What is step 1 - polar form of the right side?

8i has modulus r = 8 and lies on the positive imaginary axis, so θ = π2: $8i = 8(cos(π)/(2) + isin(π)/(2)).$

What are not reducing arguments?

Final arguments should usually be given as principal values in (-π, π]; an argument like (3π)/(2) may be expected as -(π)/(2).

SASpecialist MathematicsSyllabus dot point

Why does a complex number have exactly n distinct nth roots, and how are they arranged?

Find the nth roots of a complex number using polar form and de Moivre's theorem, and represent them geometrically on the Argand plane.

Finding the n distinct nth roots of a complex number with de Moivre's theorem, why they are equally spaced on a circle, and solving polynomial equations such as z^n = w.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

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

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What this dot point is asking
The formula for nth roots
Geometry on the Argand plane
Roots of unity

What this dot point is asking

You need to find the $n$ th roots of a complex number using de Moivre's theorem, solve equations of the form $z^{n} = w$ , and represent the roots geometrically on the Argand plane.

The formula for nth roots

To solve $z^{n} = w$ , write $w$ in polar form $w = r(\cos\theta + i\sin\theta)$ . Because adding any whole number of full turns leaves $w$ unchanged, write the argument as $\theta + 2\pi k$ . Taking the $n$ th root with de Moivre's theorem gives

Only $k = 0$ to $n-1$ give distinct roots; $k = n$ repeats $k = 0$ (the angle increases by a full $2\pi$ ).

Geometry on the Argand plane

Because all $n$ roots share the modulus $r^{1/n}$ , they lie on a circle of radius $r^{1/n}$ centred at the origin. The equal angular spacing of $\dfrac{2\pi}{n}$ means they form the vertices of a regular $n$ -sided polygon. Sketching this circle and marking one root lets you read off the rest by rotation.

Cube roots of a complex number

Solve $z^{3} = 8i$ , giving roots in polar form

Step 1 - polar form of the right side. $8i$ has modulus $r = 8$ and lies on the positive imaginary axis, so $\theta = \dfrac{\pi}{2}$ :

8i = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right).

Step 2 - apply the roots formula with $n = 3$ . The modulus of each root is $8^{1/3} = 2$ , and the arguments are

\frac{\frac{\pi}{2} + 2\pi k}{3}, \quad k = 0, 1, 2.

$k = 0$ : argument $\dfrac{\pi}{6}$ , so $z_0 = 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right)$ .
$k = 1$ : argument $\dfrac{\pi/2 + 2\pi}{3} = \dfrac{5\pi}{6}$ , so $z_1 = 2\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right)$ .
$k = 2$ : argument $\dfrac{\pi/2 + 4\pi}{3} = \dfrac{3\pi}{2}$ , equivalently $-\dfrac{\pi}{2}$ , so $z_2 = 2\left(\cos\dfrac{3\pi}{2} + i\sin\dfrac{3\pi}{2}\right) = -2i$ .

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

Roots of unity

The special case $z^{n} = 1$ gives the $n$ th roots of unity:

z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \quad k = 0, 1, \ldots, n-1.

They lie on the unit circle, one of them is always

z = 1

, and their sum is

0

for

n \ge 2

(the vertices of a regular polygon centred at the origin balance out). If

\omega

denotes the root with smallest positive argument, the full set is

1, \omega, \omega^2, \ldots, \omega^{n-1}

Exam-style practice questions

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

2017 SACE Stage 23 marksSolve z^5 = 1. Write your solutions in polar form.

Show worked answer →

Write the right-hand side in polar form: 1 = cis(0), but more generally 1 = cis(2 k pi) for any integer k, since adding full turns does not change the number. [1 mark]

Take fifth roots using de Moivre's theorem: z = cis((0 + 2 k pi)/5) = cis(2 k pi / 5), for k = 0, 1, 2, 3, 4 (five distinct roots before they repeat). [1 mark]

The five solutions are:
z = cis(0) = 1, cis(2 pi / 5), cis(4 pi / 5), cis(6 pi / 5), cis(8 pi / 5).
Equivalently the arguments may be written as 0, 2 pi/5, 4 pi/5, -4 pi/5, -2 pi/5. They are equally spaced by 2 pi/5 around the unit circle. [1 mark]

2025 SACE Stage 22 marksUse De Moivre's theorem to show that all the solutions to z^5 = 32 are 2 cis(2 pi/5), 2 cis(4 pi/5), 2 cis(-4 pi/5), 2 cis(-2 pi/5) and 2 cis(0).

Show worked answer →

Write 32 in polar form: 32 = 32 cis(2 k pi) for any integer k, since 32 is a positive real with argument 0 (plus full turns). [1 mark]

Take fifth roots. The modulus of each root is 32^(1/5) = 2. By de Moivre's theorem the arguments are (2 k pi)/5 for k = 0, 1, 2, 3, 4:
arguments 0, 2 pi/5, 4 pi/5, 6 pi/5, 8 pi/5.
Writing the last two in the equivalent range (-pi, pi] gives 6 pi/5 = -4 pi/5 and 8 pi/5 = -2 pi/5.

Hence the solutions are 2 cis(0), 2 cis(2 pi/5), 2 cis(4 pi/5), 2 cis(-4 pi/5), 2 cis(-2 pi/5), the five fifth roots equally spaced on the circle of radius 2. [1 mark]