Modulus and argument | Year 12

01 / LEARN

Understand the idea.

Modulus and principal argument

The modulus measures distance from the origin; an argument measures direction. For a nonzero number, rotating through any whole number of turns leaves its direction unchanged. All arguments therefore differ by multiples of $2 π$ .

The principal argument selects one representative. Here $Arg z \in (- π, π]$ : add or subtract full turns until the angle lies in this interval. Its negative endpoint is excluded, so the negative real axis has principal argument $π$ .

∣ x + i y ∣ = \sqrt{x^{2} + y^{2}}

Conjugation and distance

Conjugating $z = x + i y$ changes the sign of its imaginary coordinate, reflecting it in the real axis. Reflection preserves distance, and $z \bar{z} = x^{2} + y^{2} = ∣ z ∣^{2}$ is real and non-negative.

To measure the distance between two complex points, first subtract them and then take the modulus: $∣ z - w ∣$ . Subtraction gives the horizontal and vertical displacement; Pythagoras combines those components into a length.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Reduce an angle to the principal interval

Find $Arg (3 cis (\frac{7 π}{4}))$ and all its arguments.

Worked solution

1
Remove one whole turn.
$\frac{7 π}{4} - 2 π = - \frac{π}{4}$
A full turn changes the angle representation without changing the point.
2
Check the principal interval.
$- π < - \frac{π}{4} \leq π$
The reduced angle belongs to the stated interval, so it is the principal argument.
3
State every equivalent direction.
$Arg z = - \frac{π}{4}, arg z = - \frac{π}{4} + 2 k π, k \in ℤ$
Each integer adds a whole number of complete turns.

Answer

Principal argument $- \frac{π}{4}$ ; all arguments $- \frac{π}{4} + 2 k π$ , $k \in ℤ$ .

EXAMPLE 02

Distance between complex points

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

Worked solution

1
Subtract the complex coordinates.
$\begin{aligned} (2 + 3 i) - (- 1 - i) & = (2 + 1) + (3 + 1) i \\ = 3 + 4 i \end{aligned}$
The difference represents the displacement between the points.
2
Take the magnitude of the displacement.
$d = ∣ 3 + 4 i ∣ = \sqrt{3^{2} + 4^{2}}$
The real and imaginary components form perpendicular sides.
3
Simplify the non-negative length.
$d = \sqrt{25} = 5$
Distance is the positive square root of the squared length.

Answer

$5$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$29$ .

The squared modulus is the sum of the squared coordinates: $∣ z ∣^{2} = {(- 2)}^{2} + 5^{2} = 4 + 25 = 29$ . No square root is needed because the question asks for the square.

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 $2 cis (- \frac{9 π}{4})$ .
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 $\frac{π}{6}$ give $- \frac{5 π}{6}$ .
$2$ and $- \frac{5 π}{6}$ .
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 $2 cis (- \frac{9 π}{4})$ .
2 marks
Add $2 π$ to $- \frac{9 π}{4}$ , obtaining $- \frac{π}{4}$ in $(- π, π]$ .
$- \frac{π}{4}$ .

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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 ↗

In this lesson

Understand the idea.

Modulus and principal argument

Conjugation and distance

See the method in action.

Reduce an angle to the principal interval

Remove one whole turn.

Check the principal interval.

State every equivalent direction.

Distance between complex points

Subtract the complex coordinates.

Take the magnitude of the displacement.

Simplify the non-negative length.

Try it yourself.

Check your understanding

Keep practising