frac{2\pi}{6}

, which is exactly

\varphi_{4}

. checked

\omega^{5} = \omega_{5} = \frac{1}{2} - \frac{\sqrt{3}}{2} i

By De Moivre,

\omega^{5}

has angle

5 \cdot \frac{2\pi}{6}

, which is exactly

\varphi_{5}

. checked

\omega^{6} = 1

After

6

steps the angle is

2\pi

: one full turn, back to

1

. This is the equation we set out to solve, seen as a journey round the circle. checked

Each root is the first one multiplied by itself again and again: one turn, repeated, walks around the whole polygon.

Prove that the roots add up to zero

Why this step

Adding the roots looks like a long calculation with cosines and sines. It is not. The sum is forced to be $0$ by symmetry alone, and seeing why is worth more than the fact: it is the same argument that makes the centre of a regular polygon its balance point.

Let $S = 1 + \omega + \omega^2 + \cdots + \omega^{5}$ , the sum of all the roots. Multiply both sides by $\omega$ :

\omega S = \omega + \omega^2 + \cdots + \omega^{5} + \omega^{6} = \omega + \omega^2 + \cdots + \omega^{5} + 1 = S

Every term moved up by one power, and the last one became

\omega^{6} = 1

. The result is the same list of terms in a different order, so it is

S

again.

\omega S = S \quad \text{so} \quad (\omega - 1) S = 0 \quad \text{so} \quad S = 0

Subtract

S

from both sides and take out the common factor

S

. A product is

0

only when one of its factors is

0

, and

\omega - 1

is not

0

because

\omega

is not

1

. So

S = 0

\omega_0 + \omega_1 + \cdots + \omega_{5} = 0

The same fact checked with the actual numbers from the table: the real parts cancel and the imaginary parts cancel. checked

The roots are evenly spaced about the origin, so as arrows they pull equally in every direction and cancel exactly. That is why they add to zero.

The n-th roots of unity