Roots of unity | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

Regular polygons on the unit circle

The nth roots of unity solve $z^{n} = 1$ and are $cis (2 k π / n)$ for $k = 0, \dots, n - 1$ . They all have modulus 1 and are equally spaced around the unit circle. One root is always 1.

If $ω = cis (2 π / n)$ , the roots are $1, ω, ω^{2}, \dots, ω^{n - 1}$ . Exponents may be reduced modulo n because $ω^{n} = 1$ .

Sums, inverses and conjugates

For $n > 1$ , the geometric-series identity gives $1 + ω + \dots + ω^{n - 1} = 0$ when $ω = cis (2 π / n)$ . Multiplying the sum by $1 - ω$ gives $1 - ω^{n} = 0$ , and $ω \neq 1$ .

On the unit circle, $\bar{z} = 1 / z$ . Thus roots occur in conjugate pairs except real roots, which equal their own conjugates. The geometric arrangement explains why the vector sum is zero.

A common trap

Every root of unity has a non-real conjugate partner.

Instead: Real roots 1 and, when n is even, -1 are their own conjugates. Other roots pair across the real axis.

02 / WORKED EXAMPLES

See the thinking

Example 1

Work with cube roots symbolically

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

Use $ω^{3} = 1$ : $ω^{8} = ω^{2}$ and $ω^{7} = ω$ .
Then $1 + ω + ω^{2} = 0$ .

0

Example 2

Measure a side of a root polygon

Find the distance between adjacent sixth roots of unity $1$ and $cis (π / 3)$ .

$cis (π / 3) = 1 / 2 + (\sqrt{3} / 2) i$ .
The difference is $- 1 / 2 + (\sqrt{3} / 2) i$ , whose modulus is $\sqrt{1 / 4 + 3 / 4} = 1$ .

1

Check your understanding

Find the sum of all fourth roots of unity.

Show answer

$0$ .

$1 + i - 1 - i = 0$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

List all fourth roots of unity in Cartesian form.
2 marks
Let $ω = cis (2 π / 3)$ . Simplify $ω^{11}$ .
2 marks
For $ω = cis (2 π / 3)$ , 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 $0, π / 2, π, 3 π / 2$ .
$1, i, - 1, - i$ .
Let $ω = cis (2 π / 3)$ . Simplify $ω^{11}$ .
2 marks
$11 = 3 \times 3 + 2$ , and every factor $ω^{3}$ is 1.
$ω^{2}$ .
For $ω = cis (2 π / 3)$ , show that $\bar{ω} = ω^{2}$ .
2 marks
Conjugation negates the argument; $- 2 π / 3$ differs from $4 π / 3$ 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 $ω = cis (2 π / 5)$ .
A regular hexagon has vertices at the sixth roots of unity. Find its perimeter.
3 marks
Adjacent vertices differ by a chord subtending $π / 3$ , of length $2 sin (π / 6) = 1$ . 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 ↗

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