Roots of unity | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

Regular polygons on the unit circle

The roots of unity solve $z^{n} = 1$ . Their common modulus is one, and their arguments are $2 k π / n$ for $k = 0, \dots, n - 1$ . They form a regular polygon on the unit circle, with $1$ as one vertex.

Set $ω = cis (2 π / n)$ . Then $ω^{n} = 1$ , so powers of $ω$ repeat every $n$ exponents. Reduce an exponent modulo $n$ to replace a large power by one of $1, ω, \dots, ω^{n - 1}$ .

Equal-scale Argand plane with the unit circle and an inscribed square joining the four labelled roots 1, i, minus 1 and minus i.

Sums, inverses and conjugates

For $n > 1$ , let $S = 1 + ω + \dots + ω^{n - 1}$ . Multiplying by $1 - ω$ makes intermediate terms cancel: $(1 - ω) S = 1 - ω^{n} = 0$ . Since $ω \neq 1$ , division gives $S = 0$ .

A unit-circle point satisfies $z \bar{z} = 1$ , so its conjugate is also its reciprocal. Non-real roots pair across the real axis. The real root $1$ , and also $- 1$ when $n$ is even, are their own conjugates.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Work with cube roots symbolically

Let $ω = cis (\frac{2 π}{3})$ . Simplify $ω^{8} + ω^{7} + 1$ .

Worked solution

1
Reduce the exponents modulo three.
$ω^{8} = {(ω^{3})}^{2} ω^{2} = ω^{2}, ω^{7} = {(ω^{3})}^{2} ω = ω$
The given cube root satisfies $ω^{3} = 1$ .
2
Establish the sum of the cube roots.
$ω^{3} - 1 = (ω - 1) (ω^{2} + ω + 1) = 0$
The difference of cubes factors exactly.
3
Use the nontrivial root.
$ω \neq 1 ⟹ ω^{2} + ω + 1 = 0$
The factor $ω - 1$ is nonzero, so the other factor must vanish.
4
Evaluate the original expression.
$ω^{8} + ω^{7} + 1 = ω^{2} + ω + 1 = 0$
The reduced powers form the complete cube-root sum.

Answer

$0$ .

EXAMPLE 02

Measure a side of a root polygon

Find the distance between adjacent sixth roots of unity $1$ and $cis (\frac{π}{3})$ .

Worked solution

1
Convert the second vertex to coordinates.
$cis \frac{π}{3} = \frac{1}{2} + \frac{\sqrt{3}}{2} i$
Exact trigonometric values locate the sixth root.
2
Find the displacement from the first vertex.
$(\frac{1}{2} + \frac{\sqrt{3}}{2} i) - 1 = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$
Distance between complex points is the modulus of their difference.
3
Calculate its modulus.
$d = \sqrt{{(- \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = 1$
The squared coordinate differences sum to one.

Answer

$1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the sum of all fourth roots of unity.

Show answer & reasoning

$0$ .

The fourth roots are $1, i, - 1, - i$ . Opposite roots cancel in pairs, giving $1 + i - 1 - i = 0$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

List all fourth roots of unity in Cartesian form.
2 marks
Let $ω = cis (\frac{2 π}{3})$ . Simplify $ω^{11}$ .
2 marks
For $ω = cis (\frac{2 π}{3})$ , show that $\bar{ω} = ω^{2}$ .
2 marks
Find the product of all fourth roots of unity.
2 marks
Show that the sum of the five fifth roots of unity is zero.
3 marks
A regular hexagon has vertices at the sixth roots of unity. Find its perimeter.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

List all fourth roots of unity in Cartesian form.
2 marks
The angles are $0, \frac{π}{2}, π, \frac{3 π}{2}$ .
$1, i, - 1, - i$ .
Let $ω = cis (\frac{2 π}{3})$ . Simplify $ω^{11}$ .
2 marks
$11 = 3 \times 3 + 2$ , and every factor $ω^{3}$ is 1.
$ω^{2}$ .
For $ω = cis (\frac{2 π}{3})$ , show that $\bar{ω} = ω^{2}$ .
2 marks
Conjugation negates the argument; $- \frac{2 π}{3}$ differs from $\frac{4 π}{3}$ by a full turn.
Both equal $- \frac{1}{2} - \frac{\sqrt{3}}{2} i$ .
Find the product of all fourth roots of unity.
2 marks
$1 \times i \times (- 1) \times (- i) = - 1$ .
$- 1$ .
Show that the sum of the five fifth roots of unity is zero.
3 marks
Multiply the sum by $1 - ω$ to get $1 - ω^{5} = 0$ . Since $ω \neq 1$ , divide by $1 - ω$ .
$1 + ω + ω^{2} + ω^{3} + ω^{4} = 0$ , where $ω = cis (\frac{2 π}{5})$ .
A regular hexagon has vertices at the sixth roots of unity. Find its perimeter.
3 marks
Adjacent vertices differ by a chord subtending $\frac{π}{3}$ , of length $2 sin (\frac{π}{6}) = 1$ . There are six sides.
$6$ units.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Regular polygons on the unit circle

Sums, inverses and conjugates

See the method in action.

Work with cube roots symbolically

Reduce the exponents modulo three.

Establish the sum of the cube roots.

Use the nontrivial root.

Evaluate the original expression.

Measure a side of a root polygon

Convert the second vertex to coordinates.

Find the displacement from the first vertex.

Calculate its modulus.

Try it yourself.

Check your understanding

Keep practising