Powers and roots | Year 12

01 / LEARN

Understand the idea.

One equation, several roots

To solve $z^{n} = R cis θ$ with $R > 0$ , write $z = r cis ϕ$ . Comparing moduli gives $r = R^{1 / n}$ . Comparing directions requires $n ϕ = θ + 2 k π$ , because directions can differ by complete turns.

Divide the entire angle $θ + 2 k π$ by $n$ . Choosing $k = 0, \dots, n - 1$ gives exactly $n$ distinct roots, evenly spaced by $2 π / n$ . If the right side is zero, zero is the only distinct root.

\begin{array}{c} z_{k} = R^{1 / n} cis (\frac{θ + 2 k π}{n}) \\ k = 0, \dots, n - 1 \end{array}

Do not divide only the principal angle

A power has one result for an integer exponent, but finding roots reverses a many-to-one operation. Dividing just one chosen argument finds only one root; the other whole-turn representations must be included before division.

Check that the list has the required number of distinct angles modulo $2 π$ . Raising any listed root to the original power must reproduce both the modulus and the direction of the right side. Reduce angles afterwards if a particular interval is requested.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Cube roots of a negative real number

Solve $z^{3} = - 8$ .

Worked solution

1
Include every argument of the right side.
$- 8 = 8 cis (π + 2 k π), k \in ℤ$
A negative real number points along the negative real axis.
2
Take cube-root moduli and divide the angles.
$z_{k} = 2 cis (\frac{π + 2 k π}{3}), k = 0, 1, 2$
The modulus satisfies $r^{3} = 8$ , and three distinct argument classes are required.
3
List the three distinct directions.
$ϕ = \frac{π}{3}, π, \frac{5 π}{3}$
The spacing is $\frac{2 π}{3}$ ; another value of $k$ would repeat a direction.
4
Convert to Cartesian form if desired.
$z = 1 + \sqrt{3} i, - 2, 1 - \sqrt{3} i$
Exact sine and cosine values give the real and imaginary coordinates.

Answer

$z = 2 cis (\frac{π}{3}), 2 cis π, 2 cis (\frac{5 π}{3})$ , or $1 + \sqrt{3} i, - 2, 1 - \sqrt{3} i$ .

EXAMPLE 02

Four roots of a non-real number

Solve $z^{4} = 16 i$ in polar form.

Worked solution

1
Write the non-real target in polar form.
$16 i = 16 cis (\frac{π}{2} + 2 k π)$
The target lies on the positive imaginary axis.
2
Solve the modulus and angle equations.
$r^{4} = 16, 4 ϕ = \frac{π}{2} + 2 k π$
De Moivre relates each root to its fourth power.
3
State four distinct roots.
$z_{k} = 2 cis (\frac{π}{8} + \frac{k π}{2}), k = 0, 1, 2, 3$
The angles differ by $\frac{π}{2}$ , giving four different points before a full turn repeats.
4
Check the fourth powers.
$z_{k}^{4} = 16 cis (\frac{π}{2} + 2 k π) = 16 i$
Each added whole turn disappears when the direction is interpreted geometrically.

Answer

$z = 2 cis (\frac{π}{8} + k π / 2)$ , $k = 0, 1, 2, 3$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

How many distinct roots does $z^{5} = 32$ have, and what is their common modulus?

Show answer & reasoning

Five roots, each of modulus 2.

A nonzero number has five distinct fifth roots with arguments spaced by $\frac{2 π}{5}$ . Each root has modulus $32^{1 / 5} = 2$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Solve $z^{2} = - 9$ .
2 marks
List the cube roots of 8 in polar form.
2 marks
Find all solutions of $z^{4} = - 16$ .
3 marks
Write the roots of $z^{3} = 8 i$ in principal-argument polar form.
3 marks
Explain why $k = 4$ gives no new root in $z_{k} = 2 cis (\frac{π}{8} + \frac{k π}{2})$ .
2 marks
The three cube roots of a nonzero number form an equilateral triangle. Explain using arguments.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $z^{2} = - 9$ .
2 marks
Use arguments $\frac{π + 2 k π}{2}$ for $k = 0, 1$ and modulus 3.
$z = 3 i$ or $z = - 3 i$ .
List the cube roots of 8 in polar form.
2 marks
The angles are $\frac{2 k π}{3}$ for $k = 0, 1, 2$ .
$2, 2 cis (\frac{2 π}{3}), 2 cis (\frac{4 π}{3})$ .
Find all solutions of $z^{4} = - 16$ .
3 marks
The modulus is 2. Arguments come from $\frac{π + 2 k π}{4}$ .
$2 cis (\frac{π}{4} + \frac{k π}{2})$ , $k = 0, 1, 2, 3$ .
Write the roots of $z^{3} = 8 i$ in principal-argument polar form.
3 marks
Angles from $\frac{π}{6} + \frac{2 k π}{3}$ are $\frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$ ; reduce the last by $2 π$ .
$2 cis (\frac{π}{6}), 2 cis (\frac{5 π}{6}), 2 cis (- \frac{π}{2})$ .
Explain why $k = 4$ gives no new root in $z_{k} = 2 cis (\frac{π}{8} + \frac{k π}{2})$ .
2 marks
The angle for $k = 4$ is $\frac{π}{8} + 2 π$ , the same direction as $\frac{π}{8}$ .
It repeats the $k = 0$ root.
The three cube roots of a nonzero number form an equilateral triangle. Explain using arguments.
3 marks
Equal central angles on one circle give equal chord lengths, so all three sides are equal.
Adjacent roots differ in angle by $\frac{2 π}{3}$ and have equal modulus.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

One equation, several roots

Do not divide only the principal angle

See the method in action.

Cube roots of a negative real number

Include every argument of the right side.

Take cube-root moduli and divide the angles.

List the three distinct directions.

Convert to Cartesian form if desired.

Four roots of a non-real number

Write the non-real target in polar form.

Solve the modulus and angle equations.

State four distinct roots.

Check the fourth powers.

Try it yourself.

Check your understanding

Keep practising