§-Syllabus dot point

WASpecialist MathematicsSyllabus dot point

Why does every nonzero complex number have exactly n distinct nth roots, evenly spaced on a circle?

Find the nth roots of a complex number and the nth roots of unity, and represent them on the Argand plane

WACE Specialist Unit 3 roots of complex numbers: solving z to the n equals w with the general root formula, the nth roots of unity, equal spacing on a circle of radius the nth root of the modulus, and their sum, with a worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The general root formula
Why the roots are equally spaced
The nth roots of unity
Roots of unity as a cyclic group
Roots of any number from the roots of unity
Plotting strategy

What this dot point is asking

SCSA wants you to solve $z^n = w$ for all roots, identify the roots of unity, and plot them as a regular polygon on the Argand plane.

The general root formula

To solve $z^n = w$ , write $w = R\,\text{cis}\,\phi$ and let $z = \rho\,\text{cis}\,\alpha$ . By de Moivre $z^n = \rho^n\,\text{cis}(n\alpha)$ , so matching moduli gives $\rho = R^{1/n}$ and matching arguments gives $n\alpha = \phi + 2\pi k$ for integer $k$ . Hence

Taking $k = 0$ to $n-1$ gives $n$ distinct roots; further values of $k$ repeat them. All roots share the modulus $R^{1/n}$ , so they lie on a circle of that radius, spaced $\tfrac{2\pi}{n}$ apart. This is why they form a regular $n$ -gon.

Why the roots are equally spaced

The equal spacing is a direct consequence of the formula. Consecutive roots differ only in the value of $k$ , and increasing $k$ by one adds $\dfrac{2\pi}{n}$ to the argument while leaving the modulus $R^{1/n}$ unchanged. Geometrically, each root is the previous one rotated by $\dfrac{2\pi}{n}$ about the origin. After $n$ such rotations the total turn is $2\pi$ , returning to the starting root, which is why exactly $n$ distinct roots arise before they repeat. This is the reason the roots always form a regular $n$ -gon inscribed in a circle of radius $R^{1/n}$ , a fact that lets you draw all the roots once you have located a single one.

The nth roots of unity

Setting $w = 1 = \text{cis}\,0$ gives the $n$ th roots of unity:

\omega_k = \text{cis}\!\left(\frac{2\pi k}{n}\right), \qquad k = 0, 1, \dots, n-1.

They all lie on the unit circle, include $z = 1$ , and are equally spaced. Writing $\omega = \text{cis}\tfrac{2\pi}{n}$ , every root is a power $\omega^k$ .

Roots of unity as a cyclic group

The $n$ th roots of unity have a tidy structure: every root is a power of the single primitive root $\omega = \text{cis}\dfrac{2\pi}{n}$ , so the full set is $\{1, \omega, \omega^2, \dots, \omega^{n-1}\}$ and multiplying any two of them gives another (with exponents added modulo $n$ ). This is why the sum of all $n$ roots is zero for $n \ge 2$ : they are the roots of $z^n - 1 = 0$ , and the coefficient of $z^{n-1}$ in $z^n - 1$ is zero, so by the sum-of-roots relation the total is zero. The factorisation $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \dots + 1)$ shows the same thing: the non-trivial roots satisfy $z^{n-1} + \dots + 1 = 0$ .

Roots of any number from the roots of unity

Once you know the $n$ th roots of unity, the $n$ th roots of any complex number follow by multiplication. If $z_0$ is any one $n$ th root of $w$ , then every $n$ th root of $w$ is $z_0 \omega^k$ for $k = 0, 1, \dots, n-1$ , because multiplying by a root of unity rotates without changing the modulus and preserves the property of being an $n$ th root. This gives a fast way to write down all the roots once a single one is found.

Plotting strategy

Find one root (usually $k = 0$ ), then rotate by $\tfrac{2\pi}{n}$ repeatedly to get the rest. You never recompute the modulus.

Solve

z^4 = -16

We apply the general root formula: write the right-hand side in polar form, read off the modulus and principal argument, then generate all four roots by adding multiples of $\tfrac{2\pi}{4}$ .

Step 1: Write $-16$ in polar form

The number $-16$ is real and negative, so its modulus is $16$ and its argument is $\pi$ . In polar form, $-16 = 16\,\text{cis}\,\pi$ .

Step 2: Find the common modulus of all roots

Every fourth root of $16\,\text{cis}\,\pi$ has the same modulus, $R^{1/4} = 16^{1/4} = 2$ . All four roots will lie on a circle of radius $2$ .

Step 3: Generate the four arguments using the general formula

The arguments of the roots are $\alpha_k = \dfrac{\pi + 2\pi k}{4}$ for $k = 0, 1, 2, 3$ , because each step adds $\tfrac{2\pi}{4} = \tfrac{\pi}{2}$ to rotate to the next root.

\alpha_k = \frac{\pi + 2\pi k}{4}, \qquad k = 0, 1, 2, 3.

Step 4: Compute each root

For $k = 0$ : $\alpha_0 = \tfrac{\pi}{4}$ , so $z_0 = 2\,\text{cis}\tfrac{\pi}{4} = \sqrt{2} + i\sqrt{2}$ .

For $k = 1$ : $\alpha_1 = \tfrac{3\pi}{4}$ , so $z_1 = -\sqrt{2} + i\sqrt{2}$ .

For $k = 2$ : $\alpha_2 = \tfrac{5\pi}{4}$ , equivalently $-\tfrac{3\pi}{4}$ , so $z_2 = -\sqrt{2} - i\sqrt{2}$ .

For $k = 3$ : $\alpha_3 = \tfrac{7\pi}{4}$ , equivalently $-\tfrac{\pi}{4}$ , so $z_3 = \sqrt{2} - i\sqrt{2}$ .

Final answer: $z = \sqrt{2} + i\sqrt{2},\ -\sqrt{2} + i\sqrt{2},\ -\sqrt{2} - i\sqrt{2},\ \sqrt{2} - i\sqrt{2}$ . The four roots lie on a circle of radius $2$ at the corners of a square.

Exam-style practice questions

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

WACE 20236 marksCalculator-free. (a) Find the three cube roots of unity in polar and Cartesian form. (b) Show that they sum to zero.

Show worked answer →

A roots-of-unity question.

(a) Solve $z^3 = 1 = \text{cis}\,0$ . The roots are $\omega_k = \text{cis}\!\left(\dfrac{2\pi k}{3}\right)$ for $k = 0, 1, 2$ : $\omega_0 = \text{cis}\,0 = 1$ ; $\omega_1 = \text{cis}\,\dfrac{2\pi}{3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ ; $\omega_2 = \text{cis}\,\dfrac{4\pi}{3} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ .

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

Markers reward the three roots in both forms and the cancellation to zero. An alternative for (b) uses the factorisation $z^3 - 1 = (z - 1)(z^2 + z + 1)$ , where the sum of all roots is the negative of the $z^2$ coefficient, namely $0$ .

WACE 20205 marksCalculator-assumed. Find all solutions of

z^3 = 8i

, giving each in the form

a + bi

where appropriate, and describe their arrangement.

Show worked answer →

Cube roots of a complex number.

Write $8i = 8\,\text{cis}\,\dfrac{\pi}{2}$ . The cube roots have modulus $8^{1/3} = 2$ and arguments $\dfrac{\frac{\pi}{2} + 2\pi k}{3}$ for $k = 0, 1, 2$ .

$k = 0$ : $\dfrac{\pi}{6}$ , so $z_0 = 2\,\text{cis}\,\dfrac{\pi}{6} = \sqrt{3} + i$ . $k = 1$ : $\dfrac{5\pi}{6}$ , so $z_1 = -\sqrt{3} + i$ . $k = 2$ : $\dfrac{3\pi}{2}$ (or $-\dfrac{\pi}{2}$ ), so $z_2 = -2i$ .

The three roots lie on a circle of radius $2$ , spaced $\dfrac{2\pi}{3}$ apart at the vertices of an equilateral triangle. Markers reward the modulus $2$ , the three arguments, the Cartesian forms, and the triangle description.