Lesson 8 · Rotation, dilation and geometry

Learning intentionTo interpret multiplication as a rotation and dilation, and use it to solve geometry problems.

Success criteria · I can…

explain why multiplying by $i$ rotates $90^\circ$ anticlockwise about $O$
multiply by $r\cis\theta$ to dilate by $r$ and rotate by $\theta$
rotate about a point other than $O$ using $c-a=(b-a)\cis\theta$
find unknown vertices of squares and equilateral triangles
prove geometric results using complex numbers

ResourcesLesson slides
Practice worksheet
Worked solutions

Lesson flow

ExamplesExample 1 · Multiplying by iSlide 3 LaunchWarm-up · Retrieval from Lessons 5 to 7Slide 2 Key ideasRotation and dilationSlide 4 ExamplesExample 2 · Rotate and enlargeSlide 5 ExamplesExample 3 · A squareSlide 6 ExamplesExample 4 · Rotating about another pointSlide 7 HSC-styleExample 5 · Equilateral triangleSlide 8 HSC-styleExample 6 · Perpendicular vectorsSlide 9 ExamplesExample 7 · Proving a quadrilateral is a squareSlide 10 Exit ticketExit ticketSlide 11

Example 1 · Multiplying by i

Slide 3

Example 1

Let

z=2+i

. Plot

z

iz

i^2z=-z

and

i^3z=-iz

. Describe what multiplying by

i

does.

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Answer

iz=-1+2i

-z=-2-i

-iz=1-2i

. Multiplying by

i

rotates a point

90^\circ

anticlockwise about

O

Why

Each multiplication by

i=\cis\frac{\pi}{2}

keeps the modulus and adds

\frac{\pi}{2}

to the argument. The four points form a square centred at

O

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Each multiplication by i is a quarter turn anticlockwise.

Warm-up · Retrieval from Lessons 5 to 7

Slide 2

(a) Find

iz

for

z=3-2i

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Answer

iz=2+3i

Why

i(3-2i)=3i-2i^2=2+3i

, using

i^2=-1

(b) Write

\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}

x+iy

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Answer

\dfrac12+\dfrac{\sqrt3}{2}i

Why

Use the exact values

\cos\frac{\pi}{3}=\frac12

and

\sin\frac{\pi}{3}=\frac{\sqrt3}{2}

(c)

Z

1+i

and

W

3+4i

. Which complex number does

\vec{WZ}

represent?

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Answer

z-w=-2-3i

Why

Head minus tail: the vector from

W

Z

represents

z-w

(d) Find the length

WZ

in part (c).

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Answer

WZ=\sqrt{13}

Why

WZ=|z-w|=|-2-3i|=\sqrt{4+9}=\sqrt{13}

Rotation and dilation

Slide 4

Key idea

Multiplying by $w=r\cis\theta$

$zw$ has modulus $r|z|$ and argument $\arg z+\theta$ : dilate by factor $r$ and rotate by $\theta$ about $O$ .

To rotate $\vec{AB}$ about $A$ through $\theta$ (keeping the length) to get $\vec{AC}$ :

c-a=(b-a)\cis\theta

Anticlockwise $90^\circ$ : multiply by $i$ . Clockwise $90^\circ$ : multiply by $-i$ .

Multiplying z by w = r cis θ turns z through θ and scales its length by r (here r > 1, so the point also moves away from O).

Working pattern

For HSC geometry: write the vector you know, rotate it, then add back the starting point:

c=a+(b-a)\cis\theta

Example 2 · Rotate and enlarge

Slide 5

Example 2

Let

z=2+2i

. Find the complex number obtained by rotating

z

anticlockwise by

\dfrac{\pi}{3}

about

O

and doubling its distance from

O

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Answer

(2-2\sqrt3)+(2+2\sqrt3)i

Why

Multiply by

2\cis\frac{\pi}{3}=1+\sqrt3\,i

(2+2i)(1+\sqrt3i)=2+2\sqrt3i+2i-2\sqrt3

. Check:

|z|=2\sqrt2

becomes

4\sqrt2

, and

\arg z=\frac{\pi}{4}

becomes

\frac{\pi}{4}+\frac{\pi}{3}=\frac{7\pi}{12}

Show diagramHide diagram

z = 2 + 2i is turned through π/3 and its distance from O is doubled.

Example 3 · A square

Slide 6

$OABC$ is a square, labelled anticlockwise, with $A$ at $2+i$ .

(a) Find

C

and

B

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Answer

C=-1+2i

and

B=1+3i

Why

Rotating

\vec{OA}

90^\circ

anticlockwise gives

\vec{OC}

c=i(2+i)=-1+2i

. Then

b=a+c=(2+i)+(-1+2i)=1+3i

since

OABC

is a parallelogram.

(b) The square is rotated

45^\circ

anticlockwise about

O

. Find the new position of

B

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Answer

-\sqrt2+2\sqrt2\,i

Why

b'=b\cis\frac{\pi}{4}=(1+3i)\cdot\frac{1+i}{\sqrt2}=\frac{-2+4i}{\sqrt2}=-\sqrt2+2\sqrt2\,i

Show diagramHide diagram

Square OABC, and the image B′ of B after a 45° rotation about O.

Example 4 · Rotating about another point

Slide 7

Example 4

In triangle

ABC

AB=BC

and

\angle ABC=90^\circ

B

5+3i

and

C

9+6i

, and

A

lies above the line

BC

. Find the complex number represented by

A

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Answer

A=2+7i

Why

\vec{BC}=c-b=4+3i

. Since

BA=BC

and

\angle ABC=90^\circ

with

A

above

BC

\vec{BA}

\vec{BC}

rotated

90^\circ

anticlockwise:

i(4+3i)=-3+4i

. So

a=b+(-3+4i)=2+7i

. (If

A

were below

BC

, multiply by

-i

to get

a=8-i

Show diagramHide diagram

Triangle ABC with a right angle at B and BA = BC.

Example 5 · Equilateral triangle

Slide 8

$O$ , $A$ and $B$ represent $0$ , $z_1$ and $z_2$ , where $z_2=\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)z_1$ .

Adapted from 2020 HSC Q14(a)(i) Explain why

\triangle OAB

is equilateral. (ii) Show that

z_1^2-z_1z_2+z_2^2=0

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Answer

(i)

|z_2|=|z_1|

, so

OA=OB

, and

\angle AOB=\frac{\pi}{3}

. An isosceles triangle with apex angle

60^\circ

is equilateral. (ii) Let

\omega=\cis\frac{\pi}{3}

. Then

\omega^3=\cis\pi=-1

, so

(\omega+1)(\omega^2-\omega+1)=0

. Since

\omega\ne-1

\omega^2-\omega+1=0

, and

z_1^2-z_1z_2+z_2^2=z_1^2(1-\omega+\omega^2)=0

Why

Alternatively,

\omega^2=\cis\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i

and

\omega-1=-\frac12+\frac{\sqrt3}{2}i

are equal, so

\omega^2=\omega-1

. (The original paper writes

z_2

in exponential form.)

Show diagramHide diagram

Triangle OAB (drawn with z1 = 2).

Example 6 · Perpendicular vectors

Slide 9

$z$ and $w$ are non-zero with $\arg\left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}$ .

Adapted from 2024 HSC Q14(c)(i) Prove that

\left|\dfrac{z-w}{z+w}\right|=1

. (ii) Interpret this result geometrically.

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Answer

(i)

\frac{z}{w}=-ri

for some

r>0

. Dividing top and bottom by

w

\dfrac{z-w}{z+w}=\dfrac{z/w-1}{z/w+1}=\dfrac{-1-ri}{1-ri}

. Both numerator and denominator have modulus

\sqrt{1+r^2}

, so the modulus is

1

. (ii)

OZ\perp OW

, so the parallelogram on

OZ

and

OW

is a rectangle and its diagonals

|z+w|

and

|z-w|

are equal.

Why

\arg\frac zw=-\frac{\pi}{2}

means

\frac zw

is a negative imaginary number. Rotating

w

-90^\circ

gives a vector in the direction of

z

, so

OZ\perp OW

(compare the exit ticket in Lesson 7).

Example 7 · Proving a quadrilateral is a square

Slide 10

The vertices $A$ , $B$ , $C$ , $D$ of a quadrilateral represent $z_1$ , $z_2$ , $z_3$ , $z_4$ .

(a) If

z_1-z_2=z_4-z_3

, show that

ABCD

is a parallelogram.

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Answer

z_1-z_2

represents

\vec{BA}

and

z_4-z_3

represents

\vec{CD}

. Equal vectors mean

BA\parallel CD

and

BA=CD

, so

ABCD

is a parallelogram.

Why

Name the vector each complex number represents before interpreting it. One pair of sides equal and parallel is enough for a parallelogram.

(b) If also

z_3-z_2=i(z_1-z_2)

, show that

ABCD

is a square.

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Answer

z_3-z_2=\vec{BC}

\vec{BA}

rotated through

90^\circ

, so

BC=BA

and

\angle ABC=90^\circ

. A parallelogram with a right angle and two equal adjacent sides is a square.

Why

Multiplying by

i

keeps the length and rotates by

90^\circ

Exit ticket

Slide 11

O

2

and

z

form an equilateral triangle with

z

in the first quadrant. Find

z

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Answer

z=1+\sqrt3\,i

Why

z=2\cis\frac{\pi}{3}

: rotate the vector

2

about

O

through

60^\circ

anticlockwise (the third vertex is at the same distance, 2, from

O

ABCD

is a square labelled anticlockwise, with

A

1+i

and

B

4+2i

. Find

C

and

D

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Answer

C=3+5i

and

D=4i

Why

\vec{AB}=3+i

\vec{BC}=i(3+i)=-1+3i

, so

C=(4+2i)+(-1+3i)=3+5i

D=A+\vec{BC}=(1+i)+(-1+3i)=4i

Show diagramHide diagram

Square ABCD for question 2.