Multiplication as rotation/dilation | Year 12

01 / LEARN

Understand the idea.

The geometry of a fixed multiplier

Multiplying by a fixed nonzero $a = ρ cis ϕ$ transforms every point about the origin. The modulus $ρ = ∣ a ∣$ is the dilation factor; the argument $ϕ$ is the signed rotation angle. Negative angles describe clockwise rotation.

The origin stays fixed. Multiplication by $i$ gives a quarter-turn anticlockwise, while $- 1$ gives a half-turn. Multiplication by zero collapses every point to the origin, so it has no inverse and no meaningful rotation angle.

Argand diagram with equal real and imaginary unit scales. Segments join the origin to two plus i and negative one plus two i. The labelled ninety-degree arc shows anticlockwise rotation without changing distance from the origin.

Translation and inverse transformations

For $w = a z + b$ , perform the multiplication first and then translate by $b$ . To undo this order, subtract $b$ first and then divide by $a$ : $z = (w - b) / a$ , provided $a \neq 0$ .

For a pure multiplication taking a known nonzero $z$ to $w$ , solve $a z = w$ to obtain $a = w / z$ . Describe the transformation using this multiplier's modulus and argument, rather than the coordinates of the image alone.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A quarter-turn

Under $w = i z$ , find the image of $z = 2 + i$ and describe the transformation.

Worked solution

1
Substitute the original point.
$w = i (2 + i) = 2 i + i^{2}$
The map multiplies the entire complex number by $i$ .
2
Simplify the image.
$w = - 1 + 2 i$
The square of $i$ is $- 1$ .
3
Describe the multiplier geometrically.
$i = 1 cis \frac{π}{2}$
Its modulus is one and its argument is $\frac{π}{2}$ , giving an anticlockwise quarter-turn about the origin.

Answer

$w = - 1 + 2 i$ ; anticlockwise rotation through $\frac{π}{2}$ about the origin.

EXAMPLE 02

Recover the multiplier

A pure multiplication maps $1 - i$ to $4$ . Find the multiplier and describe it.

Worked solution

1
Solve for the multiplier.
$a (1 - i) = 4 ⟹ a = \frac{4}{1 - i}$
The input is nonzero, so division determines one multiplier.
2
Rationalise the denominator.
$a = \frac{4 (1 + i)}{(1 - i) (1 + i)} = 2 + 2 i$
The conjugate product in the denominator is $2$ .
3
Find its scale and angle.
$∣ a ∣ = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}, Arg a = \frac{π}{4}$
Equal positive coordinates place the multiplier halfway through the first quadrant.
4
State the transformation.
$a = 2 \sqrt{2} cis \frac{π}{4}$
The map dilates by $2 \sqrt{2}$ and rotates anticlockwise by $\frac{π}{4}$ about the origin.

Answer

$a = 2 + 2 i = 2 \sqrt{2} cis (\frac{π}{4})$ ; scale by $2 \sqrt{2}$ , rotate anticlockwise by $\frac{π}{4}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Describe multiplication by $2 cis (- \frac{π}{3})$ .

Show answer & reasoning

Scale by 2 and rotate clockwise by $\frac{π}{3}$ about the origin.

The modulus $2$ doubles distances from the origin. The argument $- \frac{π}{3}$ decreases each direction angle by $\frac{π}{3}$ , so the rotation is clockwise about the origin.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the image of $3 - 2 i$ under multiplication by $i$ .
2 marks
Describe the map $w = - 3 z$ .
2 marks
A robot step is represented by $z = 2 + i$ . A turn-and-scale instruction multiplies it by $1 + i$ . Find the new step.
3 marks
Find the multiplier taking $2$ to $- 2 + 2 \sqrt{3} i$ .
2 marks
Find the inverse of $w = 2 i z + 3$ .
3 marks
Two instructions multiply successively by $2 i$ and $- \frac{i}{2}$ . What is the combined transformation?
2 marks

Read the worked solutions

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

Find the image of $3 - 2 i$ under multiplication by $i$ .
2 marks
$i (3 - 2 i) = 3 i - 2 i^{2} = 2 + 3 i$ .
$2 + 3 i$ .
Describe the map $w = - 3 z$ .
2 marks
$- 3 = 3 cis π$ .
Dilation of scale factor 3 and rotation through $π$ about the origin.
A robot step is represented by $z = 2 + i$ . A turn-and-scale instruction multiplies it by $1 + i$ . Find the new step.
3 marks
$(1 + i) (2 + i) = 2 + 3 i + i^{2} = 1 + 3 i$ . The instruction scales by $\sqrt{2}$ and rotates by $\frac{π}{4}$ .
$1 + 3 i$ .
Find the multiplier taking $2$ to $- 2 + 2 \sqrt{3} i$ .
2 marks
Divide image by original: $a = \frac{- 2 + 2 \sqrt{3} i}{2}$ . Its modulus is 2 and quadrant-II angle is $\frac{2 π}{3}$ .
$- 1 + \sqrt{3} i = 2 cis (\frac{2 π}{3})$ .
Find the inverse of $w = 2 i z + 3$ .
3 marks
Subtract 3 first, then divide by $2 i$ .
$z = \frac{w - 3}{2 i} = - \frac{i}{2} (w - 3)$ .
Two instructions multiply successively by $2 i$ and $- \frac{i}{2}$ . What is the combined transformation?
2 marks
The combined multiplier is $(2 i) (- \frac{i}{2}) = - i^{2} = 1$ .
The identity transformation.

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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.

The geometry of a fixed multiplier

Translation and inverse transformations

See the method in action.

A quarter-turn

Substitute the original point.

Simplify the image.

Describe the multiplier geometrically.

Recover the multiplier

Solve for the multiplier.

Rationalise the denominator.

Find its scale and angle.

State the transformation.

Try it yourself.

Check your understanding

Keep practising