What is the nth roots formula?

The n distinct nth roots of w = r e^iθ are

What is uneven spacing?

The roots are spaced exactly (2π)/(n) apart; an uneven set signals an arithmetic slip.

SEAB Original-style practice: Solve z^4 = -16, giving the roots in exponential form.

-16 = 16 e^i(π + 2kπ). The fourth roots are 16^1/4 e^i(π + 2kπ)/4 = 2 e^i(π/4 + kπ/2) for k = 0, 1, 2, 3. k = 0: 2e^iπ/4. k = 1: 2e^i 3π/4. k = 2: 2e^i 5π/4 (or 2e^-i 3π/4). k = 3: 2e^i 7π/4 (or 2e^-iπ/4). The four roots have modulus 2 and arguments spaced π2 apart. Markers reward writing -16 with the general argument, taking the fourth root of the modulus, dividing the argument, and listing all four roots.

SingaporeMathsSyllabus dot point

How do we find all the nth roots of a complex number and the roots of higher polynomial equations?

Find the nth roots of a complex number using de Moivre's theorem, and solve polynomial equations with complex roots, interpreting the roots geometrically

A focused answer to the H2 Mathematics outcome on roots of complex numbers. Finding the nth roots via de Moivre's theorem, their symmetric arrangement on a circle, the roots of unity, and solving polynomial equations.

Generated by Claude Opus 4.810 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
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Try this

What this dot point is asking

SEAB wants you to find all the $n$ th roots of a complex number using de Moivre's theorem, recognise that they are equally spaced on a circle, work with the roots of unity, and solve polynomial equations whose roots are complex, interpreting the solution set geometrically.

The answer

The key idea: add multiples of 2 pi

A complex number $w = r e^{i\theta}$ has the same value if you add any multiple of $2\pi$ to its argument: $w = r e^{i(\theta + 2k\pi)}$ . This is what produces several distinct $n$ th roots from a single number.

The nth roots formula

The $n$ distinct $n$ th roots of $w = r e^{i\theta}$ are

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

All have the same modulus $r^{1/n}$ and arguments differing by $\dfrac{2\pi}{n}$ .

Geometric arrangement

The $n$ roots lie on a circle of radius $r^{1/n}$ centred at the origin, equally spaced $\dfrac{2\pi}{n}$ apart, forming the vertices of a regular $n$ -gon. The roots of unity (the $n$ th roots of $1$ ) are the special case with one root at $1$ and the rest spread evenly around the unit circle.

Solving polynomial equations

An equation such as $z^n = w$ is solved directly by the roots formula. More general polynomials are solved by factoring (using known or conjugate roots) into linear and quadratic factors, then solving each. The fundamental theorem guarantees a degree- $n$ polynomial has exactly $n$ roots counted with multiplicity.

Worked example

Solve $z^3 = 1 + i$ , giving the roots in exponential form.

Step 1: Write the right side in polar form with the general argument

$1 + i$ has modulus $\sqrt{2}$ and argument $\dfrac{\pi}{4}$ , so $1 + i = \sqrt{2}\, e^{i(\pi/4 + 2k\pi)}$ .

Step 2: Apply the roots formula with n = 3

z_k = (\sqrt{2})^{1/3} e^{i(\pi/4 + 2k\pi)/3} = 2^{1/6} e^{i(\pi/12 + 2k\pi/3)}, \quad k = 0, 1, 2

Step 3: List the three roots

$k = 0$ : $2^{1/6} e^{i\pi/12}$ . $k = 1$ : $2^{1/6} e^{i(\pi/12 + 2\pi/3)} = 2^{1/6} e^{i 3\pi/4}$ . $k = 2$ : $2^{1/6} e^{i(\pi/12 + 4\pi/3)} = 2^{1/6} e^{i 17\pi/12}$ (or $2^{1/6} e^{-i 7\pi/12}$ ).

Step 4: Describe the geometry

All three have modulus $2^{1/6} \approx 1.12$ and are spaced $\dfrac{2\pi}{3} = 120^\circ$ apart, forming an equilateral triangle on the circle of that radius.

Examples in context

Example 1. Roots of unity in signal processing. The $n$ th roots of unity $e^{2\pi i k/n}$ are the sampling points of the discrete Fourier transform, equally spaced around the unit circle, which is why complex roots underpin digital signal analysis.

Example 2. Designing a regular polygon. Because the $n$ th roots of any complex number form a regular $n$ -gon, you can generate the vertices of a regular hexagon by taking the sixth roots of a chosen number, scaling and rotating the standard pattern.

Try this

Q1. How many distinct fifth roots does a nonzero complex number have, and how are they arranged? [2 marks]

Cue. Five, equally spaced $72^\circ$ apart on a circle of radius $r^{1/5}$ .

Q2. Find the square roots of $i$ in exponential form. [3 marks]

Cue. $i = e^{i(\pi/2 + 2k\pi)}$ , roots $e^{i\pi/4}$ and $e^{i 5\pi/4}$ .

Q3. State the modulus of each cube root of $27$ . [1 mark]

Cue. $27^{1/3} = 3$ .

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

8

, giving them in the form

a + bi

where appropriate, and describe their arrangement on an Argand diagram.

Show worked answer →

Write $8 = 8e^{i(0 + 2k\pi)}$ for integer $k$ . The cube roots are $8^{1/3} e^{i(2k\pi/3)} = 2e^{i(2k\pi/3)}$ for $k = 0, 1, 2$ .

$k = 0$ : $2e^{0} = 2$ .
$k = 1$ : $2e^{i 2\pi/3} = 2(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}) = 2(-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i) = -1 + \sqrt{3}i$ .
$k = 2$ : $2e^{i 4\pi/3} = -1 - \sqrt{3}i$ .

The three roots lie on a circle of radius $2$ , equally spaced $120^\circ$ apart.

Markers reward adding $2k\pi$ to the argument, taking the cube root of the modulus, the three roots, and the symmetric arrangement.

Original5 marksSolve

z^4 = -16

, giving the roots in exponential form.

Show worked answer →

$-16 = 16 e^{i(\pi + 2k\pi)}$ . The fourth roots are $16^{1/4} e^{i(\pi + 2k\pi)/4} = 2 e^{i(\pi/4 + k\pi/2)}$ for $k = 0, 1, 2, 3$ .

$k = 0$ : $2e^{i\pi/4}$ . $k = 1$ : $2e^{i 3\pi/4}$ . $k = 2$ : $2e^{i 5\pi/4}$ (or $2e^{-i 3\pi/4}$ ). $k = 3$ : $2e^{i 7\pi/4}$ (or $2e^{-i\pi/4}$ ).

The four roots have modulus $2$ and arguments spaced $\dfrac{\pi}{2}$ apart.

Markers reward writing $-16$ with the general argument, taking the fourth root of the modulus, dividing the argument, and listing all four roots.