Modulus and argument | Year 12

01 / LEARN

The idea, step by step

Modulus and principal argument

The modulus $∣ z ∣$ is a non-negative real distance. For $z \neq 0$ , an argument is any angle that represents its direction. This chapter uses the principal argument $Arg z \in (- π, π]$ . State a different convention if a task uses one.

All arguments of a nonzero number are $Arg z + 2 k π$ , where $k \in ℤ$ . The angles $π$ and $- π$ indicate the same direction, but our principal interval includes only $π$ .

∣ x + i y ∣ = \sqrt{x^{2} + y^{2}}

Conjugation and distance

The conjugate $\bar{z} = x - i y$ reflects a point in the real axis, so $∣ \bar{z} ∣ = ∣ z ∣$ . Also $z \bar{z} = x^{2} + y^{2} = ∣ z ∣^{2}$ . These facts support division and proofs.

The modulus $∣ z - w ∣$ measures the distance between the points $z$ and $w$ . Do not confuse it with $∣ z ∣ - ∣ w ∣$ , which compares their distances from the origin and can be negative.

A common trap

$Arg (- 4) = - π$ under the interval $(- π, π]$ .

Instead: Under this convention the negative real axis has principal argument $π$ ; $- π$ is still an argument.

02 / WORKED EXAMPLES

See the thinking

Example 1

Reduce an angle to the principal interval

Find $Arg (3 cis (7 π / 4))$ and all its arguments.

Subtract $2 π$ from $7 π / 4$ to obtain $- π / 4$ .
Adding any whole turn gives the same point.

Principal argument

- π / 4

; all arguments

- π / 4 + 2 k π

k \in ℤ

Example 2

Distance between complex points

Find the distance between $2 + 3 i$ and $- 1 - i$ .

Subtract: $(2 + 3 i) - (- 1 - i) = 3 + 4 i$ .
The distance is $∣ 3 + 4 i ∣ = \sqrt{9 + 16} = 5$ .

5

Check your understanding

Find $∣ z ∣^{2}$ for $z = - 2 + 5 i$ .

Show answer

$29$ .

$z \bar{z} = {(- 2)}^{2} + 5^{2} = 29$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the modulus and principal argument of $- \sqrt{3} - i$ .
2 marks
Give all arguments of $- 2$ .
2 marks
If $z = 3 - 4 i$ , calculate $z \bar{z}$ and $∣ z ∣$ .
2 marks
Find $∣ (1 + 2 i) - (4 - 2 i) ∣$ .
2 marks
Is $∣ z + w ∣ = ∣ z ∣ + ∣ w ∣$ always true? Use $z = 1$ , $w = - 1$ to justify.
3 marks
Find the principal argument of $2 cis (- 9 π / 4)$ .
2 marks

Read the worked solutions

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

Find the modulus and principal argument of $- \sqrt{3} - i$ .
2 marks
The modulus is $\sqrt{3 + 1} = 2$ ; quadrant III and reference angle $π / 6$ give $- 5 π / 6$ .
$2$ and $- 5 π / 6$ .
Give all arguments of $- 2$ .
2 marks
The negative real direction is a half-turn plus any integer number of full turns.
$π + 2 k π$ , $k \in ℤ$ .
If $z = 3 - 4 i$ , calculate $z \bar{z}$ and $∣ z ∣$ .
2 marks
$(3 - 4 i) (3 + 4 i) = 9 + 16 = 25$ and $∣ z ∣ = \sqrt{25} = 5$ .
$25$ and $5$ .
Find $∣ (1 + 2 i) - (4 - 2 i) ∣$ .
2 marks
The difference is $- 3 + 4 i$ , whose modulus is $\sqrt{9 + 16} = 5$ .
$5$ .
Is $∣ z + w ∣ = ∣ z ∣ + ∣ w ∣$ always true? Use $z = 1$ , $w = - 1$ to justify.
3 marks
$z + w = 0$ , so $∣ z + w ∣ = 0$ ; $∣ 1 ∣ + ∣ - 1 ∣ = 2$ . Equality holds only in suitable aligned directions.
No: the left side is 0 but the right side is 2.
Find the principal argument of $2 cis (- 9 π / 4)$ .
2 marks
Add $2 π$ to $- 9 π / 4$ , obtaining $- π / 4$ in $(- π, π]$ .
$- π / 4$ .

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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 ↗

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