Lesson 7 · Complex numbers as vectors

Learning intentionTo represent complex numbers as vectors and interpret addition and subtraction geometrically.

Success criteria · I can…

represent $z$ as a vector with length $|z|$ and direction $\arg z$
add (parallelogram, tip to tail) and subtract vectors on the Argand diagram
use $|z-w|$ as the distance $WZ$ and $\arg(z-w)$ as the direction of $\vec{WZ}$
prove and use the triangle inequality $|z_1+z_2|\le|z_1|+|z_2|$

ResourcesLesson slides
Practice worksheet
Worked solutions
Complex Locus Lab applet

Lesson flow

LaunchWarm-up · Retrieval from Lessons 4 to 6Slide 2 Key ideasVectors on the Argand diagramSlide 3 ExamplesExample 1 · A vector between two pointsSlide 4 ExamplesExample 2 · Adding on the Argand diagramSlide 5 ExamplesExample 3 · Subtraction, tip to tailSlide 6 ExamplesExample 4 · Distance and directionSlide 7 Key ideasThe triangle inequalitySlide 8 ExamplesExample 6 · Using the triangle inequalitySlide 9 HSC-styleHSC practice · Adding unit vectorsSlide 10 Exit ticketExit ticketSlide 11

Warm-up · Retrieval from Lessons 4 to 6

Slide 2

(a) Find

|z|

z=(3+4i)(1-i)

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Answer

|z|=5\sqrt2

Why

The modulus of a product is the product of the moduli:

|3+4i||1-i|=5\cdot\sqrt2

(b) Find

\Arg(z_1z_2)

\arg z_1=\dfrac{3\pi}{4}

and

\arg z_2=\dfrac{\pi}{2}

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Answer

\Arg(z_1z_2)=-\dfrac{3\pi}{4}

Why

Arguments add:

\frac{3\pi}{4}+\frac{\pi}{2}=\frac{5\pi}{4}

. This is outside

(-\pi,\pi]

, so subtract

2\pi

to get the principal argument

-\frac{3\pi}{4}

(c) For

\tilde{a}=3\tilde{i}+2\tilde{j}

and

\tilde{b}=\tilde{i}-2\tilde{j}

, find

\tilde{a}+\tilde{b}

and

\tilde{a}-\tilde{b}

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Answer

\tilde{a}+\tilde{b}=4\tilde{i}

and

\tilde{a}-\tilde{b}=2\tilde{i}+4\tilde{j}

Why

Add or subtract components. These are the same numbers as

u+v=4

and

u-v=2+4i

in Example 3, which you met in the Extension 1 vectors work.

(d) Plot

z=-1+2i

and

iz

. Through what angle has

z

turned about

O

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Answer

iz=-2-i

, and

z

has turned through

90^\circ

anticlockwise.

Why

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

. Multiplying by

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

adds

\frac{\pi}{2}

to the argument.

Vectors on the Argand diagram

Slide 3

Key idea

Complex numbers as vectors

$z=x+iy$ is represented by the vector $\vec{OZ}$ , or by any vector with the same length $|z|$ and the same direction $\arg z$ .

The vector from $W$ to $Z$ represents $z-w$ : “head minus tail”.

$z+w$ is the diagonal $\vec{OC}$ of the parallelogram with sides $\vec{OA}=z$ and $\vec{OB}=w$ . Equivalently, place $w$ at the tip of $z$ (“tip to tail”).

The parallelogram on z and w. One diagonal is z + w, the other (from W to Z) is z − w.

Example 1 · A vector between two points

Slide 4

Example 1

Plot

z=3+4i

and

\conj{z}

as vectors from

O

. Draw the vector from

Z

\conj{Z}

. Which complex number does it represent?

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Answer

\conj{z}-z=(3-4i)-(3+4i)=-8i

: a vector pointing straight down, of length 8.

Why

The vector from

Z

\conj{Z}

is “head minus tail”,

\conj{z}-z

. The same vector

-8i

can be drawn starting anywhere: only length and direction matter.

Show diagramHide diagram

The vector from Z to Z̄ is straight down, length 8.

Example 2 · Adding on the Argand diagram

Slide 5

(a) Show

(1+2i)+(3+i)

on the Argand diagram.

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Answer

(1+2i)+(3+i)=4+3i

. With

\vec{OA}=1+2i

and

\vec{OB}=3+i

, the diagonal

\vec{OC}=4+3i

of parallelogram

OACB

Why

Add real parts and imaginary parts. Geometrically, slide the vector OB so its tail is at A (tip to tail): it ends at C.

(b) If

z=1+3i

and

w=4+i

, plot

z

w

-z

and

z+w

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Answer

z+w=5+4i

and

-z=-1-3i

Why

-z

z

rotated by

180^\circ

about

O

(same length, opposite direction).

z+w

is the diagonal of the parallelogram on

z

and

w

Show diagramHide diagram

Part (b): z, w, z + w and −z.

Example 3 · Subtraction, tip to tail

Slide 6

The diagram shows $u=3+2i$ and $v=1-2i$ . On the same diagram draw the vectors representing $u+v$ , $u-v$ and $v-u$ .

u = 3 + 2i and v = 1 − 2i.

u+v

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Answer

u+v=4

(a horizontal vector of length 4).

Why

Place the tail of v at the tip of u, or take the long diagonal of the parallelogram on u and v.

u-v

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Answer

u-v=u+(-v)=2+4i

: the vector from

V

U

Why

Head minus tail: the vector pointing from the tip of v to the tip of u. It is the other diagonal of the parallelogram.

v-u

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Answer

v-u=-2-4i

: the vector from

U

V

Why

This is the opposite of the vector in part 2: same length, reversed direction.

Show diagramHide diagram

u + v along the real axis, and u − v drawn from the tip of v to the tip of u.

Example 4 · Distance and direction

Slide 7

$|z-w|$ is the distance $WZ$ , and $\arg(z-w)$ is the angle $\vec{WZ}$ makes with the positive real direction.

OABC with A at 5 + i and C at 2 + 3i.

Example 4

OABC

is a parallelogram with

A

5+i

and

C

2+3i

. Find the complex numbers represented by (a)

B

, (b)

\vec{AC}

, (c)

\vec{CA}

, and (d) the lengths of both diagonals.

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Answer

(a)

b=7+4i

. (b)

\vec{AC}=-3+2i

. (c)

\vec{CA}=3-2i

. (d)

OB=\sqrt{65}

and

AC=\sqrt{13}

Why

(a)

\vec{OB}=\vec{OA}+\vec{OC}=(5+i)+(2+3i)

. (b) head minus tail:

c-a=(2+3i)-(5+i)

. (c) the opposite vector,

a-c

. (d)

|7+4i|=\sqrt{49+16}

and

|-3+2i|=\sqrt{9+4}

The triangle inequality

Slide 8

Theorem

Triangle inequality

|z_1+z_2|\le|z_1|+|z_2|

Equality holds when $z_1$ and $z_2$ point in the same direction (or one of them is $0$ ).

In triangle OAC the side OC is shorter than OA + AC.

Example 5

Prove the triangle inequality (a) geometrically, using the parallelogram, and (b) algebraically, using

|z|^2=z\conj{z}

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Answer

(a) Let

A

represent

z_1

and

C

represent

z_1+z_2

, so

AC=|z_2|

. In

\triangle OAC

OC\le OA+AC

because one side of a triangle is at most the sum of the other two. That is

|z_1+z_2|\le|z_1|+|z_2|

. (b)

\begin{aligned}|z_1+z_2|^2&=(z_1+z_2)(\conj{z_1}+\conj{z_2})=|z_1|^2+|z_2|^2+z_1\conj{z_2}+\conj{z_1}z_2\\&=|z_1|^2+|z_2|^2+2\Re(z_1\conj{z_2})\\&\le|z_1|^2+|z_2|^2+2|z_1\conj{z_2}|=(|z_1|+|z_2|)^2.\end{aligned}

Both sides are non-negative, so take square roots.

Why

In (b),

z_1\conj{z_2}+\conj{z_1}z_2

is a number plus its conjugate, so it equals

2\Re(z_1\conj{z_2})

. Then use

\Re(w)\le|w|

and

|z_1\conj{z_2}|=|z_1||z_2|

. In (a), the triangle is degenerate (equality) when

O

A

C

are collinear with

A

between

O

and

C

Example 6 · Using the triangle inequality

Slide 9

Example 6(a)

Exam practiceGiven

|z|<\dfrac12

, show that

\left|(1+i)z^3+iz\right|<\dfrac34

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Answer

\left|(1+i)z^3+iz\right|\le|1+i||z|^3+|i||z|=\sqrt2|z|^3+|z|<\dfrac{\sqrt2}{8}+\dfrac12<\dfrac34

Why

Apply the triangle inequality, then

|z_1z_2|=|z_1||z_2|

. Since

\frac{\sqrt2}{8}\approx0.177

, the bound is about

0.677<0.75

Example 6(b)

Adapted from 2020 HSC Q9

z

moves on the circle

|z|=1

. Find the greatest value of

|z+2|+|z-2|

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Answer

The greatest value is

2\sqrt5

, at

z=\pm i

Why

With

z=\cos\theta+i\sin\theta

|z\pm2|=\sqrt{5\pm4\cos\theta}

. Then

\left(\sqrt{5+4c}+\sqrt{5-4c}\right)^2=10+2\sqrt{25-16c^2}

, which is greatest when

c=\cos\theta=0

, giving

20

and so

2\sqrt5

. The triangle inequality alone only gives the bound

6

, which is never reached: a good point to discuss.

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Part (b): |z + 2| + |z − 2| is the sum of the distances from Z to −2 and to 2.

HSC practice · Adding unit vectors

Slide 10

Adapted from 2023 HSC Q14(a)Let

z=\dfrac{\sqrt3}{2}+\dfrac12i

and

w=-\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}i

. (i) Write

z

and

w

in polar form. (ii) Show that

|z+w|^2=\dfrac{4+\sqrt2-\sqrt6}{2}

. (iii)

A

and

C

represent

z

and

z+w

. Explain why

\angle AOC=\dfrac{7\pi}{24}

, and hence find

\cos\dfrac{7\pi}{24}

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Answer

(i)

z=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}

and

w=\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}

. (ii)

|z+w|^2=2+2\cos\dfrac{7\pi}{12}=2+\dfrac{\sqrt2-\sqrt6}{2}=\dfrac{4+\sqrt2-\sqrt6}{2}

. (iii)

|z|=|w|=1

, so

OACB

is a rhombus (with

B

representing

w

) and the diagonal

OC

bisects

\angle AOB=\frac{3\pi}{4}-\frac{\pi}{6}=\frac{7\pi}{12}

, giving

\angle AOC=\frac{7\pi}{24}

. In

\triangle OAC

the perpendicular from

A

meets

OC

at its midpoint, so

\cos\dfrac{7\pi}{24}=\dfrac{|z+w|/2}{1}=\dfrac{\sqrt{8+2\sqrt2-2\sqrt6}}{4}\approx0.6088

Why

(ii)

|z+w|^2=|z|^2+|w|^2+2\Re(z\conj{w})

, and

z\conj{w}=\cis\left(\frac{\pi}{6}-\frac{3\pi}{4}\right)

has real part

\cos\frac{7\pi}{12}=\frac{\sqrt2-\sqrt6}{4}

. (iii) The diagonals of a rhombus bisect its angles and meet at right angles, so the foot of the perpendicular from

A

is the midpoint of

OC

. Then

\cos\frac{7\pi}{24}=\frac{|z+w|}{2}=\frac12\sqrt{\frac{4+\sqrt2-\sqrt6}{2}}=\frac{\sqrt{8+2\sqrt2-2\sqrt6}}{4}

. (The original paper uses the exponential form.)

Exit ticket

Slide 11

Z

2+i

and

W

-1+5i

. Find the complex number represented by

\vec{WZ}

and the length

WZ

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Answer

\vec{WZ}=3-4i

and

WZ=5

Why

Head minus tail:

z-w=(2+i)-(-1+5i)=3-4i

, and

|3-4i|=\sqrt{9+16}=5

2. Explain geometrically why

|z+w|=|z-w|

exactly when

OZ\perp OW

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Answer

z+w

and

z-w

are the two diagonals of the parallelogram on

OZ

and

OW

. The diagonals are equal exactly when the parallelogram is a rectangle, that is, when

OZ\perp OW

Why

A parallelogram with equal diagonals is a rectangle, so adjacent sides are perpendicular. This idea is used again in Lesson 8.