Modulus | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

Distance, not a complex component

The modulus $∣ z ∣$ is the distance from the origin to the point $z = a + b i$ . Pythagoras gives $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ . It is a non-negative real number, with value zero only when $z = 0$ .

The squared modulus is $∣ z ∣^{2} = a^{2} + b^{2} = z \bar{z}$ . If a question asks for the squared modulus, no square root is needed. The two components are perpendicular, so adding their absolute values does not generally give the distance.

Distances and products

The distance between points $z$ and $w$ is $∣ z - w ∣$ . Find their displacement first, then its modulus. Reversing subtraction reverses the displacement direction but preserves its length, so $∣ z - w ∣ = ∣ w - z ∣$ .

Moduli multiply: $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , and $∣ z / w ∣ = ∣ z ∣ / ∣ w ∣$ for $w \neq 0$ . There is no corresponding equality for every sum. Points may face different directions, so $∣ z + w ∣$ need not equal $∣ z ∣ + ∣ w ∣$ .

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find an exact modulus

Find $∣ - 2 + 3 i ∣$ .

Worked solution

1
Identify the real components.
$a = - 2, b = 3$
The imaginary coefficient is three, not three i.
2
Calculate the squared distance.
$a^{2} + b^{2} = {(- 2)}^{2} + 3^{2} = 4 + 9 = 13$
Both squared components are non-negative even when a component is negative.
3
Take the non-negative square root.
$∣ - 2 + 3 i ∣ = \sqrt{13}$
Distance uses the non-negative root, so no plus-or-minus sign is attached.

Answer

$\sqrt{13}$ .

EXAMPLE 02

Use a geometric condition

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

Worked solution

1
Use the given real part.
$z = 3 + b i, b \in ℝ$
The imaginary coefficient is the only unknown component.
2
Square the modulus equation.
$\begin{aligned} ∣ z ∣^{2} & = 3^{2} + b^{2} = 25 \\ b^{2} & = 16 \end{aligned}$
The distance from the origin is five, so its square is twenty-five.
3
Retain both imaginary coordinates.
$b = 4 or b = - 4$
Both signs give the same squared distance.
4
State and verify the two numbers.
$z = 3 \pm 4 i, ∣ 3 \pm 4 i ∣ = \sqrt{9 + 16} = 5$
Both points have real coordinate three and the required modulus.

Answer

$z = 3 + 4 i$ or $z = 3 - 4 i$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find $∣ 1 + 2 i ∣^{2}$ .

Show answer & reasoning

$5$ .

The question asks for the square of the modulus: $∣ 1 + 2 i ∣^{2} = 1^{2} + 2^{2} = 5$ . Taking a square root would instead calculate $∣ 1 + 2 i ∣$ .

Keep practising

6 worksheet questions · 16 marks

Open worksheet 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 $| \frac{z}{w} |$ .
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 $| \frac{z}{w} |$ .
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 ↗

In this lesson

Understand the idea.

Distance, not a complex component

Distances and products

See the method in action.

Find an exact modulus

Identify the real components.

Calculate the squared distance.

Take the non-negative square root.

Use a geometric condition

Use the given real part.

Square the modulus equation.

Retain both imaginary coordinates.

State and verify the two numbers.

Try it yourself.

Check your understanding

Keep practising