What is the roots as powers of one root?

Writing ω = e^i\,2π/n (the first primitive root), the full set is 1, ω, ω^2, , ω^n-1. Each root is a power of ω, which makes algebra with them compact.

What is the sum of the roots of unity?

The n roots of unity sum to zero for n ≥ 2:

What is the nth roots of a general complex number?

To solve z^n = w where w = r\,e^i, write w = r\,e^i( + 2kπ) and take nth roots:

What is uneven spacing?

The arguments must step by exactly 2πn; an arithmetic slip breaks the regular polygon.

What modulus does each fourth root of 81e^iθ have? [1 mark]

SingaporeFurther MathsSyllabus dot point

What are the nth roots of unity and of a general complex number, and how are they arranged geometrically?

Find the nth roots of unity and the nth roots of a general complex number, and describe their geometric arrangement on the Argand diagram

A focused answer to the H2 Further Mathematics outcome on roots of unity. The n nth roots of unity, their arrangement as a regular polygon, finding the nth roots of a general complex number, and the sum of the roots.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to solve $z^n = 1$ to find the $n$ roots of unity, to solve $z^n = w$ for the $n$ th roots of a general complex number $w$ , and to describe how these roots are arranged: equally spaced on a circle, at the vertices of a regular polygon. You should also know that the roots of unity sum to zero.

The answer

The nth roots of unity

The equation $z^n = 1$ has exactly $n$ solutions. Writing $1 = \mathrm{e}^{i(0 + 2k\pi)}$ and taking $n$ th roots,

z_k = \mathrm{e}^{i\,2k\pi/n}, \qquad k = 0, 1, 2, \dots, n-1.

All have modulus $1$ , so they lie on the unit circle, equally spaced $\dfrac{2\pi}{n}$ apart, forming the vertices of a regular $n$ -gon with one vertex at $z = 1$ .

The roots as powers of one root

Writing $\omega = \mathrm{e}^{i\,2\pi/n}$ (the first primitive root), the full set is $1, \omega, \omega^2, \dots, \omega^{n-1}$ . Each root is a power of $\omega$ , which makes algebra with them compact.

The sum of the roots of unity

The $n$ roots of unity sum to zero for $n \geq 2$ :

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

This follows from the geometric series $\dfrac{\omega^n - 1}{\omega - 1} = 0$ (since $\omega^n = 1$ but $\omega \neq 1$ ), or geometrically because the equally spaced vectors cancel by symmetry.

The nth roots of a general complex number

To solve $z^n = w$ where $w = r\,\mathrm{e}^{i\phi}$ , write $w = r\,\mathrm{e}^{i(\phi + 2k\pi)}$ and take $n$ th roots:

z_k = r^{1/n}\,\mathrm{e}^{i(\phi + 2k\pi)/n}, \qquad k = 0, 1, \dots, n-1.

Every root has modulus $r^{1/n}$ , and the arguments step by $\dfrac{2\pi}{n}$ , so the roots again sit at the vertices of a regular $n$ -gon, now on a circle of radius $r^{1/n}$ .

Worked example

Solve $z^3 = 8i$ , giving the roots in Cartesian form.

Step 1: Write the right side in exponential form

$8i$ has modulus $8$ and argument $\dfrac{\pi}{2}$ , so $8i = 8\,\mathrm{e}^{i(\pi/2 + 2k\pi)}$ .

Step 2: Take cube roots

Modulus $8^{1/3} = 2$ ; arguments $\dfrac{1}{3}\left(\dfrac{\pi}{2} + 2k\pi\right) = \dfrac{\pi}{6} + \dfrac{2k\pi}{3}$ for $k = 0, 1, 2$ :

z_0 = 2\,\mathrm{e}^{i\pi/6}, \quad z_1 = 2\,\mathrm{e}^{i\,5\pi/6}, \quad z_2 = 2\,\mathrm{e}^{i\,3\pi/2}.

Step 3: Convert each to Cartesian form

$z_0 = 2\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right) = 2\left(\tfrac{\sqrt{3}}{2} + \tfrac{1}{2}i\right) = \sqrt{3} + i$ .

$z_1 = 2\left(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\right) = 2\left(-\tfrac{\sqrt{3}}{2} + \tfrac{1}{2}i\right) = -\sqrt{3} + i$ .

$z_2 = 2\left(\cos\tfrac{3\pi}{2} + i\sin\tfrac{3\pi}{2}\right) = 2(0 - i) = -2i$ .

Step 4: Check the arrangement

The three roots have modulus $2$ and are spaced $\dfrac{2\pi}{3}$ apart, the vertices of an equilateral triangle on the circle of radius $2$ .

Examples in context

Example 1. Factorising $z^n - 1$ . Because the roots of unity are exactly the solutions of $z^n - 1 = 0$ , the polynomial factors as $z^n - 1 = (z - 1)(z - \omega)\cdots(z - \omega^{n-1})$ , linking roots of unity to the factor theorem for polynomials.

Example 2. Sampling and the discrete Fourier transform. The roots of unity are the sample points used in the discrete Fourier transform; their symmetry and the sum-to-zero property are what make the transform decompose a signal into frequencies, a cornerstone of digital signal processing.

Try this

Q1. Write the general form of the $n$ th roots of unity. [1 mark]

Cue. $z_k = \mathrm{e}^{i\,2k\pi/n}$ for $k = 0, 1, \dots, n-1$ .

Q2. State the sum of the five fifth roots of unity. [1 mark]

Cue. $0$ (the roots of unity sum to zero for $n \geq 2$ ).

Q3. What modulus does each fourth root of $81\mathrm{e}^{i\theta}$ have? [1 mark]

Cue. $81^{1/4} = 3$ .

Exam-style practice questions

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

Original6 marksFind the three cube roots of unity in exponential form and show that their sum is zero.

Show worked answer →

Solve $z^3 = 1 = \mathrm{e}^{i(0 + 2k\pi)}$ for $k = 0, 1, 2$ . Taking cube roots, $z = \mathrm{e}^{i\,2k\pi/3}$ :

z_0 = \mathrm{e}^{i0} = 1, \quad z_1 = \mathrm{e}^{i\,2\pi/3}, \quad z_2 = \mathrm{e}^{i\,4\pi/3}.

In Cartesian form

z_1 = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i

and

z_2 = -\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i

Sum: $1 + \left(-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) + \left(-\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i\right) = 1 - 1 + 0i = 0$ .

Markers reward the general form $\mathrm{e}^{i\,2k\pi/3}$ , listing the three roots, and showing the imaginary parts cancel and the real parts sum to zero.

Original7 marksFind the fourth roots of

z = 16\,\mathrm{e}^{i\pi/3}

, giving them in exponential form, and describe their arrangement on the Argand diagram.

Show worked answer →

Write $z = 16\,\mathrm{e}^{i(\pi/3 + 2k\pi)}$ for integer $k$ . The fourth roots have modulus $16^{1/4} = 2$ and arguments $\dfrac{1}{4}\left(\dfrac{\pi}{3} + 2k\pi\right) = \dfrac{\pi}{12} + \dfrac{k\pi}{2}$ for $k = 0, 1, 2, 3$ :

2\,\mathrm{e}^{i\pi/12}, \quad 2\,\mathrm{e}^{i\,7\pi/12}, \quad 2\,\mathrm{e}^{i\,13\pi/12}, \quad 2\,\mathrm{e}^{i\,19\pi/12}.

(The last two may be written with arguments in

(-\pi, \pi]

2\mathrm{e}^{-i\,11\pi/12}

and

2\mathrm{e}^{-i\,5\pi/12}

Geometrically the four roots lie on a circle of radius $2$ centred at the origin, equally spaced $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$ apart, forming the vertices of a square.

Markers reward the modulus $2$ , the arguments stepping by $\tfrac{\pi}{2}$ , listing all four roots, and describing the square on a circle of radius $2$ .