Roots of unity - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

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

TCESpecialist MathematicsUnit 38 min readUpdated 2 Jun 2026

Quick answer

The $n$ th roots of unity are the solutions of $z^n = 1$ . They are $z_k = \operatorname{cis}\tfrac{2\pi k}{n}$ for $k = 0, 1, \dots, n-1$ , and they sit equally spaced around the unit circle, forming the vertices of a regular $n$ -sided polygon with one vertex at $z = 1$ .

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.

Primitive root and the standard notation

Write $\omega = \operatorname{cis}\tfrac{2\pi}{n}$ . Then every $n$ th root of unity is a power of $\omega$ , namely $1, \omega, \omega^2, \dots, \omega^{n-1}$ . The number $\omega$ is called a primitive $n$ th root of unity because its powers generate them all.

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

Sum of the roots of unity

For $n \ge 2$ , the $n$ th roots of unity sum to zero:

1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0.

This follows from the geometric series formula, since the sum equals

\dfrac{\omega^n - 1}{\omega - 1} = \dfrac{1 - 1}{\omega - 1} = 0

. Geometrically the vectors to the vertices of a regular polygon cancel by symmetry.

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.

Stopping after finding only the real roots

Students often solve $z^n = 1$ by writing $z = 1$ (and $z = -1$ when $n$ is even) and stop, missing the complex roots. There are always exactly $n$ roots, counted with the full range $k = 0$ to $n-1$ . Use De Moivre to list every argument, and check you have $n$ distinct points equally spaced on the unit circle before finishing.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 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)

Practice6 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)

Practice4 marks

Solve

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.

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC

What this dot point is asking

Setting up the roots

The geometry

Worked example

The sum and product of the roots

Connection to factorisation

Properties of the cube roots: a worked staple

Step 1: Recall the two key identities

Step 2: Expand and simplify the product (1+ω)(1+ω2)(1 + \omega)(1 + \omega^2)(1+ω)(1+ω2)

Step 3: Simplify 1−ω+ω21 - \omega + \omega^21−ω+ω2

Why this matters

Exam-style questions

Practise this

Sources & how we know this

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

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