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 4.78 min answerUpdated 2026-05-30

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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.

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.

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.

2024 TASC4 marks(a) Write down the three roots of z^3 = 1, giving the non-real roots as e^(i theta) with -pi < theta <= 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.

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(a) The cube roots of unity are equally spaced around the unit circle at angles 0, 2pi/3 and -2pi/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 2pi 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 are 1, omega, omega^2 and they 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 = (omega^3 - 1)/(omega - 1) = 0 because omega^3 = 1 and omega is not 1. (1 mark)

2022 TASC6 marks(a) Write down the sum of the geometric series sum from n = 0 to 6 of 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.

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(a) The geometric series with first term 1, ratio z and 7 terms sums to (z^7 - 1)/(z - 1), valid for z not equal to 1. (1 mark)

(b) The polynomial equals (z^7 - 1)/(z - 1), so setting it to zero requires z^7 = 1 with z not equal to 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(2 pi k / 7) z + 1. So the polynomial factors as [z^2 - 2 cos(2pi/7) z + 1][z^2 - 2 cos(4pi/7) z + 1][z^2 - 2 cos(6pi/7) z + 1]. (2 marks)

2021 TASC4 marksSolve z^5 + 1 = 0. Give your answers in polar form r e^(i theta) with -pi < theta <= 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 pi/5, 3pi/5, pi (= 5pi/5), and for the remaining two, 7pi/5 and 9pi/5, which reduce into (-pi, pi] as -3pi/5 and -pi/5.

So the five solutions are z = e^(i pi/5), e^(3 i pi/5), e^(i pi) = -1, e^(-3 i 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.