Mex n1 2 Complex Numbers Showcloze 201027

MATHEMATICS EXTENSION 2
Topic summary and exercises:

With references to
X2 Complex numbers
Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Initial version by H. Lam, August 2012. With major changes in October 2019 by I. Ham. Updated October 27, 2020 for latest
syllabus.
Various corrections by students & members of the Mathematics Departments at North Sydney Boys High School and
Normanhurst Boys High School.
Acknowledgements Pictograms in this document are a derivative of the work originally by Freepik at
http://www.flaticon.com, used under CC BY 2.0.
Symbols used Syllabus outcomes addressed

! Beware! Heed warning. MEX12-4 uses the relationship between algebraic and
geometric representations of complex numbers and
F Provided on NESA Reference Sheet complex number techniques to prove results, model and
solve problems
M Facts/formulae to memorise.
L Literacy: note new word/phrase.

Syllabus subtopics
MEX-N1 Introduction to Complex Numbers
N Further reading/exercises to enrich your understanding
and application of this topic. MEX-N2 Using Complex Numbers
S Syllabus specified content
U Facts/formulae to understand, as opposed to blatant

memorisation.
N the set of natural numbers
Z the set of integers
Q the set of rational numbers
R the set of real numbers
C the set of complex numbers
∀ for all
V Gentle reminder
• For a thorough understanding of the topic, every blank space/example question
in this handout is to be completed!
• Additional questions from CambridgeMATHS Extension 2 (Sadler & Ward,
2019) and other selected texts will be completed at the discretion of your teacher.
• Remember to copy the question into your exercise book!
Contents
1 A new number system 5

1.1 Review of number systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 The “imaginary” numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4 Basic operations with complex numbers . . . . . . . . . . . . . . . . . . . . . 9
1.4.1 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4.2 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4.3 Complex conjugate pairs . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.5 Properties of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5.1 Equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5.2 Solutions to equations . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Further arithmetic & algebra of complex numbers 16

2.1 Vector representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.1.1 Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.1.2 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.1.3 Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.1.4 Scalar multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2 Modulus/argument of a complex number . . . . . . . . . . . . . . . . . . . . 20
2.2.1 Natural ordering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.2 Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.3 (Principal) Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.2.4 Triangle inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.3.1 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.3.2 Multiplication/Division . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.3.3 Rotations/Enlargement . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3.4 Conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.3.5 Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3 Curves and regions in the complex plane 40

3.1 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.1.1 Lines/rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.1.2 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.2 Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3
4 Contents –
4 Applications to polynomials 52
4.1 Polynomial theorems for equations with complex roots . . . . . . . . . . . . 52
4.2 Trigonometric identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.3 Further exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.4 Roots of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.4.1 Factorisations of higher powers . . . . . . . . . . . . . . . . . . . . . 60
4.4.2 Graphical solutions and consequent factorisations . . . . . . . . . . . 62
4.4.3 Roots of unity: reduction from higher powers . . . . . . . . . . . . . 69
A Past HSC questions 74

A.1 2001 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
A.2 2002 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
A.3 2003 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
A.4 2004 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
A.5 2005 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
A.6 2006 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
A.7 2007 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
A.8 2008 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
A.9 2009 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
A.10 2010 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
A.11 2011 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
A.12 2012 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
A.13 2013 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
A.14 2014 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
A.15 2015 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
A.16 2016 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
A.17 2017 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
A.18 2018 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
A.19 2019 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
A.20 2020 Extension 2 HSC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
References 109
NORMANHURST BOYS’ HIGH SCHOOL

Section 1
A new number system
1.1 Review of number systems

• . . . . . . Natural
. . . . . . . . . . . . . numbers. N = {1, 2, 3 · · · }
Example 1
Solve x + 1 = 5 and x + 3 = 0 over N. Answer: x = 4, no solution
• . . . . . . .Integers
. . . . . . . . . . . . . . Z = {· · · , −3, −2, −1, 0, 1, 2, 3, · · · }
Example 2
Solve x + 3 = 0 and 2x + 4 = 7 over Z. Answer: x = −3, no solution

p
• . . . . . . .Rational
. . . . . . . . . . . . . . numbers. Q = : p, q ∈ Z, q 6= 0
q
Example 3
Solve 2x + 4 = 7 and x2 − 2 = 0 over Q. Answer: x = 3
2
, no solution
• . . . .Real
. . . . . . . numbers. R
Example 4 √
Solve x2 − 2 = 0 and x2 + 5 = 0 over R. Answer: x = ± 2, no solution
5
6 A new number system – Rotation
• N⊆Z⊆Q⊆R
√
5 4
log3 8
11 4 R
e7 23 7
Q
−9 3 √
1.1 Z 5 2
−8 −10
2 1
5π
2 −7
3 N 6 75
5.45 4 5 −4 9
2
−6 −5 73
−1+ 50.4453
√ 6 16
2 1
2
cosh 7
sin 1.25
1.2 Rotation
• From x = 1, go to x = −1 by rotating π radians in the usual direction.
– Multiply 1 by −1 to obtain −1 corresponds to rotating by .π. . radians.
• Stop halfway whilst rotating? Quarter of way whilst rotating?
−2 −1 0 1 2

A new number system – The “imaginary” numbers 7
1.3 The “imaginary” numbers

Definition 1
Imaginary number The imaginary number i to be the “quantity” to multiply with
a real number when rotating anti-clockwise by π2 about x = 0.
• “Jump off” the real number line.
Definition 2
The imaginary number i has property such that
i × i = i2 = −1
• Why?
Definition 3
The set of all imaginary numbers, called the
. . . . . . . . . . . . . complex
. . . . . . . . . .numbers
. . . . . . . . . . . . . . . . . . . . . , is defined to be
C = {z : z = x + iy; x, y ∈ R}
Example 5
Find the values of i2 , i3 , i4 and i5 .
• i2 = . . . . . . . . . . • i3 = . . . . . . . . . . • i4 = . . . . . . . . . . • i5 = . . . . . . . . . .

8 A new number system – The “imaginary” numbers
Definition 4
Complex number A complex number z has . . .real
. . . . . . and . . . . . . . .imaginary
.................
parts and is defined by z = x + iy.
• The . . .real
. . . . . . part of z: Re(z) = x.
• The . . . . . . . .imaginary
. . . . . . . . . . . . . . . . . part of z: Im(z) = y.
• Treat real and imaginary parts as . . . . . . . . . .components

. . . . . . . . . . . . . . . . . . . . of a complex number.
• z = x + iy is known as . . . . . . . .Cartesian
. . . . . . . . . . . . . . . . form.
• Plot on . . . . . .Argand
. . . . . . . . . . . . . . . . . . . diagram
. . . . . . . . . . . . . . , similar to plotting points coordinate
geometry.
Example 6
On the following diagram, plot the location of:
• z1 = 3 + 4i. • z2 = 2 − i. • z3 = −1 − 3i. • z4 = − 21 + 32 i.
Im
4
0 Re
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
V Important note
Looks like another familiar topic from the Extension 1 course?

A new number system – Basic operations with complex numbers 9
1.4 Basic operations with complex numbers

Example 7
Find the value of
√ √ √ √
1. 2+ 3 − 5−4 3 2. 2+ 3 5−4 3
1.4.1 Addition
• Operations similar to surds (group rational parts with rational parts, irrational parts
with irrational parts).
• Group . . . .real
. . . . . . parts with . . .real
. . . . . . parts
• Group . . . . . . . .imaginary
. . . . . . . . . . . . . . . . . parts with . . . . . . . .imaginary
. . . . . . . . . . . . . . . . . parts.
1.4.2 Multiplication
• Use distributive law.
• Beware that i2 = . .−1
. . . . , which becomes . . . .real
.....
Example 8
If z1 = 2 + 3i and z2 = −1 + 5i, find the value of
(a) z1 + z2 (b) z1 − z2 (c) 3z1 (d) 3iz1 (e) z1 z2
Example 9
Find z ∈ C such that Re(z) = 2 and z 2 is imaginary.

10 A new number system – Basic operations with complex numbers
1.4.3 Complex conjugate pairs

Definition 5
If z = x + iy, then its . . . . . . complex
. . . . . . . . . . . . . . . . . . . . . .conjugate
. . . . . . . . . . . . . . . . is denoted z
such that
z = x − iy
√ √
• Analogous to conjugate surds, where the conjugate of a + b c is . . . . . a. .−
. . . . . .c. . . . .
b
• Geometrically,
Im
b
z = x + iy
Re
Example 10
If z1 = 2 + i and z2 = 1 − 3i, evaluate in Cartesian form:
(a) z1 + z2 (c) −z2 (e) z1 − z2

1
(b) z2 − z1 (d) z1 (f)
z2

A new number system – Basic operations with complex numbers 11
© Laws/Results
Summary of complex number properties These involve the Cartesian form:
1. z1 + z2 = . . . .z.1. .+. .z. 2. . . .
2. z1 z2 = . . .z.1.z.2. .

3. z+z = 2 Re(z)
....................

4. z−z = 2i Im(z)
......................
Proof
1. Let z1 = x + iy and z2 = a + ib
2. Let z1 = x + iy and z2 = a + ib
3. Let z = x + iy
4. Let z = x + iy

12 A new number system – Basic operations with complex numbers
History
Gerolamo Cardano (1501-1576), Mathematician (gambler and

chess player!), published solutions to the cubic ax3 + bx + c = 0
in Ars Magna. Cardano was one of the first to acknowledge the
existence of imaginary numbers. Given during the Renaissance,
negative numbers were treated suspiciously, imaginary numbers
would have been almost heretical.
Cardano did not avoid (as most contemporaries did) nor did he immediately provide
solutions to these imaginary numbers (possibly 200 years away). With the equations
containing complex conjugate pairs, Cardano multiplied them together and obtained
real numbers: √
Putting
√ aside the mental tortures involved, multiply 5 + −15 with
5 − −15, making 25 − (−15), which √ is −15. Hence the product is 40.
Cardano, remarked in another work, that −9 is neither +3 or −3, but some “obscure
sort of thing”.
Source:
• Wikipedia
(http://en.wikipedia.org/wiki/Gerolamo Cardano)
• Complex and unpredictable Cardano, Artur Ekert, Mathematical Institute, University of Oxford,
United Kingdom
(http://www.arturekert.org/Site/Varia files/NewCardano.pdf)
Î Further exercises
Ex 1A (Sadler & Ward, 2019)
• All questions
Other references
• Lee (2006, Ex 2.3)

A new number system – Properties of complex numbers 13
1.5 Properties of complex numbers

1.5.1 Equality
Definition 6
Two complex numbers z1 and z2 are equal iff the real and imaginary parts are equal.
Proof
• Let z1 = a + ib, z2 = c + id
• If z1 = z2 , then . . . . . . . . .a. .+. . ib
. . .=. . c. .+ . . . . . . . . . . , or
. . id
• . . . . . . . . . .(a
. . .−. .c)
. . .+. .i(b
. . .− . . . . . . . . . . . = 0 + 0i
. . .d)
1.5.2 Solutions to equations
Example 11
Solve z 2 + 1 = 0 for z ∈ C. Answer: z = ±i
Example 12
Solve z 2 + 2z + 10 = 0 for z ∈ C. Answer: z = −1 ± 3i

14 A new number system – Properties of complex numbers
Example 13
Solve 2z 2 + (1 − i)z + (1 − i) = 0 for z ∈ C. Answer: z = i, z = − 21 − 21 i

A new number system – Properties of complex numbers 15
Example 14
Find the square roots of −3 + 4i in Cartesian form.
L Fill in the spaces

Observations
• Equations with . . . real
...... coefficients will have . . . . . . .complex
.............
. . . . . . . . . . . . . . . . . . . . . . . . roots.
conjugate
• Equations with . . . . . . complex . . . . . . . . . . . . . . coefficients do not necessarily have
. . . . . . . . . . . . . . . . . . . . . . . . . . . conjugate
complex . . . . . . . . . . . . . . . . . roots.
Ex 1B (Sadler & Ward, 2019)
• All questions
Other sources
• Fitzpatrick (1991, Ex 31(a), (b), (c))

• Lee (2006, Ex 2.1, 2.2)
• Arnold and Arnold (2000, Ex 2.1)

Section 2
Further arithmetic & algebra of complex

numbers
2.1 Vector representation

2.1.1 Equivalence
Definition 7
Two vectors p and q on the Argand diagram are equal iff both
• Modulus ( . . . . . . . .magnitude
. . . . . . . . . . . . . . . . . . ), and
• Argument ( . . . . . . .direction
............... )
are equal.
• The starting point ( . . .tail

. . . . . ) is irrelevant for a vector.
16
Further arithmetic & algebra of complex numbers – Vector representation 17
2.1.2 Addition
• Place vectors, head-to-tail.
Im
p q
p+q
Re
O
q
• Parallelogram of vectors when adding two vectors.
Example 15
If z = −3 + 2i and w = 2 + 4i, draw z + w on the Argand diagram.
Im Im
6 6
4 4
2 2
0 Re 0 Re
−3 −2 −1 1 2 3 −3 −2 −1 1 2 3
−2 −2
2.1.3 Subtraction
• For z1 − z2 , add −z2 to z1 .
Im
−q
p + (−q)
p p−q
Re
O p+q
q
• Alternatively, “what vector get you to z1 from z2 ?”

18 Further arithmetic & algebra of complex numbers – Vector representation
• Quick & easy parallelogram of vectors:

1. z1 + z2 starts from tails of z1 & z2 ,
2. z2 − z1 starts from head of z1 , goes to head of z2 .
Example 16
If z = 3 − 2i and w = 2 − 5i, draw z − w on the Argand diagram
Im Im
3 3
2 2
1 1
0 Re 0 Re
−1−1 1 2 3 −1−1 1 2 3
−2 −2
−3 −3
−4 −4
−5 −5
2.1.4 Scalar multiplication

• For kz1 where k ∈ R, stretch z1 by factor of k.
• If k < 0, direction of new vector is opposite to original vector.
Example 17
If z = 1 + 2i, draw 3z and −2z on separate Argand diagrams.
Im Im
6 6
5 5
4 4
3 3
2 b
z 2 b
z
1 1
0 Re 0 Re
−3 −2 −1 1 2 3 4 −3 −2 −1 1 2 3 4
−2 −2
−3 −3
−4 −4

Further arithmetic & algebra of complex numbers – Vector representation 19
Example 18 √ √
[2011 HSC Q2] On the Argand diagram, the complex numbers 0, 1 + i 3, 3 + i,
and z form a rhombus. Im
z b
√
1+i 3
b
θ
√b
3+i
b
Re
O
(i) Find z in the form a + ib, where a and b are real numbers.
(ii) An interior angle, θ, of the rhombus is marked on the diagram. Find the value
of θ.
√ √ 5π
Answer: z = ( 3 + 1) + i( 3 + 1), θ = 6
Ex 1C Ex 1E
• All questions • Q1-21

20 Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
2.2 Modulus/argument of a complex number

2.2.1 Natural ordering
• N, Z, Q, R are well ordered :
22 5 5
.. 1
– 6 .> – 7 .>
.. π – 4 .>
.. 7
• However:
. . . . 3 + 2i
– 6 + 4i . .??? – 3 − 3i . .???
....2 + i
• Natural ordering does not exist with complex numbers.
2.2.2 Modulus
Definition 8
The modulus of a complex number, denoted |z| (where z = x + iy) is the magnitude
( . . . . .length
. . . . . . . . . . ) of the vector from O to z on the Argand diagram.
Im
z = x + iy
b
|
|z
Re
x
i.e. p
|z| = x2 + y 2

Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number 21
2.2.3 (Principal) Argument

Definition 9
The argument of a complex number, measured in . . . . . .radians
. . . . . . . . . . . . , is denoted
arg(z)
(where z = x + iy) is the . . . .angle

. . . . . . . . . that the vector from O to z makes with the
positive real axis on the Argand diagram, with angles increasing in the anticlockwise
direction. Im
z = x + iy
b
arg(z)
Re
i.e. y
arg(z) = tan−1
x
• Duplicate argument(s)?
Example 19
Evaluate arg(z), where z = 1 + i.
Definition 10
The principal argument of a complex number, denoted
Arg(z)
lies within the domain −π < Arg(z) ≤ π.
V Important note
• The principal argument is generally quoted henceforth.
y
• Be aware of the quadrant which z lies. Inputting tan−1 x
on the calculator
mindlessly may give an erroneous result.
• The complex number z = 0 + 0i has . . .no
. . . argument . . . . . defined
............ .

Example 20
Find the modulus and principal argument of the following:
√
(a) 2 + 2i Answer: modulus: 2 2, argument π4
√
(b) −1 − i 3 Answer: modulus: 2, argument − 2π
3
Example 21
[2011 HSC Q2] Let w = 2 − 3i and z = 3 + 4i.
(a) Find w + z. Answer: 5 + i
√
(b) Find |w|. Answer: 13
w
(c) Express in the form a + ib, where a, b ∈ R. 6
Answer: − 25 − 17
25
i
z

Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number 23
2.2.4 Triangle inequality
º Theorem 1
For every complex number z1 and z2 ,
|z1 + z2 | ≤ |z1 | + |z2 |
1
Proof
³ Steps
1. Let p and q (with P and Q being the head of the arrow) represent the complex
numbers z1 and z2 respectively, p + q with R being the head of the arrow.
2. On the Argand diagram:
Im
Re
3. . . . . . . . . . . . .|z
. .1. +
. . .z.2.| .=
. . .|z. 1. |. .+. .|z
. .2.|. . . . . . . . . . . iff O, P and Q are collinear (which
implies OP k OQ k OR)
• Conclusion: z1 = kz2 , where k ∈ R as vectors are parallel.
4. Otherwise, . . . . . . . . . . . .|z. .1.+ . . .z.2.| .<
. . .|z. 1. |. .+. .|z. .2.| . . . . . . . . . . . .
5. Hence, . . . . . . . . . . . .|z. 1. .+
. . z. 2. |. .≤. . |z
. .1.|. +
. . .|z. .2 .| . . . . . . . . . . .
1
Never attempt to prove this algebraically!

Example 22
If z1 = 3 + 4i and |z2 | = 13, find the greatest value of |z1 + z2 |. If |z1 + z2 | is at its
greatest value, find the value of z2 in Cartesian form.
39 52
Answer: |z1 + z2 | = 18 at its greatest; z2 = 5
+ 5
i
Other references
• Lee (2006, Ex 2.6 Q1-7) • Arnold and Arnold (2000, Ex 2.3)
• Patel (2004, Ex 4K)

Further arithmetic & algebra of complex numbers – Euler’s Formula 25
2.3 Polar form (Euler’s formula)

2.3.1 Arithmetic
Definition 11
The modulus-argument form of a complex number z is
Im
z = x + iy
b
z = x + iy (Cartesian form)
y = |z| sin θ
|
|z
= |z| cos θ + i |z| sin θ
= |z| (cos θ + i sin θ)
= r (cos θ + i sin θ) (Mod-arg form)
θ
Re
where Arg(z) = θ. x = |z| cos θ
2
• . . . . . . . . . . .z. .= . . . . . θ. .+
. . .r(cos . . i. .sin . . . . . . . . . . . . often abbreviated to z = r cis θ.
. . .θ)
iθ
• Better to abbreviate . . . . . . . . . . .z. .=. . r(cos
. . . . . .θ. .+. .i. sin . . . . . . . . . . . . to z = re (for reasons
. . . .θ)
that will be made obvious later)
Definition 12
Polar form: Euler’s formula
eiθ = cos θ + i sin θ
where e ≈ 2.71828 · · ·
(Leonhard Euler, 1707-1783.

http://en.wikipedia.org/wiki/Leonhard Euler)
(All index laws in R also apply to the complex exponential).
2
cis θ does very little to assist your understanding of the rules for multiplying complex numbers!

26 Further arithmetic & algebra of complex numbers – Euler’s Formula
Example
√
23

Write z = 2 cos 3π
4
+ i sin 3π
4
in Cartesian form. Answer: z = −1 + i
Example 24 √ 3π
Write z = −2 − 2i in polar form. Answer: z = 2 2e−i 4
Example 25
10
Write z = √ in polar form, and hence write in simplest Cartesian form.
3+i iπ √
5 3
Answer: z = 5e− 6 = 2
− 25 i
Example 26 2iπ √
Write z = 6e− 3 in Cartesian form. Answer: −3 − 3i 3

Example 27 √
Evaluate the product (1 + i) 1 − i 3 in Cartesian form and polar form, to show
√
π 1+ 3
that cos = √ .
12 2 2

Example 28
Use Euler’s formula to write cos θ and sin θ in terms of e.
Ex 3D
• Q1-15
Other references
• Patel (2004, Ex 4C, Q1-10)

• Arnold and Arnold (2000, Ex 2.2, Q1-4)
• Lee (2006, Ex 2.6 Q8 onwards)

2.3.2 Multiplication/Division
© Laws/Results
Multiplication of z1 and z2 : moduli . . . . . . multiply
. . . . . . . . . . . . . . . , arguments . . . .add
.....
z1 z2 = r1 r2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) = r1 r2 ei(θ1 +θ2 )
Division of z1 and z2 : moduli . . . . . divide

. . . . . . . . . . , arguments . . . . . . .subtract
..............
z1 r1
= (cos (θ1 − θ2 ) + i sin (θ1 − θ2 )) = r1 r2 ei(θ1 −θ2 )
z2 r2
Proof Let z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 )
z1 z2 =
Proof (via index laws and complex exponential) Let z1 = r1 eiθ1 and z2 = r2 eiθ2 .
z1 z2 =

2.3.3 Rotations/Enlargement
© Laws/Results
Multiplication of z by another complex number ω = reiθ = r (cos θ + i sin θ):
• . . . . . . .Enlarges
. . . . . . . . . . . . . . the modulus of z by factor r
• . . . . . .Rotates
. . . . . . . . . . . . . z . . . . . . . . . .anticlockwise
. . . . . . . . . . . . . . . . . . . . . . about the origin by θ.
• (Multiplication by ω = re−iθ ) : . . . . . .Rotates
. . . . . . . . . . . . . z . . . . . . . .clockwise
...............
about the origin by θ.
Example
√
29
π
Let z = 3 + i. Multiply z by another complex number ω = ei 2 . Find zω in
Cartesian form. Plot z and iz on an Argand diagram.
© Laws/Results
Multiplication of z by i and −i respectively:
π
• . . . . . .Rotates
. . . . . . . . . . . . . z . . . . . . . . . .anticlockwise
. . . . . . . . . . . . . . . . . . . . . . about the origin by 2
π
• . . . . . .Rotates
. . . . . . . . . . . . . z . . . .π. . .clockwise
. . . . . . . . . . . . . . . . about the origin by 2
±i 2
(Essentially, multiply by . . . e. . . . . . . )
Example 30
In the Argand diagram below, intervals AB, OP and OQ are equal in length, OP is
π
parallel to AB and ∠P OQ = .
2
Im
B
A
P
Re
(a) If A and B represent the complex numbers 3 + 5i and 9 + 8i respectively, find

the complex number which is represented by P . NORMANHURST BOYS’ HIGH SCHOOL
(b) Hence find the complex number which is represented by Q.

Example 31
(Sadler & Ward, 2019) Let z = 1 + i.
(a) Find, in Cartesian form, the complex number w such that wz is a rotation of
z by π3 anticlockwise about the origin.
(b) Evaluate wz in Cartesian form.
(c) Verify |wz| = |z|, then plot z and wz on an Argand diagram.

Example 32
[UNSW MATH1131 exercises, Problems 1.7, Q35]
(a) Explain why multiplying a complex number z by eiθ rotates the point
represented by z anticlockwise about the origin, through an angle θ.
(b) The point represented by the complex number 1 + i is rotated anticlockwise
π
about the origin through an angle of . Find the resultant complex number
6
in polar and Cartesian form.
(c) Find the complex number (in Cartesian form) obtained by rotating 6 − 7i
3π
anticlockwise about the origin through an angle .
√ 5π
√ 4 √
Answer: (a) Explain (b) 2ei 12 = 1
2
3−1 +i 3+1 (c) 1
√ (1 + 13i)
2

2.3.4 Conjugates
© Laws/Results
If z = r (cos θ + i sin θ), then
z = r (cos(−θ) + i sin(−θ)) = re−iθ
Proof Let z = r (cos θ + i sin θ).
Example 33
[2020 Ext 2 Sample Q4] The Argand diagram shows the complex number eiθ .
Which of the following could be the complex number −ie−iθ ?
A. B. C. D.
(A) (C)
O O
(B) (D)
O O O

2.3.5 Powers
º Theorem 2
De Moivre’s Theorem For n ∈ Z,
(cos θ + i sin θ)n = cos(nθ) + i sin(nθ)
(Abraham De Moivre, 1667–1754.

http://en.wikipedia.org/wiki/Abraham de Moivre)
Proof (via complex exponential)
Proof (by induction – for later in the course)

Example 34
Simplify the following, 3 expressing the answer in polar form:
(a) cos π3 + i sin π3 . Answer: −1

3π 3
(b) 2 cos 3π4
+ i sin 4
. Answer: 8ei 4
π
√ 8
(c) 2 cos π6 − i sin π6 . Answer: 16ei
2π
3
1 3π

3π 4
1 π

π 5 243 −i 39π
(d) 2
cos 5
+ i sin 5
÷ 3
cos 8
− i sin 8
Answer: 16
e 40
Example 35
If |z1 | = 3, Arg(z1 ) = 2, |z2 | = 2 and Arg(z2 ) = 3, find the modulus and argument of
2z1 2
. 9
Answer: |z| = 20 , Arg(z) = 2π − 5
5z2 3

Example 36
[1988 4U HSC Q4(a)]
√ √
(a) Express z = 2 − i 2 in modulus-argument form.
π π

Answer: z = 2 cos 4
− i sin 4
(b) Hence write z 22 in the form a + ib, a, b ∈ R. Answer: z 22 = 222 i

Example 37
√
(a) If z1 = 1 + i and z2 = 3 − i, find the moduli and principal arguments of z1 ,
z1 √
z2 and .
z1
Answer: z1 = 2 exp iπ , z2 = 2 exp − iπ , z = √1 exp 5iπ

4 6 12
.
z2 2 2
1+i
(b) If z = √ , find the smallest positive integer n such that z n is real, and
3−i
evaluate z n for this integer n. 1
Answer: n = 12, z 12 = − 64

© Laws/Results
Summary of complex number properties These involve the modulus-argument
form:
1. z1 + z2 = . . . . .z.1. + . . .z.2. . . .
2. z1 z2 = . . . z. .1 .z.2. .

3. z + z = . . . . . . .2. Re(z) ............

4. z − z = . . . . . . . 2i . . .Im(z)
............
5. |z1 z2 | = . . . . |z
. . 1.|. |z
. .2.|. . . . , arg (z1 z2 ) = . . . . . . . .arg . . . .z.1. .+. .arg
. . . .z.2. . . . . . . .

z1 |z1 | z1
6. =
z2 . . .|. . , arg z
. . .|z = . . . . . . . . . .arg
. . . .(z
. .1.). −
. . .arg
. . . .(z. .2.). . . . . . . . .
2 2
n
7. |z n | = . . .|z| n
. . . . . , arg (z ) = . . . . .n. .arg(z)
..........

1 1 1
8. =
zn . .n. . . , arg z n = . . . . . −n
. . .|z| . . . .arg(z)
...........
Proof
1. Let z1 = r1 (cos θ + i sin θ) and z2 = r2 (cos φ + i sin φ)
3. Let z = r (cos θ + i sin θ)

Note all uses of ‘cis θ’ should really be replaced with eiθ .
Ex 1D Ex 3A
• Q1-22
• Q1-17
Other references
• Patel (2004, Ex 4C, Q11 onwards),

• Patel (2004, Ex 4D)
• Arnold and Arnold (2000, Ex 2.2, Q6 onwards)
• Lee (2006, Ex 2.5, 2.9)

Section 3
Curves and regions in the complex plane
3.1 Curves
3.1.1 Lines/rays
• |z| = r
Derivation of Cartesian equation: Diagram:
Description: . . . . .Circle
. . . . . . . . . . , . . . . .centre
. . . . . . . . . . (0, 0), . . . . .radius
.......... r
Write the equation that represents the . . . . .circle

........ with . . . . .centre
. . . . . . . . . . z1 and
radius
. . . . . . . . . . . . . . . r:
40
Curves and regions in the complex plane – Curves 41
• |z − z1 | = |z − z2 |
Diagram:
Description: . . . . . . . . . . .Perpendicular
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .bisector
. . . . . . . . . . . . . of the interval joining
z1 to z2 .

42 Curves and regions in the complex plane – Curves
• Arg(z − z1 ) = α, α ∈ R
Diagram:
Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
• Arg(z − z1 ) = Arg(z − z2 )
Diagram:
Description: Line through z1 and z2 but . . . . . . . .excluding

. . . . . . . . . . . . . . . . the . . . . . . .segment
.............
from z1 to z2
3.1.2 Curves
• |z − z1 | = r
Derivation of Cartesian equation: Diagram:
Description: . . . . . . . . . . . . . . . . . . . . . Circle,
. . . . . . . .centred
. . . . . . . . at
. . . z. .1 ,. .radius
. . . . . . .r. . . . . . . . . . . . . . . . . . . . . .

• ! Arg(z − z1 ) − Arg(z − z2 ) = α, 0 < α < π

Diagrams:
Origin: circle geometry theorem – Angle at the circumference subtended by the same
arc/chord
Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Special note: Arc is taken in the anticlockwise direction from z1 to z2 :
...................................................................................
π
• ! 2 Arg(z − z1 ) = Arg(z − z2 ) = α, 0 < α < 2
Diagram:
Origin: circle geometry theorem – Angle at the centre is double the angle at the circumference
subtended by the same arc/chord
Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Example 38
For the following:
i. Describe the path traced out by the conditions on z.
ii. Draw a sketch.
iii. Give the Cartesian equation of the line/curve.
(a) |z| = 2 (c) |z + 2| = 1

(b) zz = 16 (d) |z + 2 + 3i| = 2
Answer: (a) x2 + y 2 = 4 (b) x2 + y 2 = 16 (c) (x + 2)2 + y 2 = 1 (d) (x + 2)2 + (y + 3)2 = 4

Example 39
[2019 NBHS Ext 2 Trial Q11]
i. Find the points of intersection on the curves given by 3
1
|z − i| = 1 and Re(z) = − √ Im(z)
3
ii. Sketch above the two curves on the Argand diagram to show the points 1
of intersection. √
Answer: 0 + 0i, − 23 + 32 i

Example 40
[2018 Independent Ext 2 Q12] z is a complex number such that

√

z − 2 2 (1 + i) = 2
i. On an Argand diagram, sketch the path which is traced by the condition 1

above.
ii. Q is a point on the path traced out where z has its smallest principal 2
argument.
Find the value of the complex number represented by Q in

modulus-argument form. √ π
Answer: 2 3ei 12
Example 41
[2012 NSGHS Ext 2 Q12] Given z is a complex number, sketch on the number
plane, the path traced out by the complex numbers z such that

arg z = arg z − (1 + i)
Example 42
Sketch the curve in the Argand diagram determined by Arg(z − 1) = Arg(z + 1) + π4 .
Find its Cartesian equation. Answer: x2 + (y − 1)2 = 2, y > 0

Example 43
[2016 Caringbah HS Ext 2, Q8] ! The complex number z satisfies

z−2 π
Arg =−
z + 2i 2
Find the
√ maximum value of |z|.
√ √ √
(A) 2 (B) 2 2 (C) 2− 2 (D) 2+ 2
Example 44
z satisfies |z − i| = Im(z)+1. Sketch the path traced out by the point P representing
z in the Argand diagram and write down its Cartesian equation. Answer: x2 = 4y

Example 45
[2003 Q2] Suppose that the complex number z lies on the unit circle, and
0 ≤ arg(z) ≤ π2 .
Prove that 2 arg(z + 1) = arg(z).
! Draw picture!

Curves and regions in the complex plane – Regions 49
3.2 Regions
R For a brief review Stage 5 (Year 10) work on regions:
Ex 3F (Pender, Sadler, Ward, Dorofaeff, & Shea, 2019)
• All questions
Example 46
[2011 CSSA Ext 2 Q2]
i. Sketch the path traced out by the complex numbers z such that 2
|z − 3 + 3i| ≤ 2
ii. Find the maximum value of |z|. 1
Example 47
Sketch the region in the Argand diagram defined simultaneously by
6 ≤ Re [(2 − 3i)z] < 12 and Re(z) Im(z) > 0

50 Curves and regions in the complex plane – Regions
Example 48
Draw a sketch of the curve or region, given z ∈ C and
π 2π π
(a) Arg(z) = 3
(b) 0 ≤ Arg(z) ≤ 3
(c) Arg(z − 2 + 3i) = 4
Example 49
z is a complex number which simultaneously satisfies
π
2 ≤ |z + 3| ≤ 3 and 0 ≤ Arg(z + 3) ≤
3
Find the area and perimeter of the region in the Argand diagram determined by these
restrictions on z. Answer: A = 5π
6
units2 , P = 2 + 5π
3
units

Curves and regions in the complex plane – Regions 51
Ex 1F
• Q1-17
Other resources
• Patel (1990, Self Testing Ex 4.9, p.127)

• Arnold and Arnold (2000, Ex. 2.5)
• Fitzpatrick (1991, Ex 31(f))
• Lee (2006, Ex 2.7, 2.8)

Section 4
Applications to polynomials
4.1 Polynomials theorems for equations with roots in C

© Laws/Results
For polynomials with . . . .real . . . . . . coefficients, the following theorems function in C,
exactly in the same way as they do in R.
• . . . . . . . . .Remainder
. . . . . . . . . . . . . . . . . . theorem.
– Added bonus: . . . . . . . .conjugate
................ roots means the
conjugate
........................ remainder can be found easily.
• . . . . .Factor
. . . . . . . . . . . theorem.
– Added bonus: . . . . . . .conjugate
. . . . . . . . . . . . . . . . roots may help!
• . . . . . .Vieta’s
........... . . . . . . .formulas
.............. :
– Sum – Triples etc
– . . . . Pairs
......... – . . . . . . .Product
.............
• Roots with . . . . . . . . .multiplicity

. . . . . . . . . . . . . . . . . . . . > 1.
Example 50
(Sadler & Ward, 2019) Let P (x) = x3 − 2x2 − x + k, k ∈ R.
(a) Show that P (i) = (2 + k) − 2i
(b) When P (x) is divided by x2 + 1, the remainder is 4 − 2x. Find the value of k.
Answer: k = 2
52
Applications to polynomials – Polynomial theorems for equations with complex roots 53
Example 51
Find all the zeros of P (x) = x4 − x3 − 2x2 + 6x − 4 over C, given 1 + i is a zero.
Hence, fully factorise P (x) over R. Answer: P (x) = (x2 − 2x + 2)(x + 2)(x − 1)
Example 52
Prove that 2 + i is a root of the equation x4 − 2x3 − 7x2 + 26x − 20 = 0, and hence
√
solve the equation completely over C. Answer: x = 2 ± i, −1 ± 5.

54 Applications to polynomials – Polynomial theorems for equations with complex roots
Example 53
Determine b, c ∈ R such that −2i is a zero of x3 + 3x2 + bx + c. Answer: b = 4, c = 12.
Example 54
P (x) is a monic polynomial of degree 4 with integer coefficients and constant term
2. P (x) has a zero i, and a rational zero. The sum of the zeros of P (x) is a positive
real number. Find P (x) factorised into irreducible factors over R.
Answer: P (x) = (x2 + 1)(x − 1)(x − 2)
Ex 1G
• Q1-16, 18

Applications to polynomials – Trigonometric identities 55
4.2 Trigonometric identities

Example 55
(a) [2003 Ext 2 HSC Q2(d)] Using De Moivre’s theorem, find an expression for
cos 5θ in terms of cos θ. Answer: cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ
(b) Hence solve 16x4 − 20x2 + 5 = 0 for x. Answer: x = cos π

10
, cos 3π
10
, cos 7π
10
, cos 9π
10
.
Solution
³ Steps
(a) • By De Moivre’s Theorem,
cos 5θ + i sin 5θ = (cos θ + i sin θ)5

................................
5
• Expand . . . . . . . . (cos
. . . . .θ. .+. .i. sin . . . . . . . . . . via binomial theorem:
. . . .θ)
• Equate real parts & simplify:
(b)

56 Applications to polynomials – Trigonometric identities
Example 56
(a) Use De Moivre’s Theorem to show that cos 3θ = 4 cos3 θ − 3 cos θ.
(b) Hence solve 8x3 − 6x − 1 = 0.
π 2π 4π
(c) Deduce that cos = cos + cos .
9 9 9 π
Answer: x = cos 9
, cos 5π
9
, cos 7π
9
.

Applications to polynomials – Trigonometric identities 57
Example 57
(a) Use De Moivre’s Theorem to express tan 5θ in terms of powers of tan θ.
(b) Hence show that x4 − 10x2 + 5 = 0 has roots ± tan π5 and ± tan 2π
5
.
(c) Deduce that tan π5 tan 2π
5
tan 3π
5
tan 4π
5
= 5.
(d) By solving x4 − 10x2 + 5 = 0 via another method, find the exact valuepof tan π5 .
5 tan θ−10 tan3 θ+tan5 θ π
√
Answer: (a) tan 5θ = 1−10 tan2 θ+5 tan4 θ
(b) Show. (c) Show. (d) tan 5
= 5 − 2 5.

58 Applications to polynomials – Further exercises
4.3 Further exercises

1. (a) Factorise z 6 − 1 into the real quadratic factors.
(b) Hence factorise z 4 + z 2 + 1.
2kπ 2kπ
2. (a) Show that the roots of y 4 + y 3 + y 2 + y + 1 = 0 are y = cos + i sin ,
5 5
π 1 2π
where k = 1, 2, 3, 4 and hence show that cos = + cos . Also prove that
√ 5 2 5
π 1+ 5
cos = .
5 4
1 2kπ
(b) By letting x = y + , show that the roots of x2 + x − 1 = 0 are 2 cos , where
y 5
π 2π 1
k = 1, 2 and deduce that cos cos =
5 5 4
(c) (Method 2)
π 2π

4 3 2 2 2
Prove that y + y + y + y + 1 = y + 2y cos + 1 y − 2y cos + 1 and
5 5
π 1 2π
hence deduce that cos = + cos
5 2 5
3. Suppose that z 7 = 1, z 6= 1
1 1 1
(a) Deduce that z 3 + z 2 + z + 1 + + 2 + 3 =0
z z z
1
(b) By letting x = z + , reduce the equation in (i) to a cubic equation in x.
z
π 2π 3π 1
(c) Hence deduce that cos cos cos =
7 7 7 8
4. (a) Express cos 3θ in terms of cos θ
(b) Use the result to solve 8x3 − 6x + 1 = 0.
(c) Deduce that
2π 4π π
i. cos + cos = cos
9 9 9
π 2π 4π
ii. sec sec sec =8
9 9 9

Applications to polynomials – Further exercises 59
5. (a) Express cos 3θ and cos 2θ in terms of cos θ

(b) Show that cos 3θ = cos 2θ can be expressed as 4x3 − 2x2 − 3x + 1 = 0, where
t = tan θ
√
2π 5−1
(c) By solving equation in (ii) for x, show that cos =
5 4
6. (a) Factorise z 6 + 1 into real quadratic factors.

π π 5π
(b) Hence deduce that cos 3θ = 4 cos θ − cos cos θ − cos cos θ − cos
6 2 6
π 5π
7. ! Show that the roots of (z − 1)6 + (z + 1)6 = 0 are ±i, ±i cot , and ±i cot
12 12
Ex 3B
• Q1-4, 6-14
Other resources
• Lee (2006, Ex 2.11 (skip Q6(iii)))

• Patel (1990, Self Testing Ex 4.7 p.109)
• Arnold and Arnold (2000, Ex 2.4)

60 Applications to polynomials – Roots of complex numbers
4.4 Roots of complex numbers

4.4.1 Factorisations of higher powers
Example 58
(a) Evaluate the following partial sum:
1 + r + r2 + r3 + r4
(b) Hence factorise x5 − 1 over Z
Example 59
(a) Evaluate the following partial sum:
1 − r + r2 − r3 + r4
(b) Hence factorise x5 + 1 over Z

Applications to polynomials – Roots of complex numbers 61
© Laws/Results
Difference of powers
xn − y n = (x − y) (xn−1 + xn−2 y + · · · + y n−1 )

..............................................................
Sum of powers
xn + y n = (x + y) (xn−1 − xn−2 y + · · · + y n−1 )

..............................................................
• Signs . . . . . . .alternate
...............

4.4.2 Graphical solutions and consequent factorisations

• To find n-th roots of complex numbers, use . . De
. . . . . . . . . . . . Moivre’s
. . . . . . . . . . . . . . . . . . . . . . .Theorem
...............
and polar form.
Example 60
Find the cube roots of unity, i.e. solve z 3 = 1.
Solution (via De Moivre’s Theorem)
³ Steps
3 roots
z }| {
3
1. z = 1 (cos 2kπ + i sin 2kπ), where k = 0, 1, 2.
1 1 1 2kπ

2. Hence, z = 1 3 (cos 2kπ + i sin 2kπ) 3 = . . . . . . . . 1. .3. . cos . . . . . . . . . . by . . .De
. . . . .3. . + ....
Moivre’s Theorem
...................... ...................... .
3. Fix up “out of range” arguments (change to principal argument):
Solution (via polar form)

Example 61
[2011 HSC Q2] Find, in modulus-argument form, all solutions of z 3 = 8.
2π 2π

Answer: z = 2, 2 cos 3
± i sin 3
Example 62 √
Solve for z: z 4 = −8 − 8 3i, and plot the solutions on the Argand diagram.
π 5π π
Answer: z = 2ei 3 , 2ei , 2ei
−2π
6 3 , 2e−i 6

Example 63
1
[2016 Ext 2 Q10] Suppose that x + = −1.
x
1
What is the value of x2016 + ?
x2016 2π 4π
(A) 1 (B) 2 (C) (D)
3 3
Example 64 √
Find the fourth roots of z = 1 + i 3 in modulus-argument form.
1 1 1 1
π π 7π 7π 11π 11π 5π 5π

Answer: 2 4 cos 12
+ i sin 12
, 2 4 cos 12
+ i sin 12
, 2 4 cos 12
− i sin 12
, 2 4 cos 12
− i sin 12

Example 65
(a) Find the five fifth roots of unity and plot them on the unit circle.
(b) If ω is a non-real fifth root of unity, show that 1 + ω + ω 2 + ω 3 + ω 4 = 0.
(c) Hence or otherwise, factorise z 5 − 1 completely over R.
2π 4π
Answer: (z − 1) z 2 − 2z cos + 1 z 2 − 2z cos

5 5
+1

Example 66
Find all the zeros of P (x) = x4 + x3 + x2√+ x + 1. Hence factorise P (x) into irreducible
2π 5−1
factors over R. Deduce that cos = . Answer: x = cos 2π5 + i sin 2π5 , cos 4π5 + i sin 4π5
5 4

Example 67
π 2kπ
(a) Find the five fifth-roots of −1. Answer: ei( 5 + 5 ), where k ∈ [0, 4]
(b) If ω is a non-real fifth root of −1 with the smallest positive argument, show
that 1 − ω + ω 2 − ω 3 + ω 4 = 0.
(c) Find the exact values of cos π5 and cos 3π
5
.
π 1
√ 3π 1
√
Answer: cos 5
= 4
1+ 5 , cos 5
= 4
1− 5

Example 68
[2014 JRAHS Trial Q15] Let α be a complex root of the polynomial z 7 = 1 with
the smallest argument. Let θ = α + α2 + α4 and φ = α3 + α5 + α6 .
(i) Show that θ + φ = −1 and θφ = 2. 3
(ii) Write a quadratic equation whose roots are θ and φ. Hence show that 2
√ √
1 i 7 1 i 7
θ=− + and φ = − −
2 2 2 2
(iii) Show that 2
2π 4π π 1
cos + cos − cos = −
7 7 7 2

4.4.3 Roots of unity: reduction from higher powers

© Laws/Results
Change subject to highest power of x:
• If x2 + x + 1 = 0, then
2
. . . . . . .x. . .=
. . .−x
. . . .−. .1. . . . . . .
• If x3 + x2 + x + 1 = 0, then
3 2
. . . . . . . . . .x. . .=
. . .−x
. . . .−
. . .x. .−. .1. . . . . . . . . .
These results can be used creatively to reduce the powers down to more manageable
powers.
Example 69
[Ex 3C Q1]
(a) Find the three cube roots of unity, expressing the complex roots in both
modulus-argument form and Cartesian form.
(b) Show that the points in the complex plane representing these three roots form
an equilateral triangle.
(c) If ω is one of the complex, non-real roots, show that the other complex root is
ω2.
(d) Write down the values of: i. ω 3 ii. 1 + ω + ω 2
(e) Show that:
3
i. (1 + ω 2 ) = −1
ii. (1 − ω − ω 2 ) (1 − ω + ω 2 ) (1 + ω − ω 2 ) = 8
iii. (1 − ω) (1 − ω 2 ) (1 − ω 4 ) (1 − ω 5 ) = 9

Example 70
[2017 BHHS Ext 2 Trial Q1] If ω is an imaginary cube root of unity, then what
2 2017
is (1 + ω − ω ) equal to?
(A) −22017 ω (B) 22017 ω (C) −22017 ω 2 (D) 22017 ω 2

Example 71
[2016 JRAHS Ext 2 Trial HSC Q11] (3 marks) Simplify

1 + 2ω + 3ω 2 1 + 3ω + 2ω 2
where ω is a complex cube root of unity.

Example 72
[2010 NSBHS Ext 2 Assessment Task 1] If w is a non-real cube root of unity, i.e.
w3 = 1,
1 1
i Prove + =1 3
1 + w 1 + w2
ii Show that ( 3
1 + wn + w2n 1 n is a multiple of 3
=
3 0 otherwise

Ex 3C
• Q1-11
Other resources • Lee (2006, Ex 2.10)

• Fitzpatrick (1991, Ex 31(a))
• Patel (2004, Ex 4D, 4E, 4I) • Arnold and Arnold (2000, Ex 2.4)

Section A
Past HSC questions
V Important note
Whilst the legacy Extension 2 (‘4 Unit’) syllabus contained Complex Numbers,
there have been several content sections that have now been removed for HSC
examinations from 2020.
If in doubt, consult your teacher regarding whether a particular part is suitable to

attempt or not.
Definition 13
Locus the path traced out by a point, subject to certain conditions.
This word was used extensively in the legacy syllabuses but has now been removed.
Simply replace any instances of locus with path traced out by the complex numbers
z. . .
A.1 2001 Extension 2 HSC

Question 2
1
(a) Let z = 2 + 3i and w = 1 + i. Find zw and in the form x + iy. 2
w
√
(b) i. Express 1 +3i in modulus-argument form. 2
√ 10
ii. Hence evaluate 1 + 3i in the form x + iy. 2
(c) Sketch the region in the complex plane where the inequalities 3
π π
|z + 1 − 2i| ≤ 3 and − ≤ arg z ≤
3 4
both hold.
(d) Find all solutions of the equation z 4 = −1. 3

Give your answers in modulus-argument form.
74
Past HSC questions – 2001 Extension 2 HSC 75
(e) In the diagram the vertices of a triangle ABC are represented by the complex numbers
z1 , z2 and z3 , respectively. The triangle is isosceles and right-angled at B.
y
D b
A b
b
C
B
x
O
i. Explain why (z1 − z2 )2 = − (z3 − z2 )2 . 2

ii. Suppose D is the point such that ABCD is a square. Find the complex 1
number, expressed in terms of z1 , z2 and z3 , that represents D.
Question 3
(b) The numbers α, β and γ satisfy the equations
1 1 1
α+β+γ =3 α2 + β 2 + γ 2 = 1 + + =2
α β γ
i. Find the values of αβ + βγ + γα and αβγ 3
Explain why α, β and γ are the roots of the cubic equation
x3 − 3x2 + 4x − 2 = 0
ii. Find the values of α, β and γ. 2
Question 7
1
(a) Suppose that z = (cos θ + i sin θ) where θ is real.
2
i. Find |z|. 1
ii. Show that the imaginary part of the geometric series 3
1
1 + z + z2 + z3 + · · · =
1−z
2 sin θ
is .
5 − 4 cos θ
iii. Find an expression for 2
1 1 1
1+ cos θ + 2 cos 2θ + 3 cos 3θ + · · ·
2 2 2
in terms of cos θ.

76 Past HSC questions – 2002 Extension 2 HSC
(b) Consider the equation x3 − 3x − 1 = 0.

p
i. Let x = where p and q are integers having no common divisors other 4
q
than +1 and −1. Suppose that x is a root of ax3 − 3x + b = 0, where
a and b are integers.
Explain why p divides b and why q divides a. Deduce that x3 −3x−1 = 0

does not have a rational root.
√
ii. Suppose that r, s and
√ d are rational3 numbers and that d is irrational. 4
Assume that r + s d is a root of x − 3x − 1 = 0.
√
Show that 3r2 s + s3 d − 3s = 0 and show that r − s d must also be a
root of x3 − 3x − 1 = 0.
3
Deduce from this result and part (i),
√ that no root of x − 3x − 1 = 0
can be expressed in the form r + s d with r, s and d rational.
π
iii. Show that one root of x3 − 3x − 1 = 0 is 2 cos . 1
9
You may assume the identity cos 3θ = 4 cos3 θ − 3 cos θ.

Question 2
(a) Let z = 1 + 2i and w = 1 + i. Find, in the form x + iy,
i. zw. 1
1
ii. . 1
w
(b) On an Argand diagram, shade in the region where the inequalities 3
0 ≤ Re(z) ≤ 2 and |z − 1 + i| ≤ 2
both hold.
(c) It is given that 2 + i is a root of
P (z) = z 3 + rz 2 + sz + 20
where r and s are real numbers.

i. State why 2 − i is also a root of P (z). 1
ii. Factorise P (z) over the real numbers. 2
(d) Prove by induction that, for all integers n ≥ 1, 3
(cos θ − i sin θ)n = cos (nθ) − i sin (nθ)

(e) Let z = 2 (cos θ + i sin θ).

i. Find 1 − z. 1
1 1 − 2 cos θ
ii. Show that the real part of is 2
1−z 5 − 4 cos θ
1
iii. Express the imaginary part of in terms of θ. 1
1−z
Question 3
(a) The equation 4x3 − 27x + k = 0 has a double root. Find the possible values 2
of k.

Question 2
(a) Let z = 2 + i and w = 1 − i. Find, in the form x + iy,
i. zw. 1
4
ii. . 1
z
(b) Let α = −1 + i.
i. Express α in modulus-argument form. 2
ii. Show that α is a root of the equation z 4 + 4 = 0. 1
iii. Hence, or otherwise, find a real quadratic factor of the polynomial z 4 +4. 2
π
|z − 1 − i| < 2 and 0 < arg(z − 1 − i) <
4
hold simultaneously.
(d) By applying De Moivre’s theorem and by also expanding (cos θ + i sin θ)5 , 3
express cos 5θ as a polynomial in cos θ.
(e) ! Suppose that the complex number z lies on the unit circle, and 2
0 ≤ arg(z) ≤ π2 .
Prove that 2 arg(z + 1) = arg(z).


Question 2
(a) Let z = 1 + 2i and w = 3 − i. Find, in the form x + iy,
i. zw. 1

10
ii. . 1
z
√
(b) Let α = 1 + i 3 and β = 1 + i.
α
i. Find in the form x + iy. 1
β
ii. Express α in modulus-argument form. 2
iii. Given that β has the modulus-argument form 1
√ π π
β = 2 cos + i sin
4 4
α
find the modulus-argument form of .
β
π
iv. Hence find the exact value of sin . 1
12
|z + z| ≤ 1 and |z − i| ≤ 1
(d) The diagram shows two distinct points A and B that represent the complex
numbers z and w respectively. The points A and B lie on the circle of radius
r centred at O. The point C representing the complex number z + w also lies
on this circle.
C
b
B b
b A
b b
i. Using the fact that C lies on the circle, show geometrically that 2
∠AOB = 2π 3
.
ii. Hence show that z 3 = w3 . 2
iii. Show that z 2 + w2 + zw = 0. 1

Question 4
(a) Let α, β and γ be the zeros of the polynomial p(x) = 3x3 + 7x2 + 11x + 51.
i. Find α2 βγ + αβ 2 γ + αβγ 2 . 1
ii. Find α2 + β 2 + γ 2 . 2
iii. Using part (ii), or otherwise, determine how many of the zeros of p(x) 1
are real. Justify your answer.
Question 7
(b) Let α be a real number and suppose that z is a complex number such that
1
z+ = 2 cos α
z
i. By reducing the above equation to a quadratic equation in z, solve for 3
z and use De Moivre’s theorem to show that
1
zn + = 2 cos nα
zn
1
ii. Let w = z + . Prove that 2
z

3 2 1 2 1 3 1
w + w − 2w − 2 = z + + z + 2 + z + 3
z z z
iii. Hence, or otherwise, find all solutions of 3
cos α + cos 2α + cos 3α = 0
in the range 0 ≤ α ≤ 2π.

Question 2
(a) Let z = 3 + i and w = 1 − i. Find, in the form x + iy,
i. 2z + iw. 1
ii. zw. 1
6
iii. . 1
w
√
(b) Let β = 1 − i 3.
i. Express β in modulus-argument form. 2
ii. Express β 5 in modulus-argument form. 2
iii. Hence express β 5 in the form x + iy. 1

|z − z| < 2 and |z − 1| ≥ 1
(d) Let ℓ be the line in the complex plane that passes through the origin and
makes an angle α with the positive real axis, where 0 < α < π2 .
ℓ
b
Q
b
P
α
b
The point P represents the complex number z1 , where 0 < arg(z1 ) < α. The
point P is reflected in the line ℓ to produce the point Q, which represents
the complex number z2 . Hence |z1 | = |z2 |.
i. Explain why arg(z1 ) + arg(z2 ) = 2α. 2
ii. Deduce that z1 z2 = |z1 |2 (cos 2α + i sin 2α). 1
π
iii. Let α = and let R be the point that represents the complex number 1
zz. 4
1 2
Describe the locus of R as z1 varies.

Question 4
(b) Suppose α, β, γ and δ are the four roots of the polynomial equation
x4 + px3 + qx2 + rx + s = 0
i. Find the values of α + β + γ + δ and αβγ + αβδ + αγδ + βγδ 2
ii. Show that α2 + β 2 + γ 2 + δ 2 = p2 − 2q. 2
iii. Apply the result in part (ii) to show that x4 − 3x3 + 5x2 + 7x − 8 = 0 1
cannot have four real roots.
iv. By evaluating the polynomial at x = 0 and x = 1, deduce that the 2
polynomial equation x4 − 3x3 + 5x2 + 7x − 8 = 0 has exactly two real
roots.

Question 6
(b) Let n be an integer greater than 2. Suppose ω is an n-th root of unity and
ω 6= 1.
i. By expanding the left, show that 2

1 + 2ω + 3ω 2 + 4ω 3 + · · · + nω n−1 (ω − 1) = n
1 z −1
ii. Using the identity = , or otherwise, prove that 1
z2 − 1 z − z −1
1 cos θ − i sin θ
=
cos 2θ + i sin 2θ − 1 2i sin θ
provided that sin θ 6= 0.
2π 2π 1
iii. Hence, if ω = cos + i sin , find the real part of . 1
n n ω−1
2π 4π 6π 8π 5
iv. Deduce that 1 + 2 cos + 3 cos + 4 cos + 5 cos =− . 1
5 5 5 5 2
v. By expressing the left hand side of the equation in part (iv) in terms of 3
π 2π π
cos and cos , find the exact value, in surd form, of cos .
5 5 5

Question 2
(a) Let z = 3 + i and w = 2 − 5i. Find, in the form x + iy,
i. z 2 . 1
ii. zw. 1
w
iii. . 1
z
√
(b) i. Express 3 − i in modulus-argument form. 2
√ 7
ii. Express 3 − i in modulus-argument form. 2
√ 7
iii. Hence express 3 − i in the form x + iy. 1
(c) Find, in modulus-argument form, all solutions of z 3 = −1. 2
Question 3
(c) Two of the zeros of P (x) = x4 − 12x3 + 59x2 − 138x + 130 are a + ib and
a + 2ib, where a and b are real and b > 0.
i. Find the values of a and b. 3
ii. Hence, or otherwise, express P (x) as the product of quadratic factors 1
with real coefficients.

Question 4
(a) The polynomial p(x) = ax3 + bx + c has a multiple zero at 1 and has a 3
remainder 4 when divided by x + 1. Find a, b, c.

Question 2
(a) Let z = 4 + i and w = z. Find, in the form x + iy,
i. w. 1
ii. w − z. 1
z
iii. . 1
w
(b) i. Write 1 + i in the form r (cos θ + i sin θ). 2

ii. Hence, or otherwise, find (1 + i)17 in the form a + ib, where a and b are 3
integers.
(c) The point P on the Argand diagram represents the complex number z, where 3
z satisfies
1 1
+ =1
z z
Give a geometrical description of the locus of P as z varies.
(d) The points P , Q and R on the Argand diagram represent the complex
numbers z1 , z2 and a respectively.
The triangles OP R and OQR are equilateral with unit sides, so

|z1 | = |z2 | = |a| = 1.
Let ω = cos π3 + i sin π3 .

Im
Q(z2 )
R(a)
Re
P (z1 )
i. Explain why z2 = ωa. 1
ii. Show that z1 z2 = a2 . 1
iii. Show that z1 and z2 are the roots of z 2 − az + a2 = 0. 2

Question 4
(d) The polynomial P (x) = x3 + qx2 + rx + s has real coefficients. It has three
distinct zeros, α, −α and β.
i. Prove that qr = s. 3
ii. The polynomial does not have three real zeros. Show that two of the 2
zeros are purely imaginary. (A number is purely imaginary if it is of
the form iy, with y real and y 6= 0.)
Question 5
(d) In the diagram, ABCDE is a regular pentagon with sides of length 1. The
perpendicular to AC through B meets AC at P .
A
1 π
5
B E
P
C D
Copy or trace this diagram into your writing booklet.

i. Let u = cos π5 . 2
Use the cosine rule in △ACD to show that 8u3 − 8u2 + 1 = 0.

ii. One root of 8x3 − 8x2 + 1 = 0 is 21 . 2
Find the other roots of 8x3 − 8x2 + 1 = 0 and hence find the exact value
of cos π5 .
Question 8
(b) i. Let n be a positive integer. Show that if z 2 6= 1, then 2
n
2 4 2n−2 z − z −n
1 + z + z + ··· + z = z n−1
z − z −1
ii. By substituting z = cos θ + i sin θ, where sin θ 6= 0 in to part (i), show 3
that
1 + cos 2θ + · · · + cos(2n − 2)θ + i [sin 2θ + · · · + sin(2n − 2)θ]

sin nθ
= [cos(n − 1)θ + i sin(n − 1)θ]
sin θ
π
iii. Suppose θ = . Using part (ii), show that 3
2n
π 2π (n − 1)π π
sin + sin + · · · + sin = cot
n n n 2n


Question 2
(a) Find real numbers a and b such that (1 + 2i)(1 − 3i)= a + ib. 2
√
1+i 3
(b) i. Write in the form x + iy, where x and y are real. 2
1+i
√
ii. By expression
√ both 1 + i 3 and 1 + i in modulus-argument form, write 3
1+i 3
in modulus-argument form.
1+i
π
iii. Hence find cos in surd form. 1
12
√ !12
1+i 3
iv. By using the result of part (ii), or otherwise, calculate . 1
1+i
(c) The point P on the Argand diagram represents the complex number 3
z = x + iy, which satisfies
z2 + z2 = 8
Find the equation of the locus of P in terms of x and y. What type of curve
is the locus?
(d) The point P on the Argand diagram represents the complex number z.
The points Q and R represent the points ωz and ωz respectively, where
ω = cos 2π
3
+ i sin 2π
3
. The point M is the midpoint of QR.
Im
Re
O
b
S
i. Find the complex number representing M in terms of z. 2
ii. The point S is chosen so that P QSR is a parallelogram. 2
Find the complex number represented by S.

Question 3
(b) Let p(z) = 1 + z 2 + z 4 .
i. Show that p(z) has no real zeros. 1
ii. Let α be a zero of p(z).
(α) Show that α6 = 1. 1
(β) Show that α2 is also a zero of p(z). 1
Question 5
(b) Let p(x) = xn+1 − (n + 1)x + n, where n in a positive integer.
i. Show that p(x) has a double zero at x = 1. 2
ii. By considering concavity, or otherwise, show that p(x) ≥ 0 for x ≥ 0. 1
iii. Factorise p(x) when n = 3. 2
Question 6
(a) Let ω be the complex number satisfying ω 3 = 1 and Im(ω) > 0. The cubic 3
polynomial, p(z) = z 3 + az 2 + bz + c, has zeros 1, −ω and −ω.
Find p(z).

Question 2
(a) Write i9 in the form a + ib where a and b are real. 1
−2 + 3i
(b) Write in the form a + ib where a and b are real. 1
2+i

(c) The points P and Q on the Argand diagram represent the complex numbers
z and w respectively.
Im
b
P (z)
b
Q(w)
Re
Copy the diagram into your writing booklet, and mark on it the following
points:
i. the point R representing iz 1
ii. the point S representing w 1
iii. The point T representing z + w. 1
(d) Sketch the region in the complex plane where the inequalities |z − 1| ≤ 2 and 2
− π4 ≤ arg(z − 1) ≤ π4 hold simultaneously.
(e) i. Find all the 5th roots of −1 in modulus-argument form. 2

ii. Sketch the 5th roots of −1 on an Argand diagram. 1
(f) i. Find the square roots of 3 + 4i. 3

ii. Hence, or otherwise, solve the equation 2
z 2 + iz − 1 − i = 0
Question 3
(c) Let P (x) = x3 + ax2 + bx + 5, where a and b are real numbers. 3
Find the values of a and b given that (x − 1)2 is a factor of P (x).

Question 6
(b) Let P (x) = x3 + qx2 + qx + 1 where q ∈ R. One zero of P (x) is −1.
1
i. Show that if α is a zero of P (x) then is a zero of P (x). 1
α
ii. Suppose that α is a zero of P (x) and α is not real.
(α) Show that |α| = 1. 2
1−q
(β) Show that Re(α) = . 2
2


Question 2
(a) Let z = 5 − i.
i. Find z 2 in the form x + iy. 1
ii. Find z + 2z in the form x + iy. 1
i
iii. Find in the form x + iy. 2
z
√
(b) i. Express − 3 − i in modulus-argument form. 2
√ 6
ii. Show that − 3 − i is a real number. 2
(c) Sketch the region in the complex plane where the inequalities 1 ≤ |z| ≤ 2 2
and 0 ≤ z + z ≤ 3 hold simultaneously.
π
(d) Let z = cos θ + i sin θ where 0 < θ < .
2
On the Argand diagram the point A represents z, the point B represents z 2
and the point C represents z + z 2 .
Im b
C
b
B
b A
θ
b
Re
O
Copy or trace the diagram into your writing booklet.

i. Explain why the parallelogram OACB is a rhombus. 1
3θ
ii. Show that arg (z + z 2 ) = 1
2
θ
iii. Show that |z + z 2 | = 2 cos . 2
2
iv. By considering the real part of z + z 2 , or otherwise, deduce that 1
θ 3θ
cos θ + cos 2θ = 2 cos cos
2 2

Question 6
(c) i. Expand (cos θ + i sin θ)5 using the binomial theorem. 1
ii. Expand (cos θ + i sin θ)5 using De Moivre’s Theorem, and hence show 3
that
sin 5θ = 16 sin5 θ − 20 sin3 θ + 5 sin θ
π
iii. Deduce that x = sin is one of the solutions to 1
10
16x5 − 20x3 + 5x − 1 = 0
iv. Find the polynomial p(x) such that 1
(x − 1)p(x) = 16x5 − 20x3 + 5x − 1

2
v. Find the value of a such that p(x) = (4x2 + ax − 1) . 1
π
vi. Hence find an exact value for sin . 1
10
Question 7
(c) Let P (x) = (n − 1)xn − nxn−1 + 1 where n is an odd integer, n ≥ 3.
i. Show that P (x) has exactly two stationary points. 1
ii. Show that P (x) has a double zero at x = 1. 1
iii. Use the graph y = P (x) to explain why P (x) has exactly one real zero 2
other than 1.
iv. Let α be the real zero of P (x) other than 1. 2
1
Using part ?? or otherwise, show that −1 < α ≤ − .
2
v. Deduce that each of the zeros of 4x5 − 5x4 + 1 has modulus less than 2
or equal to 1.

See Examples 21 on page 22, 18 on page 19 and 61 on page 63.
Question 2
(d) i. Use the binomial theorem to expand (cos θ + i sin θ)3 . 1
ii. Use De Moivre’s theorem and your result from part (i) to prove that 3
1 3
cos3 θ = cos 3θ + cos θ
4 4
iii. Hence, or otherwise, find the smallest positive solution of 2
4 cos3 θ − 3 cos θ = 1


1. Let z = 5 − i and w = 2 + 3i. 1
What is the value of 2z + w?
(A) 12 + i (B) 12 + 2i (C) 12 − 4i (D) 12 − 5i
2. The complex number z is shown on the Argand diagram below. 1

y
b
z
x
O
Which of the following best represents iz?

(A) (C)
y y
b
iz
x x
O O
iz
(B) (D)
y y
iz b
x x
O O
iz

5. The equation 2x3 − 3x2 − 5x − 1 = 0 has roots α, β and γ. 1

1
What is the value of ?
α3 β 3 γ 3
1
(A) 8
(B) − 18 (C) 8 (D) −8
8. The following diagram shows the graph y = P ′ (x), the derivative of a 1

polynomial P (x).
y
x
− 45 1
Which of the following expressions could be P (x)?

(A) (x − 2)(x − 1)3 (C) (x − 2)(x + 1)3
(B) (x + 2)(x − 1)3 (D) (x + 2)(x + 1)3
Question 11
√
2 5+i
(a) Express √ in the form x + iy, where x and y are real. 2
5−1
(b) Sketch the region in the complex plane where the inequalities 2
|z + 2| ≥ 2 and |z − i| ≤ 1
both hold.
√
(d) i. Write z = 3 − i in modulus-argument form. 2
ii. Hence express z 9 in the form x + iy, where x and y are real. 1

Question 12
(d) On the Argand diagram the points A1 and A2 correspond to the distinct
complex numbers u1 and u2 respectively. Let P be a point corresponding to
a third complex number z.
Points B1 and B2 are positioned so that △A1 P B1 and △A2 B2 P , labelled in

an anti-clockwise direction, are right-angled and isosceles with right angles
at A1 and A2 respectively. The complex numbers w1 and w2 correspond to
B1 and B2 respectively.
y
B1 (w1 )
b
B2 (w2 )
b
P (z)
b
A1 (u1 )
b
A2 (u2 )
i. Explain why w1 = u1 + i(z − u1 ). 1

ii. Find the locus of the midpoint of B1 B2 as P varies. 2
Question 15
(b) Let P (z) = z 4 − 2kz 3 + 2k 2 z 2 − 2kz + 1, where k ∈ R.
Let α = x + iy, where x, y ∈ R.
Suppose that α and iα are roots of P (z), where α 6= iα.

i. Explain why α and −iα are zeros of P (z). 1
ii. Show that P (z) = z 2 (z − k)2 + (kz − 1)2 . 1
iii. Hence show that if P (z) has a real zero then 2

P (z) = z 2 + 1 (z + 1)2 or P (z) = z 2 + 1 (z − 1)2
iv. Show that all zeros of P (z) have modulus 1. 2
v. Show that k = x − y. 1
√ √
vi. Hence show that − 2 ≤ k ≤ 2. 2


π π
5. Which region on the Argand diagram is defined by ≤ |z − 1| ≤ ?
4 3
(A) (C)
y y
π
3 π
π π 3
bc 4 x bc 4 x
1 1
(B) (D)
y y
1− π
4 1− π
4
b b b b
x b b b b
x
1− π
1+ π
1+ π
1− π
1+ π
1+ π
3 1 4 3 3 1 4 3
Question 11
√ √
(a) Let z = 2 − i 3 and w = 1 = i 3.
i. Find z + w. 1
ii. Express w in modulus-argument form. 2
iii. Write w24 in its simplest form.
(c) Factorise z 2 + 4iz + 5. 2
(e) Sketch the region on the Argand diagram defined by z 2 + z 2 ≤ 8. 2
Question 14
(b) ! Let z2 = 1 + i and, for n > 2, let 3

i
zn = zn−1 1 +
|zn−1 |
√
Use mathematical induction to prove that |zn | = n for all integers n ≥ 2.

Question 15
(a) The Argand diagram shows complex numbers w and z with arguments φ and 3
θ respectively, where φ < θ. The area of the triangle formed by O, w and z
is A.
Im
b
z
b
w
θ
φ
Re
Show that zw − wz = 4iA

2. The polynomial P (z) has real coefficients, and z = 2 − i is a root of P (z). 1
Which quadratic polynomial must be a factor of P (z).
(A) z 2 − 4z + 5 (B) z 2 + 4z + 5 (C) z 2 − 4z + 3 (D) z 2 + 4z + 3

π π
4. Given z = 2 cos + i sin , which expression is equal to (z)−1 ? 1
3 3
1 π π 1 π π
(A) cos − i sin (C) cos + i sin
2 3 3 2 3 3
π π π π
(B) 2 cos − i sin (D) 2 cos + i sin
3 3 3 3
8. The Argand diagram shows the complex numbers w, z and u, where w lies in 1
the first quadrant, z lies in the second quadrant and u lies on the negative real
axis.
Im
z w
b b
b
Re
u O
Which statement could be true?
(A) u = zw and u = z + w (C) z = uw and u = z + w
(B) u = zw and u = z − w (D) z = uw and u = z − w

Question 11
(a) Consider the complex numbers z = −2 − 2i and w = 3 + i.
i. Express z + w in modulus-argument form. 2
z
ii. Express in the form x + iy, where x and y are real numbers. 2
w
(c) Sketch the region in the Argand diagram where |z| ≤ |z − 2| and 3
π π
− ≤ arg z ≤ .
4 4
Question 12
(b) It can be shown that 4 cos3 θ − 3 cos θ = cos 3θ. (Do NOT prove this.)
√
Assume that x = 2 cos θ is a solution of x3 − 3x = 3.
√
3
i. Show that cos 3θ = . 1
2
√
ii. Hence, or otherwise, find the three real solutions of x3 − 3x = 3. 1
Question 14
(a) Let P (x) = x5 − 10x2 + 15x − 6.
i. Show that x = 1 is a root of P (x) of multiplicity three. 2
ii. Hence, or otherwise, find the two complex roots of P (x). 2

2. What value of z satisfies z 2 = 7 − 24i? 1
(A) 4 − 3i (C) 3 − 4i
(B) −4 − 3i (D) −3 − 4i

5. Given that z = 1 − i, which expression is equal to z 3 ? 1

√ 3π 3π
(A) z = 2 cos − + i sin −
4 4

√ 3π 3π
(B) z = 2 2 cos − + i sin −
4 4

√ 3π 3π
(C) z = 2 cos + i sin
4 4

√ 3π 3π
(D) z = 2 2 cos + i sin
4 4
9. The complex number z satisfies |z − 1| = 1. 1
What is the greatest distance that z can be from the point i on the Argand
diagram?
√ √ √
(A) 1 (B) 5 (C) 2 2 (D) 2 + 1
Question 11
4 + 3i
(a) Express in the form x + iy, where x and y are real. 2
2−i
√ π π
(b) Consider the complex numbers z = − 3 + i and w = 3 cos + i sin .
7 7
i. Evaluate |z|. 1
ii. Evaluate arg(z). 1
z
iii. Find the argument of . 1
w
Question 12
π
(a) The complex number z is such that |z| = 2 and arg(z) = .
4
Plot each of the following complex numbers on the same half-page Argand
diagram.
i. z. 1
ii. u = z 2 . 1
iii. v = z 2 − z. 1
(b) The polynomial P (x) = x4 − 4x3 + 11x2 − 14x + 10 has roots a + ib and a + 2ib
where a and b are real and b 6= 0.
i. By evaluating a and b, find all the roots of P (x). 3
ii. Hence or otherwise, find one quadratic polynomial with real coefficients 1
that is a factor of P (x).


4. The Argand diagram shows the complex numbers z and w, where z lies in the 1
first quadrant and w lies in the second quadrant.
Im
Which complex number could lie in the
b
z 3rd quadrant?
(A) −w
w b
Re (B) 2iz
O
(C) z
(D) w − z
1−i
5. Multiplying a non-zero complex number by results in a rotation about 1
1+i
the origin on an Argand diagram.
What is the rotation?

π π
(A) Clockwise by (C) Anticlockwise by
4 4
π π
(B) Clockwise by (D) Anticlockwise by
2 2
Question 11
√
(a) Let z = 3 − i.
i. Express z in modulus-argument form. 2
ii. Show that z 6 is real. 2
iii. Find a positive integer n such that z n is purely imaginary. 1
Question 12
(a) Let z = cos θ + i sin θ.
i. By considering the real part of z 4 , show that cos 4θ is 2
cos4 θ − 6 cos2 θ sin2 θ + sin4 θ

ii. Hence, or otherwise, find an expression for cos 4θ involving only powers 1
of cos θ.
Question 13
(d) Suppose p(x) = ax3 + bx2 + cx + d with a, b, c and d ∈ R, a 6= 0.
i. Deduce that if b2 − 3ac < 0 then p(x) cuts the x axis only once. 2
b

ii. If b2 − 3ac = 0 and p − 3a = 0, what is the multiplicity of the root 2
b
x = − 3a ?

Question 16
(a) i. The complex numbers z = cos θ + i sin θ and w = cos α + i sin α, where 3
−π < θ ≤ π and −π < α ≤ π satisfy
1+z+w =0
By considering the real and imaginary parts of 1 + z + w, or otherwise,

show that 1, z and w form the vertices of an equilateral triangle in the
Argand diagram
ii. Hence, or otherwise, show that if the three non-zero complex numbers 2
2i, z1 and z2 satisfy
|2i| = |z1 | = |z2 | AND 2i + z1 + z2 = 0
then they form the vertices of an equilateral triangle in the Argand

diagram.
(b) i. The complex numbers 0, u and v form the vertices of an equilateral 2

triangle in the Argand diagram.
Show that u2 + v 2 = uv
ii. Give an example of non-zero complex numbers u and v, so that 0, u and 1
v form the vertices of an equilateral triangle in the Argand diagram.

1. The complex number z is chosen so that 1, z, . . ., z 7 form vertices of the regular 1
polygon as shown.
z2
z3 z
z4
1
z5 z7
z6
Which polynomial equation has all of these complex numbers as roots?
(A) x7 − 1 = 0 (C) x8 − 1 = 0
(B) x7 + 1 = 0 (D) x8 + 1 = 0

3. Which complex number lies in the region 2 < |z − 1| < 3? 1

√
(A) 1 + 3i (B) 1 + 3i (C) 2 + i (D) 3 − i
6. It is given that z = 2 + i is a root of z 3 + az 2 − 7x + 15 = 0, where a ∈ R. 1
What is the value of a?
(A) −1 (B) 1 (C) 7 (D) −7

Question 11
√
(a) Let z = 1 − 3i and w = 1 + 1.
i. Find the exact value of the argument of z. 1
z
ii. Find the exact value of the argument of . 2
w
(c) Sketch the region in the Argand diagram where 2
π
− ≤ arg z ≤ 0 and |z − 1 + i| ≤ 1
4
Question 12
(b) Solve the quadratic equation z 2 + (2 + 3i)z + (1 + 3i) = 0, giving your answers 3
in the form a + bi, where a and b are real numbers.
Question 13
(e) The points A, B, C and D on the Argand diagram represents the complex 2
numbers a, b, c and d respectively. The points form a square as shown on
the diagram.
B
D
By using vectors, or otherwise, show that c = (1 + i)d − ia.

Question 16
(a) Let α = cos θ + i sin θ, where 0 < θ < 2π.
i. Show that αk + α−k = 2 cos kθ, for any integer k. 1
Let C = α−n + · · · + α−1 + 1 + α + · · · + αn , where n is a positive integer.
ii. By summing the series, prove that 3

αn + α−n − αn+1 + α −(n+1)
C=
(1 − α) (1 − α)
iii. Deduce, from parts (i) and (ii), that 2
cos nθ − cos(n + 1)θ
1 + 2 (cos θ + cos 2θ + · · · + cos nθ) =
1 − cos θ
π 2π nπ
iv. Show that cos + cos + · · · + cos is independent of n. 1
n n n

6. Which complex number is a 6th root of i? 1
1 1 √ √
(A) − √ + √ i (C) − 2 + 2i
2 2
1 1 √ √
(B) − √ − √ i (D) − 2 − 2i
2 2
7. Which diagram best represent the solutions to the equation 1
arg(z) = arg(z + 1 − i)
bc
(A) bc (C) bc
bc
bc
(B) bc (D) bc
bc

Question 11
(a) Let z = 2 + 3i and w = 1 − i.
i. Find zw. 1
2
ii. Express z − in the form x + iy, where x and y are real numbers. 2
w
(d) The points A, B and C on the Argand diagram represent the complex
numbers u, v and w respectively.
The points O, A, B and C form a square as shown on the diagram.
B(v)
b
C(w) b
b
A(u)
O
It is given that u = 5 + 2i.
i. Find w. 1
ii. Find v. 1
w
iii. Find arg . 1
v
Question 13
(b) Let z = 1 − cos 2θ + i sin 2θ, where 0 < θ ≤ π.
i. Show that |z| = 2 sin θ. 2
π
ii. Show that arg(z) = − θ. 2
2
Question 15
(b) i. Use De Moivre’s theorem and the expansion of (cos θ + i sin θ)8 to show 2
that

8 7 8 5 3 8 3 5 8
sin 8θ = cos θ sin θ− cos θ sin θ+ cos θ sin θ− cos θ sin7 θ
1 3 5 7
ii. Hence, show that 3
sin 8θ
= 4 1 − 10 sin2 θ + 24 sin4 θ − 16 sin6 θ
sin 2θ


1. What is the value of (3 − 2i)2 ? 1
(A) 5 − 12i (C) 13 − 12i
(B) 5 + 12i (D) 13 + 12i

8. Let z be a complex number such that z 2 = −iz. 1
Which of the following is a possible value for z?

√ √
1 3 3 1
(A) − i (C) − i
2 2 2 2
√ √
1 3 3 1
(B) + i (D) + i
2 2 2 2
Question 11
(a) Let z = 1 + 3i and w = 2 − i.
i. Find z + w. 1
z
ii. Express in the form x + iy, where x and y are real numbers. 2
w
√
(e) Let z = −1 + i 3.
i. Write z in modulus-argument form. 2
ii. Find z 3 , giving your answer in the form x + iy, where x and y are real 2
numbers.
Question 12
π π
(a) Sketch the region defined by ≤ arg(z) ≤ and Im(z) ≤ 1. 2
4 2
Question 16
(b) Let P (z) = z 4 − 2kz 3 + 2k 2 z 2 + mz + 1, where k and m are real numbers.
The roots of P (z) are α, α, β, β.
It is given that |α| = 1 and |β| = 1.

2 2
i. Show that Re(α) + Re(β) = 1. 2

ii. The diagram shows the position of α. 2
b
α
O 1
Copy or trace the diagram into your writing booklet.
On the diagram, accurately show all possible positions of β.

2. Given that z = 3 + i is a root of z 2 + pz + q = 0, where p and q are real, what 1
are the values of p and q?
√ √
(A) p = −6, q = 10 (C) p = 6, q = 10
(B) p = −6, q = 10 (D) p = 6, q = 10

4. The diagram shows the complex number z on the Argand diagram. 1
b z
z2
Which of the following diagrams best shows the position of ?
|z|
(A) (C)
O O
(B) (D)
O O
b

9. What is the maximum value of eiθ − 2 + eiθ + 2 for 0 ≤ θ ≤ 2π? 1
√ √
(A) 5 (B) 4 (C) 2 5 (D) 10
Question 11
(a) Consider the complex numbers w = −1 + 4i and z = 2 − i
i. Evaluate |w|. 1

ii. Evaluate wz. 2
(e) Solve z 2 + 3z + (3 − i) = 0, giving your answer(s) in the form a + bi, where 4

a and b are real.
Question 13
(d) i. Show that for any integer n, einθ + e−inθ = 2 cos (nθ). 1
4
ii. By expanding einθ + e−inθ , show that 3
1
cos4 θ =(cos (4θ) + 4 cos (2θ) + 3)
8
iii. Note: For later, after Topic 27 Further Integration 2
Z π
2
Hence, or otherwise, find cos4 θ dθ.
0
Question 14
iπ
(a) Let z1 be a complex number and z2 = e 3 z1 .
The diagram shows points A and B which represent z1 and z2 respectively,

in the Argand plae.
Im
b
B (z2 )
b
A (z1 )
b
Re
O
i. Explain why triangle OAB is an equilateral triangle. 2

ii. Prove that z1 2 + z2 2 = z1 z2 . 3

NESA Reference Sheet – calculus based courses
NSW Education Standards Authority
2020 HIGHER SCHOOL CERTIFICATE EXAMINATION
Mathematics Advanced
Mathematics Extension 1
Mathematics Extension 2
–1–
–2–
–3–
–4– © 2018 NSW Education Standards Authority
References
Arnold, D., & Arnold, G. (2000). Cambridge Mathematics 4 Unit (2nd ed.). Cambridge University
Press.
Fitzpatrick, J. B. (1991). New Senior Mathematics – Four Unit Course for Years 12. Rigby
Heinemann.
Lee, T. (2006). Advanced Mathematics: A complete HSC Mathematics Extension 2 Course (2nd
ed.). Terry Lee Enterprise.
Patel, S. K. (1990). Excel 4 Unit Maths. Pascal Press.
Patel, S. K. (2004). Maths Extension 2 (2nd ed.). Pascal Press.
Pender, W., Sadler, D., Ward, D., Dorofaeff, B., & Shea, J. (2019). CambridgeMATHS Stage 6
Mathematics Extension 1 Year 12 (1st ed.). Cambridge Education.
Sadler, D., & Ward, D. (2019). CambridgeMATHS Stage 6 Mathematics Extension 2 (1st ed.).
Cambridge Education.
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MATHEMATICS EXTENSION 2

 Topic summary and exercises:

Symbols used Syllabus outcomes addressed

L Literacy: note new word/phrase.

S Syllabus specified content

U Facts/formulae to understand, as opposed to blatant

N the set of natural numbers

Z the set of integers

Q the set of rational numbers

R the set of real numbers

C the set of complex numbers

1 A new number system 5

2 Further arithmetic & algebra of complex numbers 16

3 Curves and regions in the complex plane 40

A Past HSC questions 74

NORMANHURST BOYS’ HIGH SCHOOL

A new number system

1.1 Review of number systems

NORMANHURST BOYS’ HIGH SCHOOL

1.3 The “imaginary” numbers

• “Jump off” the real number line.

NORMANHURST BOYS’ HIGH SCHOOL

• Treat real and imaginary parts as . . . . . . . . . .components

NORMANHURST BOYS’ HIGH SCHOOL

1.4 Basic operations with complex numbers

(a) z1 + z2 (b) z1 − z2 (c) 3z1 (d) 3iz1 (e) z1 z2

NORMANHURST BOYS’ HIGH SCHOOL

1.4.3 Complex conjugate pairs

(a) z1 + z2 (c) −z2 (e) z1 − z2

NORMANHURST BOYS’ HIGH SCHOOL

NORMANHURST BOYS’ HIGH SCHOOL

Gerolamo Cardano (1501-1576), Mathematician (gambler and

• Lee (2006, Ex 2.3)

NORMANHURST BOYS’ HIGH SCHOOL

1.5 Properties of complex numbers

1.5.2 Solutions to equations

NORMANHURST BOYS’ HIGH SCHOOL

NORMANHURST BOYS’ HIGH SCHOOL

L Fill in the spaces

• Fitzpatrick (1991, Ex 31(a), (b), (c))

NORMANHURST BOYS’ HIGH SCHOOL

Further arithmetic & algebra of complex

2.1 Vector representation

• The starting point ( . . .tail

• Parallelogram of vectors when adding two vectors.

• Alternatively, “what vector get you to z1 from z2 ?”

NORMANHURST BOYS’ HIGH SCHOOL

• Quick & easy parallelogram of vectors:

2.1.4 Scalar multiplication

NORMANHURST BOYS’ HIGH SCHOOL

• All questions • Q1-21

NORMANHURST BOYS’ HIGH SCHOOL

2.2 Modulus/argument of a complex number

• Natural ordering does not exist with complex numbers.

NORMANHURST BOYS’ HIGH SCHOOL

2.2.3 (Principal) Argument

(where z = x + iy) is the . . . .angle

lies within the domain −π < Arg(z) ≤ π.

NORMANHURST BOYS’ HIGH SCHOOL

NORMANHURST BOYS’ HIGH SCHOOL

Topic summary and exercises: