Modulus | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

Distance, not a complex component

The modulus of $z = a + b i$ is its distance from the origin: $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ . It is a non-negative real number. Only the zero complex number has modulus zero.

The identity $∣ z ∣^{2} = z \bar{z}$ is often quicker to use than a square root. Modulus is not found by simply adding the absolute values of the components.

Distances and products

The distance from $z$ to $w$ is $∣ z - w ∣$ . First find the displacement, then its modulus. Subtraction order changes the direction but not the distance.

Complex numbers satisfy $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and, for $w \neq 0$ , $∣ z / w ∣ = ∣ z ∣ / ∣ w ∣$ . In contrast, $∣ z + w ∣$ need not equal $∣ z ∣ + ∣ w ∣$ : vectors can point in different directions.

A common trap

Writing $∣ 3 - 4 i ∣ = 3 + 4 = 7$ .

Instead: The horizontal and vertical components are perpendicular, so use Pythagoras: $∣ 3 - 4 i ∣ = 5$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Find an exact modulus

Find $∣ - 2 + 3 i ∣$ .

Square the real and imaginary components: ${(- 2)}^{2} + 3^{2} = 13$ .
Take the non-negative square root, giving $\sqrt{13}$ .

\sqrt{13}

Example 2

Use a geometric condition

Find all complex numbers $z$ with $Re (z) = 3$ and $∣ z ∣ = 5$ .

Write $z = 3 + b i$ with real $b$ .
$∣ z ∣^{2} = 9 + b^{2} = 25$ , so $b^{2} = 16$ .
Both $b = 4$ and $b = - 4$ satisfy the conditions.

z = 3 + 4 i

z = 3 - 4 i

Check your understanding

Find $∣ 1 + 2 i ∣^{2}$ .

Show answer

$5$ .

Squared modulus is $1^{2} + 2^{2} = 5$ ; do not take a square root when the square is requested.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find $∣ 5 - 12 i ∣$ .
2 marks
Find $∣ - 3 + 2 i ∣^{2}$ .
2 marks
Find the distance between $2 - i$ and $- 1 + 3 i$ .
3 marks
Given $∣ z ∣ = 6$ and $∣ w ∣ = 2$ , find $∣ z w ∣$ and $∣ z / w ∣$ .
3 marks
Find all $z$ such that $Im (z) = 5$ and $∣ z ∣ = 13$ .
3 marks
Use $z = 1$ and $w = - 1$ to show that $∣ z + w ∣ = ∣ z ∣ + ∣ w ∣$ is not always true.
3 marks

Read the worked solutions

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

Find $∣ 5 - 12 i ∣$ .
2 marks
$∣ 5 - 12 i ∣ = \sqrt{25 + 144} = \sqrt{169}$ .
$13$ .
Find $∣ - 3 + 2 i ∣^{2}$ .
2 marks
Squared modulus is $9 + 4 = 13$ .
$13$ .
Find the distance between $2 - i$ and $- 1 + 3 i$ .
3 marks
The difference is $3 - 4 i$ .
Its modulus is $\sqrt{9 + 16} = 5$ .
$5$ .
Given $∣ z ∣ = 6$ and $∣ w ∣ = 2$ , find $∣ z w ∣$ and $∣ z / w ∣$ .
3 marks
Use the product and quotient modulus identities.
$12$ and $3$ .
Find all $z$ such that $Im (z) = 5$ and $∣ z ∣ = 13$ .
3 marks
Write $z = a + 5 i$ .
$a^{2} + 25 = 169$ , so $a^{2} = 144$ and $a = \pm 12$ .
$z = 12 + 5 i$ or $z = - 12 + 5 i$ .
Use $z = 1$ and $w = - 1$ to show that $∣ z + w ∣ = ∣ z ∣ + ∣ w ∣$ is not always true.
3 marks
$z + w = 0$ , whereas each of $1$ and $- 1$ is one unit from the origin.
Oppositely directed vectors cancel.
Left side $0$ ; right side $2$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, real and complex numbers: Cartesian arithmetic, conjugates, quadratic roots and Argand geometry. Polar form and roots of unity are developed in Year 12.

Official curriculum source ↗

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