What are properties of the cube roots?

The cube roots of unity 1, ω, ω^2 appear so often that their two key identities are worth memorising: 1 + ω + ω^2 = 0 and ω^3 = 1. These collapse otherwise messy expressions in a single line.

TASC TCE 2024-style practice: (a) Write down the three roots of z^3 = 1, giving the non-real roots as e^iθ with -π < θ π. (b) If one non-real root is ω, show that the other non-real root is ω^2. (c) Verify that 1 + ω + ω^2 = 0.

(a) The cube roots of unity are equally spaced around the unit circle at angles 0, 2π3 and -2π3. They are z = 1, z = e^2π i/3 and z = e^-2π i/3. (2 marks) (b) Let ω = e^2π i/3. Then ω^2 = e^4π i/3 = e^-2π i/3 (subtracting 2π to bring the argument into (-π, π]). This is exactly the other non-real root, so the two non-real roots are ω and ω^2. (1 mark) (c) The three roots 1, ω, ω^2 are the roots of z^3 - 1 = (z - 1)(z^2 + z + 1) = 0. Since ω and ω^2 are roots of z^2 + z + 1 = 0, the sum of all three roots equals the negative of the z^2 coefficient of z^3 - 1, which is 0. More directly, 1 + ω + ω^2 = ω^3 - 1ω - 1 = 0 because ω^3 = 1 and ω 1. (1 mark)

TASC TCE 2021-style practice: Solve z^5 + 1 = 0. Give your answers in polar form r e^iθ with -π < θ π.

Write -1 in polar form: -1 = e^iπ, and more generally -1 = e^i(π + 2π k) for integer k. Then z^5 = e^i(π + 2π k). Taking fifth roots, z = e^i(π + 2π k)/5 for k = 0, 1, 2, 3, 4, all with modulus r = 1. The arguments are π5, 3π5, π (from 5π5), and for the remaining two, 7π5 and 9π5, which reduce into (-π, π] as -3π5 and -π5. So the five solutions are z = e^iπ/5, e^3iπ/5, e^iπ = -1, e^-3iπ/5 and e^-iπ/5. They are equally spaced on the unit circle. Markers reward writing -1 with the general argument and listing all five roots with arguments in the required range.

§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

What are the nth roots of unity and why do they lie equally spaced on the unit circle?

Find the nth roots of unity and use their symmetry and algebraic properties.

Solving z to the n equals one, the geometry of equally spaced roots on the unit circle, and the sum and product properties of roots of unity, for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point is a focused special case of finding roots of complex numbers, but it deserves its own treatment because the roots of unity have a beautiful structure that recurs throughout the course. Solving $z^n = 1$ is the cleanest possible application of De Moivre's theorem, and the answers carry symmetry that simplifies sums, products and factorisations.

Setting up the roots

Write $1$ in polar form. Its modulus is $1$ and its argument is $0$ , but we must allow all coterminal arguments $0 + 2\pi k$ to capture every root. So $1 = \operatorname{cis}(2\pi k)$ for integer $k$ . Then

z^n = \operatorname{cis}(2\pi k) \quad\Longrightarrow\quad z = \operatorname{cis}\!\left(\frac{2\pi k}{n}\right), \qquad k = 0, 1, \dots, n-1.

Each root has modulus $1$ , so all $n$ roots lie on the unit circle. The arguments increase in equal steps of $\tfrac{2\pi}{n}$ , so the roots are evenly spaced, like the numbers on a clock face.

The geometry

Because the roots are vertices of a regular polygon centred at the origin, they have a strong rotational symmetry. Multiplying any root by $\omega$ rotates it to the next root anticlockwise. This polygon picture is often the quickest way to write down all roots without separate calculation for each.

The fifth roots of unity

Worked example

Solve $z^5 = 1$ and describe the roots geometrically.

The roots are $z_k = \operatorname{cis}\tfrac{2\pi k}{5}$ for $k = 0, 1, 2, 3, 4$ :

z_0 = \operatorname{cis}0 = 1, \quad z_1 = \operatorname{cis}\tfrac{2\pi}{5}, \quad z_2 = \operatorname{cis}\tfrac{4\pi}{5}, \quad z_3 = \operatorname{cis}\tfrac{6\pi}{5}, \quad z_4 = \operatorname{cis}\tfrac{8\pi}{5}.

All five have modulus $1$ , so they lie on the unit circle, spaced $\tfrac{2\pi}{5} = 72^\circ$ apart. They form a regular pentagon with one vertex at $z = 1$ . The non-real roots occur in two conjugate pairs: $z_4 = \overline{z_1}$ and $z_3 = \overline{z_2}$ , since the polygon is symmetric about the real axis.

The sum and product of the roots

This sum result is a favourite exam tool. For example, the real parts of the roots sum to zero, which gives identities such as $\cos\tfrac{2\pi}{5} + \cos\tfrac{4\pi}{5} + \cos\tfrac{6\pi}{5} + \cos\tfrac{8\pi}{5} = -1$ once you separate the root $z_0 = 1$ .

Connection to factorisation

Because the $n$ th roots of unity are exactly the solutions of $z^n - 1 = 0$ , the polynomial factorises as

z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}).

Dividing by $z - 1$ gives the useful identity $z^{n-1} + z^{n-2} + \dots + z + 1 = (z - \omega)\cdots(z - \omega^{n-1})$ , which links the geometric series to the non-trivial roots.

Properties of the cube roots: a worked staple

The cube roots of unity $1, \omega, \omega^2$ appear so often that their two key identities are worth memorising: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$ . These collapse otherwise messy expressions in a single line.

Simplifying with the cube roots of unity

If $\omega$ is a non-real cube root of unity, evaluate $(1 + \omega)(1 + \omega^2)$ and $(1 - \omega + \omega^2)$ .

Step 1: Recall the two key identities

Because $\omega$ is a cube root of unity, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$ . The second identity rearranges to $\omega + \omega^2 = -1$ , which is the substitution that collapses products and sums involving all three cube roots.

Step 2: Expand and simplify the product $(1 + \omega)(1 + \omega^2)$

Expand the brackets in the usual way:

(1 + \omega)(1 + \omega^2) = 1 + \omega^2 + \omega + \omega^3.

Now substitute: $\omega + \omega^2 = -1$ and $\omega^3 = 1$ :

= 1 + (-1) + 1 = 1.

Step 3: Simplify $1 - \omega + \omega^2$

Rewrite to reveal the sum-to-zero pattern. Add and subtract $2\omega$ :

1 - \omega + \omega^2 = (1 + \omega + \omega^2) - 2\omega = 0 - 2\omega = -2\omega.

Final answer: $(1 + \omega)(1 + \omega^2) = 1$ and $1 - \omega + \omega^2 = -2\omega$ .

Why this matters

Roots of unity model anything periodic and symmetric, and their cancellation property turns awkward trigonometric sums into one-line answers. Recognising a hidden $z^n = 1$ inside a problem is a powerful shortcut that connects this topic to factorisation, geometric series, and the multiple-angle identities of complex powers.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20244 marks(a) Write down the three roots of

z^3 = 1

, giving the non-real roots as

e^{i\theta}

with

-\pi < \theta \le \pi

. (b) If one non-real root is

\omega

, show that the other non-real root is

\omega^2

. (c) Verify that

1 + \omega + \omega^2 = 0

Show worked answer →

(a) The cube roots of unity are equally spaced around the unit circle at angles $0$ , $\dfrac{2\pi}{3}$ and $-\dfrac{2\pi}{3}$ . They are $z = 1$ , $z = e^{2\pi i/3}$ and $z = e^{-2\pi i/3}$ . (2 marks)

(b) Let $\omega = e^{2\pi i/3}$ . Then $\omega^2 = e^{4\pi i/3} = e^{-2\pi i/3}$ (subtracting $2\pi$ to bring the argument into $(-\pi, \pi]$ ). This is exactly the other non-real root, so the two non-real roots are $\omega$ and $\omega^2$ . (1 mark)

(c) The three roots $1, \omega, \omega^2$ are the roots of $z^3 - 1 = (z - 1)(z^2 + z + 1) = 0$ . Since $\omega$ and $\omega^2$ are roots of $z^2 + z + 1 = 0$ , the sum of all three roots equals the negative of the $z^2$ coefficient of $z^3 - 1$ , which is $0$ . More directly, $1 + \omega + \omega^2 = \dfrac{\omega^3 - 1}{\omega - 1} = 0$ because $\omega^3 = 1$ and $\omega \ne 1$ . (1 mark)

TCE 20226 marks(a) Write down the sum of the geometric series

\displaystyle\sum_{n=0}^{6} z^n

. (b) Hence determine the solutions of

z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0

. (c) Write this polynomial as a product of three quadratic polynomials with real coefficients.

Show worked answer →

(a) The geometric series with first term $1$ , ratio $z$ and $7$ terms sums to $\dfrac{z^7 - 1}{z - 1}$ , valid for $z \ne 1$ . (1 mark)

(b) The polynomial equals $\dfrac{z^7 - 1}{z - 1}$ , so setting it to zero requires $z^7 = 1$ with $z \ne 1$ . The solutions are the six non-trivial seventh roots of unity: $z = e^{2\pi i k/7}$ for $k = 1, 2, 3, 4, 5, 6$ . (3 marks)

(c) These six roots form three conjugate pairs ( $k$ with $7 - k$ ). Each pair gives a real quadratic $z^2 - 2\cos\dfrac{2\pi k}{7}\,z + 1$ . So the polynomial factors as $\left(z^2 - 2\cos\dfrac{2\pi}{7}z + 1\right)\left(z^2 - 2\cos\dfrac{4\pi}{7}z + 1\right)\left(z^2 - 2\cos\dfrac{6\pi}{7}z + 1\right)$ . (2 marks)

TCE 20214 marksSolve

z^5 + 1 = 0

. Give your answers in polar form

r e^{i\theta}

with

-\pi < \theta \le \pi

Show worked answer →

Write $-1$ in polar form: $-1 = e^{i\pi}$ , and more generally $-1 = e^{i(\pi + 2\pi k)}$ for integer $k$ . Then $z^5 = e^{i(\pi + 2\pi k)}$ .

Taking fifth roots, $z = e^{i(\pi + 2\pi k)/5}$ for $k = 0, 1, 2, 3, 4$ , all with modulus $r = 1$ .

The arguments are $\dfrac{\pi}{5}$ , $\dfrac{3\pi}{5}$ , $\pi$ (from $\dfrac{5\pi}{5}$ ), and for the remaining two, $\dfrac{7\pi}{5}$ and $\dfrac{9\pi}{5}$ , which reduce into $(-\pi, \pi]$ as $-\dfrac{3\pi}{5}$ and $-\dfrac{\pi}{5}$ .

So the five solutions are $z = e^{i\pi/5}, e^{3i\pi/5}, e^{i\pi} = -1, e^{-3i\pi/5}$ and $e^{-i\pi/5}$ . They are equally spaced on the unit circle. Markers reward writing $-1$ with the general argument and listing all five roots with arguments in the required range.