Expanding the number
system: complex numbers
3
So
far
you
numbers.
have
In
Concepts
only
this
worked
chapter,
with
you
will
real
begin
■
Systems
■
Patterns
to
Microconcepts
work
with
number
of
real
complex
that
rst
a
stage,
since
real
but
When
discovered
numbers
nd
complies
numbers,
dimension.
they
areas,
complex
numbers.
the
complex
also
theorems
we
of
about
in
of
into
their
will
could
real
not
this
of
mathematics:
At
to
a
was
nd
the
■
Quadratic equations and inequalities
■
Discriminant
■
omplex numbers
■
Modulus of a complex number
■
Operations with complex numbers
■
Powers and roots of complex numbers
■
Polynomial functions and their graphs
■
Operations on polynomials
■
Linear combination of two polynomials
■
Factor and remainder theorem
■
The fundamental theorem of algebra
■
Polynomial equations
■
Sum and product of the roots of polynomial
later
a
in
many
changed
that
you
to
many
solution
actually
most
Quadratic function and graph
not
use
chapter
the
were
■
imaginary
discover
numbers
In
one
name
of
another
could
them.
put
type
properties
numbers
called
were
You
solutions.
learn
theorem
set
and
for
new
the
mathematicians
practical
within
were
a
all
complex
numbers
that
with
expands
application
these
equations
numbers:
to
have
will
important
fundamental
algebra.
equations
■
Polynomial inequalities
■
Simultaneous equations
Are real numbers the only
numbers that "exist"?
How do physicists at the Large Hadron Collider
at CERN model the smashing of par ticles?
146
/
Your
plan
city
of
park.
the
council
a
roller
There
height
The
using
kinds
modelling
Is
it
If
it
is
like
new
to
prepare
a
amusement
regarding
model
kind
given
roller
to
a
to
coaster.
functions
the
you
constraints
What
have,
of
possible
using
would
to
for
roller
functions.
impossible
What
some
the
designer
asked
coaster
are
of
has
of
this
are
the
one
function
only?
not
possible,
what
track
features
are
restriction?
suitable
coaster
model
the
and
for
why?
entire
track
conditions
by
must
Developing inquiry
the
piecewise
functions
have
at
their
joining
points?
skills
How
the
many
points
equation
of
a
are
necessary
curved
part
to
of
determine
the
Write
track?
you
down
might
example,
ask
the
computer
might
any
similar
to
you
model
path
game.
need
inq u iry
of
a
Wha t
to
qu estions
a nother
cha ra cter
d if f erent
pa th—f or
in
a
2D
qu estions
ask?
Think about the questions in this opening
problem and answer any you can. As you work
through the chapter, you will gain mathematical
knowledge and skills that will help you to answer
them all.
Before you start
Click here for help
with this skills check
You should know how to:
1
Solve
simple
quadratic
Skills check
equations,
1
eg
Solve
the
following
quadratic
equations.
2
a
x
b
7x
c
11x
=
169
2
a
x

225 
x


225

15
2
2
b
2
3x
Use
a
=
196
2
 72
GDC

to
0

x

x
graph
functions,
and
to
Given
functions


24

24
linear
solve
2

2
and
+
13
=
1102
6
quadratic
equations,
eg
2
For
the
functions
x
such
that
a
f(x)
=
2x
b
f(x)
=
4x
f(x)
=
f
and
g,
nd
the
values
of
g(x).
2
2
the
g(x)
=
-2x
that
f(x)
Graph
of
=
+
3
nd
f(x)
the
=
x
-
values
2x
of
x
+
2
and
such
-
x
-
1
and
g(x)
=
-3x
-
x
-
1
and
g(x)
=
x
2
+
3
2
+
x
g(x).
both
functions
and
nd
the
points
intersection.
Continued
on
next
page
147
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
y
10
8
g(x)
=
–2x
+
3
6
2
(–1, 5)
f(x)
=
–2x
x
+
2
2
(1, 1)
x
0
–2
–4
x
3
=
-1
Use
a
or
x
=
GDC
1
to
inequalities,
solve
linear
or
3
quadratic
eg
Find
the
a
2x
b
x
c
3x
-1
values
of
>
2
3x
+
x
such
that
2
Find
the
Rewrite
the
function
which
values
x
such
inequality
and
the
of
nd
graph
all
is
as
the
on
that
or
a
x
-
3x
≤10.
2
+
quadratic
values
of
x
below
the
2x
≤
8
2
2
-
2x
≤
2
+
5x
-
x
for
x-axis.
y
12
9
2
f(x)
=
x
–
3x
–
10
6
3
(–2, 0)
(5, 0)
x
0
5
6
7
–3
–6
–12
–15
x
4
∈[-2,
5].
Solve
simultaneous
3 x
4 y

5
 3y

1
equations,
4
eg
Solve
the
following
equations

by
a
simultaneous
method
of
your
choice.


2x
7 x
 4 y
2 x
 1
a
b

Use
one
of
the
methods
method
of
elimination.
to
solve,
eg

4 y

5
6 x



2x
 3y
-y
=
13
⇒
2x
=

1
⇒
y
=
38
⇒


6x

2

3  9y
y

4
x
d
 10
2x
 3y

5y
 1

 1

2x
 3
 10 y

6 x
-13
=
2 x




x
 8y
 3y
 1
the
c
3 x
5x

 3y
⇒
19
 9y
2x
⇒
+

3
(x, y)
3
×
=
(-13)
(19,
=
-1
-13)
148
/
3.1
Quadratic equations and
International-
mindedness
inequalities
The abylonians
(2000–1600 ) used
Solving quadratic equations by factorization
quadratics to nd
When
linear
a
quadratic
factors,
equation
you
can
can
factorize
be
expressed
the
equation
as
the
and
product
solve
of
two
the area of elds for
it.
agriculture and taxation.
arbegla dna rebmuN
3 .1
Example 1
Solve
the
following
quadratic
equations
2
a
x
factorization.
2
+
3x
-
40
=
b
0
2
a
by
2x
-
7x
-
4
=
0
40
=
0
2
x
+
3x
-
40
=
0
⇒
x
⇒
x(x
⇒
+
(x
8x
+
+
-
8)
5x
-
8)(x
-
5(x
-
5)
+
=
8)
=
Split
the
middle
factorize
a
rst
terms,
term,
common
so
that
factor
you
from
can
the
0
0
two
from
Use
the
last
and
two
distribution
a
common
factor
terms.
to
factorize
the
expression.
x
+
8
=
0orx
-
5
=
0
⇒
x
=
-
8orx
=
Use
5
the
2
b
2x
2
-
7x
-
4
=
0
⇒
2x
-
8x
+
x
-
4
=
the
split
factorize
2x(x
-
4)
+
(x
-
4)
=
0
⇒
(2x
+
1)(x
-
4)
=
Use
0
product
theorem
to
nd
solutions.
Again,
0
zero
in
the
middle
term
and
pairs.
distribution
to
factorize
the
expression.
1
2x
+
1
=
0orx
-
4
=
0
⇒
x
=
-
orx
=
Use
4
the
zero
product
theorem
to
nd
2
the
solutions.
Exercise 3A
2
Solve
the
following
equations
by
factorization.
3
3x
4
4x
5
5x
-
7x
+
-
20x
-
4x
2
=
0
2
1
x
+
8x
+
15
=
0
2
+
25
=
0
2
2
x
+
5x
-
14
=
0
2
-
12
=
0
Solving equations by completing the square
Most
a
quadratic
perfect
square.
expression
already
However,
it
Completing
quadratic
has
real
the
does
not
the
you
they
involve
a
will
can
come
easily
perfect
be
across
do
not
transformed
square
by
a
involve
into
process
an
called
square
know
is
that
However,
which
completing
You
equations
how
to
always
square
equation.
is
You
solve
a
quadratic
possible
to
another
method
can
use
this
equation
factorize
a
that
method
by
factorization.
quadratic
you
can
whenever
equation.
use
the
to
solve
a
equation
solutions.
149
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
2
First
you
perfect
The
will
nd
the
value
of
c
that
can
be
added
to
x
+
bx
to
form
a
square.
following
diagrams
x
will
show
you
how
The
6
to
area
do
of
this.
the
region
shaded
in
2
green
is
region
x
and
shaded
the
in
area
pink
of
is
the
6 x.
Thus,
2
x
6 x
x
the
area
of
the
whole
rectangle
is
2
x
x
3
+
6x
Divide
3
equal
the
pink
region
into
two
parts.
2
x
3x
x
x
3x
Rearrange
3
The
large
(x
3)
the
parts.
square
formed
has
area
2
+
2
x
2
x
3x
So
the
expression
becomes
you
3
3x
add
a
perfect
x
+
6x
square
when
9.
9
Investigation 1
For each of the following diagrams, nd the value of
c which makes the expression into a perfect square, and
then write the expression as a perfect square.
2
1
x
2
+ 4x + c
2
x
x
+ 10x +
c
x
2
2
2
x
x
2
2x
2x
x
x
150
/
2
3
2
+
x
arbegla dna rebmuN
3 .1
3x +
4
c
x
+ bx + c
2
5
Fill in the blank: To complete the square for
6
Factual
7
Conceptual
+ bx, add ____________________________.
x
What are the steps of the completing the square method?
How does the area model help you understand the process of completing the square and
why is the process called completing the square?
Use
your
ndings
following
from
the
investigation
to
algebraically
solve
the
examples.
Example 2
For
each
square.
quadratic
Hence,
expression,
write
each
in
a
2
a
x
nd
the
value
factorized
of
k
which
8x
+
b
k
c
9x
the
expression
a
perfect
form.
2
+
makes
2
x
-
7x
+
k
k
8 x
Use





+
3x
the
+
k
binomial
expansion
and
express
2
2
a
2
x
 8x
 k

x
2

2 
x
 4

4

x

4 
the
term
as
twice
the
constant
term.
Use
with
a
term
to
in
x
product
this
of
x
constant
2
⇒
k
=
4
=
16
nd
the
value
of
k.
k

7 x




2
7
2
b
 7 
2
x
 7x
 k

x
2 
x



2
2

x

2

 7 



2

2
7 


k

49






2
4

k

3x



2
1
2
 1 
2
c
9x
 3x
 k

3 x 
 2 
3 x  


2
2
3x



 1 


2

2

2
1 


k

1






2

4
151
/
3
E X PA N D I N G
Once
you
involves
have
a
THE
NUMBER
transformed
perfect
square,
a
S YS T E M :
quadratic
you
can
C O M PL E X
equation
easily
solve
to
the
a
NUMBERS
form
which
equation.
Example 3
Solve
the
following
equations
by
2
a
the
square.
Give
your
answers
in
exact
independent
of
x
form.
2
x
+
5x
-
24
b
=
0
24
=
2
a
completing
x
2x
-
7x
+
=
0
2
+
5x
-
0
⇒
x
+
5x
=
24
2
5
x

x
2 

2
2

the
term
hand
side.
to
the
right-
 5 



Move
2
 5 
2

1
Complete

24




the
square
on
the
left-hand
side,
and

2

add
the
side
so
same
that
constant
you
don’t
term
to
change
the
the
right-hand
equation.
2
5 


x
121

5



121

x

Factorize

the
complete
square
and
solve
the

2

4

2
4
equation.
5

x

11


2

x

x

8
7
2x

3

0
2
2
b
x
or
1
2
 7x
 1 
0
2

x

2
Since
the
2

2

x

2 
x

7


4



is
not
1,
divide
the
equation
by
2.
7

2

Move
the
term
independent
of
x
to
the
right-



4
x
2
1


of
2
whole
7
coefcient


4
hand
side.

Complete
the
square
on
the
left-hand
side,
2
and
7 


x
41



7


x


hand

4

16

add
the
4
constant
term
to
the
right-
side.
16
Factorize
7 
same
41
the
expression
and
solve
the
41
equation.

x

4
There
since
is
no
they
need
are
to
separate
written
in
the
the
solutions
surd
conjugate
form.
In
Example
solving
you
real-life
answers
values
3b,
in
can
context
of
your
problems,
decimal
then
the
left
form
be
used
problem
it
to
in
answer
is
an
much
in
surd
more
appropriate
making
form.
However,
practical
degree
to
of
measurements,
give
when
your
accuracy.
or
These
whatever
the
is.
Exercise 3B
2
1
Solve
by
the
following
completing
your
answers
the
in
equations
square.
exact
c
x
d
3x
e
4x
f
5x
Give
-
x
-
1
=
0
2
-
7x
+
2
=
0
+
12x
+
5
=
0
-
10x
+
2
=
0
form.
2
2
a
x
+
6x
-
7
=
0
2
2
b
x
-
7x
-
30
=
0
152
/
3 .1
Solve
the
following
2
equations
a
x
+
2x
b
(p
-
4)x
c
2x
-
1
=
0
HINT
by
completing
Give
your
the
square.
answers
correct
to
+
(p
+
2)y
=
-3
When solving quadratic
3
2
signicant
-
x
-
3
=
0
gures.
equations the solutions
2
d
3x
+
9x
+
5
=
0
are called roots since in
the formula you use a
square root to nd them.
The quadratic formula
Notice
when
by
that
its
you
When using quadratic
generally
solve
roots
are
integers,
but
completing
the
square
even
disadvantage
is,
can
therefore,
quadratic
is
that
you
completing
helpful
equation
–
to
have
with
real
arbegla dna rebmuN
2
a
a
quadratic
can
when
the
always
its
solve
roots
square
quicker
equation
way
can
to
are
be
a
surds.
the
factorization
quadratic
long
nd
by
equation
However,
and
the
repetitive.
solutions
to
It
any
roots.
functions, the points
of intersection with the
x-axis have x values that
are obtained by solving
the corresponding
quadratic equation. In
this case you call them
zeros because the value
Investigation 2
of the function for those
In this investigation, you will star t with the general form of a quadratic
x values is 0.
2
equation, ax
+ bx + c = 0,
a,
b,
c
∈
ℝ,
a
≠ 0, and use completing the square
to nd its solutions in terms of a, b and c
HINT
opy the following table and ll in the missing steps or explanations.
You could take the
Mathematical working
Explanation
used in Example 3b,
2
ax
same approach that we
+ bx + c = 0
by dividing the whole
1
Multiply both sides of the
1
equation by a, since
equation by a
a ≠ 0. However, this
2
Subtract ac from both
2
approach will guide you
sides of the equation.
through using a slightly
2
b
2
a
 b
2
x

2  ax

 b










2
2

2
ac
dierent method which

2
will give the same result.

2
2
b 

4
ax
b



 ac

2

TOK
4
4

How can you deal with the
ethical dilemma of using
2
2
b 

5
ax
b


4ac



mathematics to cause
5

2
4

harm, such as plotting the
course of a missile?
6
Take the square root of
6
both sides.
2
b
7
ax
=
-
b
- 4ac
7
±
2
2
8
Factual
8
Divide by a to nd x
What do a, b and c stand for in your formula in line 8?
Conceptual
How does the quadratic formula generalize the solutions to
quadratic equations?
153
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
The quadratic formula
2
For a quadratic equation in the form ax
+ bx + c = 0,
a,
b,
c
∈ℝ,
a ≠ 0, the
2
-b
solutions or roots are given by
x
±
b
-
4ac
=
2a
Example 4
Use
the
quadratic
formula
to
solve
2
a
the
following
equations.
4
c
2
x
+
11x
+
24
=
b
0
x
+
11x
+
24
=
0
⇒
a
=
+
1,
your
answers
2
3x
8x
+
=
0
5x
2
a
Leave
b
=
11,
c
=
exact
form.
2
-
2x
-
1
Identify
24
in
apply
=
the
the
d
0
2x
coefcients
+
x
+
1
=
0
and
a,
b
and
c
a,
b
and
c
a,
b
and
c
a,
b
and
c
formula.
2
11 

x
11
11 
 4  1  24

121  96

2 1
2
11  5


x

8
=
0
⇒
or
x

Separate
3
the
solutions.
2
2
b
3x
+
8x
+
4
a
=
3,
b
=
8,
c
=
Identify
4
and
the
apply
coefcients
the
formula.
2
8 

x
8
8 
 4  3  4

64
 48

2  3
8 
4
6
12


x

4

6
2
or
x
2

6

Separate

6
the
solutions.
3
2
c
5x
-
2x
-
1
=
0
⇒
a
=
5,
b
=
-
2,
c
=
-
1
Identify
the
coefcients
and
2
  2 

x

 2 
 4  5 
apply
 1
2 


the
2
6
1 
10
6
1 
6
Separate


x

or
10
x
2x
+
x
5
+
1
=
0
⇒
a
=
the
solutions.

5
2
d
formula.
20

2  5
2 
4
2,
b
=
1,
c
=
Identify
1
and
the
apply
coefcients
the
formula.
2
1 

x
1
1 
 4  2 1

1  8

2  2
1 
4
7


x

Since

the
expression
under
the
4
square
real
When
either
special
cases
of
the
that
coefcients
we
solve
by
b
and/or
using
c
are
simpler
equal
to
zero,
root
is
negative,
there
is
no
solution.
we
have
methods.
154
/
arbegla dna rebmuN
3 .1
Case 1
b = 0, c ≠ 0
c
2
2
ax

c

0

ax
c
2

c

x



x


a
 
if
a
c
c
x
, where



x
0 and
 
if


0
x

a
a
Case 2
b ≠ 0,
= 0
c
b
2
ax
 bx

0

x
 ax
 b 

0

0
or
ax
 b

0

x


, so you
a
have two real solutions, one of which is always
If
one
or
more
unknown
solutions
of
the
constant,
in
terms
coefcients
you
of
can
that
use
in
a
the
0
quadratic
quadratic
equation
is
an
formula
to
nd
the
x,
by
using
the
quadratic
Rewrite
the
equation
constant.
Example 5
2
Solve
the
Express
quadratic
x
in
terms
equation
of
the
2
px
real
px
+
3
=
3px
parameter
+
p
3
=
3px
+
x
⇒
px
-
(3p
+
1)x
+
3
0,
=
0
quadratic
⇒
a
=
p,
b
=
-(3p
+
1),
c
=
formula.
p
2
+
≠
Identify
3
in
the
general
form.
the
coefcients.
2


x

  3p
 1



 3p
 1

 4  p  3
Apply

the
quadratic
formula.
2  p
2
3p
9p
+ 1 ±
+ 6p
+ 1 - 12p
Simplify.
=
2p
2
3p
+ 1 ±
9p
- 6p
+ 1
=
2p
2
3p
+ 1 ±
(3p
- 1)
Factorize
=
2p
3p
 1 
square
 3p
 1

3p

x
 1  3p

3p
x
2p
+ 1 - 3p
+ 1
2
=
=
2p
under
the
root.

3
Simplify
again.
2p
1
Write
=
2p
expression
6p

2p
or
 1
the
both
solutions
separately.
p
155
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Exercise 3C
1
Solve
the
following
quadratic
formula.
the
form.
exact
equations
Give
your
x
c
x
using
answers
2
the
Solve
in
by
using
2
+
9x
-
x
+
18
=
0
x
d
2x
-
x
-
30
=
0
a
x
c
x
2
1
=
0
x
the
the
in
following
quadratic
terms
of
quadratic
formula.
the
real
3x
=
2
Give
your
2
+
3x
=
ax
+
kx
=
2k
2
-
equations
parameter.
2
b
2
-
for
answers
2
a
by
+
3a
b
2x
d
p
2
2
x
-
b
=
x
-
2bx
2
+
3px
=
px
+
3
2
e
2x
+
3
= 11x
Discriminant of a quadratic equation
Investigation 3
1
Use the quadratic formula to solve the following equations.
2
a
x
c
x
e
6x
2
- 3x - 10 = 0
b
3x
- 12x + 36 = 0
d
4x
f
x
2
2
2
2
- 2x + 5 = 0
- 20x + 25 = 0
2
- 13x + 6 =
Factual
0
- 5x + 7 = 0
How many real solutions can dierent quadratic equations
have?
2
-b
3
Conceptual
x
y considering the quadratic formula
±
b
-
4ac
=
,
2a
explain why quadratic equations have dierent numbers of solutions.
2
The expression b
4
- 4ac
is called the discriminant of a quadratic.
Sketch graphs of y = f(x) for each equation given in question 1. On your
International-
sketch, label any zeros of each function.
mindedness
5
Factual
What do you notice about the number of zeros of each graph,
Frenchman Nicole
and the number of solutions to the quadratic equation?
Oresme was one of the
6
Factual
What does the graph of a quadratic equation tell you about the
rst mathematicians to
consider the concept
value of the discriminant?
of functions in the
7
Conceptual
What can the discriminant be used for?
14th century, the
term "function" was
introduced by the
German mathematician
2
Given a quadratic equation of the form
+ bx + c = 0,
ax
a,
b,
c
∈
ℝ
,
a ≠ 0,
Gottfried Wilhelm
the discriminant is the expression in the formula that is under the square root
Leibniz in the 17th
2
and is denoted by the Greek letter Δ, Δ = b
- 4ac
century and the
notation was coined by
Swiss Leonard Euler in
As
you
found
in
the
investigation,
the
sign
of
the
the
quadratic.
discriminant
the 18th century.
determines
the
number
of
roots
of
156
/
arbegla dna rebmuN
3 .1
Case 1: Δ > 0
2
-b
If the discriminant is positive, then
x
±
b
-
4ac
and there are t wo
=
2a
distinct real roots.
Case 2: Δ = 0
-b
x
If the discriminant is equal to zero, then
=
. This is regarded as one
2a
repeated real root.
Case 3: Δ < 0
2
b
-
nature
of
If the discriminant is less than zero, then
4ac
is not real. In this case,
the
roots
there is no real solution
Example 6
Without
solving,
determine
7x
0
the
of
each
equation.
7
2
a
+
-
1
=
b
1
2
2
5x
25x
+
49
=
c
70x
2x
+
x
+
4
2
a
5x
+
Δ
7
7x
-
1
=
0
⇒
a
=
5,
b
4
×
5
×
(-1)
=
69
=
7,
c
=
2
=
-
There
are
two
distinct
real
2
0
roots.
the
coefcients.
Calculate
the
discriminant.
Since
the
discriminant
there
are
two
Write
in
distinct
2
25 x
b
>
0
2
Identify
-1
=

49

70 x

25 x

a

is
greater
real
roots.
+
+
than
0
2
 70 x
25,
b


49

0
c

70,
Identify
49
the
the
form
ax
bx
c
=
0.
equal
to
coefcients.
2
Δ
=
(-70)
There
is
-
4
one
7
×
25
×
49
repeated
=
4900
real
root.
-
4900
7
1
=
0
Calculate
the
Since
the
discriminant
there
is
one
discriminant.
repeated
is
real
0
root.
1
2
c
2x

x


4
0

a

2,
b

,
c
Identify

4
2

7

 4  2 


might
that
the
49

4
2

are
also
no
be
equation
real
the
discriminant.

4



Since
the
discriminant
there
are
no
0
16
is
less
than
0
16
real
roots.
roots.
asked
has
Calculate
15

There
You
1

coefcients.
2
2

the
to
determine
different
types
the
of
value
of
a
real
parameter
so
roots.
Example 7
Find
the
value(s)
of
the
real
parameter
p
so
that
2
a
2x
c
(p
2
-
3x
+
p
=
0
has
two
real
b
roots
px
+
p
=
13x
has
one
real
repeated
root
2
+
2)x
+
2px
=
1
-
p
has
no
real
roots.
Continued
on
next
page
157
/
3
E X PA N D I N G
THE
NUMBER
0
Δ
2
a
S YS T E M :
C O M PL E X
2
2x
-
3x
+
p
=
⇒
=
(-3)
-
4
×
2
×
Find
p
NUMBERS
the
discriminant.
9
⇒
9  8p

0

p
Set

the
discriminant
to
be
greater
than
0
8
and
2
b
2
px
+
p
=
13x
⇒
Δ
=
(-13)
-
4
×
p
×
solve
Find
p
the
the
inequality.
discriminant.
2
=
169
-
4p
13
2
 169  4 p

0

p

Set

the
discriminant
to
be
equal
be
less
to
0
and
2
solve
2
c
(p
+
2)x
+
2px
-
Find
=
1
p
=
(2p)
=
4p
-
=
8 -
4p
the
the
equation.
discriminant.
2
⇒
Δ
-
4(p
2
⇒
8
-
4p
<
0
⇒
p
>
+
2)(p
-
1)
2
4p
-
4p
+ 8
Set
2
the
solve
discriminant
the
to
than
0
and
inequality.
Example 8
2
Find
a
the
two
values
distinct
of
r
for
real
which
equation
b
roots
2
x
the
one
x
real
+
3rx
repeated
2
+
3rx
+
1
=
0
⇒
Δ
=
(3r)
⇒
Δ
=
9r
-
4
×
1
×
+
1
0
has
c
root
Find
1
=
the
no
real
discriminant
roots.
and
simplify
it.
2
a
Two
distinct
real
-
4
When
roots:
4
2


0

9r

0


r

r
One
r
repeated
0

real
the
No
real
9r

0

r

0
r

Solve

9r
2
All
of
the
both
sides,
r
=
r .
inequality.
=
0
there
is
one
repeated
real
root.
the
equation.
Δ
<
0
there
is
no
real
root.

0

r

both
sides,
2

r
2
Take

the
square
root
of
r
=
r .
3
the
inequality.
2

3
of
2
 4
Solve

Δ
When
9

root
3
roots:

roots.
2

4

real
2
 4
2

square
the
When
root:
9
c
distinct
2
Take

4

two
3
2

are

3
b
there
2
or

0
2

Solve
r
>
2
 4
2

Δ
r

3
inequalities
can
be
solved
by
graphing
on
your
calculator.
158
/
Exercise 3D
1
Without
the
solving
nature
of
these
their
equations,
2
determine
Find
roots.
2
the
values
of
which
the
i
two
distinct
ii
one
real
iii
no
the
equation
given
parameter
for
has
2
a
x
c
x
e
2x
+
3x
-
7
=
0
b
x
+
2x
+
1
=
0
d
5x
f
2.25x
+
x
+
2
=
0
3x
+
2
49
=
real
roots
2
2
+
2
=
0
repeated
root
2
=
πx
-
1
+
21x
real
roots.
2
a
mx
c
(2s
arbegla dna rebmuN
3 .1
2
+
2x
-
5
=
b
0
4x
=
3x
+
4
-
t
2
+
1)x
=
s(3x
-
1)
Solving quadratic inequalities
TOK
Apart
from
using
a
calculator
and
identifying
the
part
of
the
parabola
We have seen the
that
is
above
two
other
or
below
the
x-axis
(shown
in
Before
you
start),
there
are
involvement of
methods
for
solving
quadratic
inequalities.
several nationalities
Method 1: Algebraic method
in the development of
In
order
to
rst
to
solve
factorize
a
quadratic
the
inequality
quadratic
that
expression
will
into
factorize,
linear
you
factors
quadratics.
need
and
then
To what extent do you
consider
the
sign
of
the
product.
if
both
You
know
that
the
product
of
two
believe that mathematics
factors
is
positive
factors
are
of
the
same
sign,
and
that
the
is a product of human
product
is
negative
if
both
factors
are
of
different
signs.
social collaboration?
A

0
and
B

0
A

A  B

0


0
and
B

0

0
and
B

0

or

or
A  B

0


or


A

0
and
B

0
A


Example 9
2
Solve
the
quadratic
inequality
-
2x
2
2x
2x
-
5x
-
3
≥
0.
2
-
5x
3
≥
0
⇒
3)
+
(x
-
6x
+
x
-
3
≥
The
0
so
⇒
2x(x
x
-
 3

0
-
and
3)
2x
≥
0
⇒
 1 
(x
-
3)(2x
+
1)
≥
quadratic
split
it
into
expression
two
linear
will
factorize,
factors.
0
0

Consider

the
signs
of
the
product.
or


x
 3
x


0
and
2x
 1 
0

1

3
and
x


Find



2



x

satisfy

or



3
and
x

values
of
x
which
the
inequality.
1
x

2

which
can
be
written

as
an


2
interval
1 

 , 
 3,


of

x

range
or

1
x
the
3
 

2

159
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Method 2: Using a “sign table”
If
a
quadratic
two
If
linear
this
is
expression
terms
the
to
case,
it
the
corresponding
the
quadratic
This
method
not
factorize,
compare
signs
is
nd
best
to
quadratic
either
will
will
side
work
of
of
you
cannot
split
it
into
factors.
any
zeros
equation,
the
then
of
the
quadratic
and
then
consider
the
quadratic
by
the
solving
signs
of
zeros.
whether
or
not
will
factorize.
Example 10
2
Solve
the
quadratic
inequality
2x
-
5x
-
3
≥
0.
Solve
5 
25 
24
the
corresponding
5  7
2
2x
 5x
 3

0

x


4
quadratic
equation,
using
formula.
eg
by
4
the
1
x
=
3
or
x
=
-
2
1
x
<
1
-
x
=
-
2
1
-
2
<
x
<
3
x
=
x
3
>
3
Order
and
2
2x
-
5x
-
3


zeros
in
the
0
-
+
investigate
quadratic
the
 , 
table
between
sign
the
1 

x
0
+
the
2
of
the
zeros.
2
 3,
The
 
coefficient
of
x
is

2


positive
so
concave
up,
zeros
the
the
parabola
is
between
the
so
expression
negative
value.
Since
the
inequality
strict,
you
zeros
in
should
your
has
is
a
not
include
the
solution.
Exercise 3E
2
Solve
the
method
following
of
your
quadratic
choice
and
inequalities
verify
your
by
the
3
x
4
4x
-
5
8
5x
6
12x
-
13x
-
30
≥
0
≤
0
solutions
2
by
a
1
x
2
x
graphical
method
on
a
11x
+
6
calculator.
2
2
+
6x
+
8
<
-
8x
+
16
<
-
6x
0
2
2
>
-
4
≥
9x
0
Developing inquiry skills
Return to the opening problem.
an you use a quadratic equation to model the roller coaster?
160
/
3.2
Complex numbers
Algebraic vs. geometric approach towards complex
TOK
numbers
Descartes showed that
geometric problems could
be solved algebraically
Investigation 4
arbegla dna rebmuN
3.2
and vice versa.
2
onsider the quadratic equation x
+ 4 = 0.
What does this tell us
1
an you nd real values that are solutions of the equation? Explain why, or
about mathematical
why not.
representation
and mathematical
2
i such that i
you can write i =
2
=
-1, therefore
knowledge?
1
an you use this fact to solve the quadratic equation, expressing your
HINT
answers in terms of i?
Recall that
2
3
Use the quadratic formula to solve the equation
x
- 2x + 5 = 0. Give the
a  b
roots in terms of i

a

b
a, b ≥ 0
Conceptual
4
What does the imaginary number i represent and how is it
used in solving quadratics?
InternationalThe letter i is used to represent the imaginary number
-1
mindedness
Greek mathematician
Heron discussed the
2
Solutions
of
x

any
equation
x
+
c
=
0,
c
>
0,
can
be
written
as
root of a negative in the



c
i d ,
d


.
We
call
these
imaginary
numbers .
rst century .
2
By
dening
algebra
i
that
=
-1,
you
you
are
have
going
to
developed
use
in
this
omplex numbers are numbers of the form
a
whole
new
avenue
of
chapter.
z
=
a
+ ib, where a,
b ∈ ℝ
a is called the real par t of z, and we write Re(z) = a.
b is called the imaginary par t of z, and we write Im(z) =
Complex
later
in
numbers
the
complex
Complex
book.
numbers
to
in
above
solutions
The
be
written
form
z
=
a
+
in
different
ib
is
called
forms,
the
as
you
will
Cartesian
see
form
of
a
number.
trying
the
can
b
solve
to
were
certain
rst
cubic
investigation,
any
quadratic
introduced
equations.
complex
equation,
when
mathematicians
However,
numbers
as
also
regardless
of
you
allow
the
have
you
value
seen
to
of
were
nd
its
discriminant.
161
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Did you know?
The rst person to propose working with the square root of a negative
number was Heron of Alexandria (c.10–c.60) when discussing the volume
of an impossible frustum of a pyramid.
Later on, in Italy, mathematical tournaments became very popular and
solving dicult cubic equations became a way in which they found the
winner. Those mathematicians discovered the formula for solutions of cubic
equations and introduced complex numbers. Niccolo Fontana Tar taglia
(1499–1557) discovered the formula rst and shared his knowledge with
Gerolamo ardano (1501–1576).

ardano introduced complex numbers of the form
a

b ,
a  ,
b  
.
They realized that those two par ts cannot be combined and the second par t
was called the imaginary or even impossible par t.
René Descar tes (1596–1650) was the rst mathematician to establish the
term “imaginary par t”, and John Wallis (1616–1703) made huge progress in
giving a geometrical interpretation to
-1
Leonhard Euler (1707–1783) was the mathematician who rst introduced
the symbol i
By
looking
every
real
has
called
0
=
In
0
a
+
this
at
the
part
real
×
Chapter
and called it an “imaginary unit”.
complex
a
0,
can
that
equal
number
be
is,
to
imaginary
seen
b
=
0.
0,
ie
a
z
as
a
In
=
number.
=
a
0,
a
+
ib,
you
complex
similar
only
The
might
number
way,
has
an
number
0
a
notice
where
complex
imaginary
can
be
that
the
number
part
seen
as
and
is
both,
0.
way,
numbers,
is
part
purely
i
-1
number
imaginary
that
=
you
and
1:
ℕ
can
view
hence
⊂
ℤ
⊂
you
ℚ
⊂
the
can
ℝ
real
numbers
expand
⊂
the
as
a
subset
subset
of
the
denition
complex
from
ℂ
.
Example 11
State
the
real
part
and
the
imaginary
part
of
the
following
2
a
2
+
b
3i
2i
-
c
5
z
=
2
b
z
=
2i
+
3i
⇒
Re(z)
=
2andIm(z)
d
7i
5
⇒
Re(z)
=
-5andIm(z)
4
=
The
3
unit
-
=
real
and
part
the
2
z

7i

Re  z


and
imaginary
imaginary
part
has
an
unit.
Im
z 


7
3
11   i
z
no
2

3
d
has
2
imaginary
c
numbers.
11   i
-
3
a
complex

4

11

Re  z


and
4
Im  z


4
162
/
3.2
Real
numbers
can
line
represents
In
similar
a
real
part
part
of
Each
of
the
the
real
on
a
number
line.
Each
point
on
the
number.
complex
numbers
plane,
number,
and
can
where
the
be
the
vertical
represented
horizontal
axis
in
axis
represents
a
two-
represents
the
the
imaginary
number.
P(x, y)
imaginary
visualized
coordinate
complex
point
one
way,
dimensional
be
arbegla dna rebmuN
Geometric approach
number
in
the
parts
of
z
plane
the
=
x
+
and
iy,
the
complex
x,
y
∈ℝ
,
is
represented
coordinates
number
are
the
by
real
a
Im
and
itself.
z
=
x
+
iy
y
Purely
real
numbers
numbers
lie
on
the
lie
on
the
x-axis,
and
purely
imaginary
y-axis.
Re
x
Modulus or absolute value of a complex number
You
already
know
that
the
modulus
or
absolute
value
of
a
real
number
Did you know?
x,
is
algebraically
dened
as
x

x

0
.


x,
x

Geometrically,
the
modulus
Jean-Rober t Argand
0
(1768–1822) rst
represents
the
distance
along
the
number
line
between
x
and
the
proposed the idea of
origin.
You
can
extend
this
idea
into
two
dimensions
in
the
complex
representing complex
plane
where
the
modulus
of
a
complex
number
represents
the
distance
numbers on a two-
between
the
complex
number
and
the
origin.
dimensional coordinate
plane. arl Friedrich
Gauss (1777–1855)
Investigation 5
independently developed
and rened the plane
1
Plot the points that represent the following complex numbers on a
model and therefore
complex plane. 3 + 4i , 1 - 2i , -5 + 4i , -2 ,
3i , -2 - 3i
the plane of complex
2
What is the distance between each point you plotted in question
1 and the
numbers is known as
the Argand diagram or
origin?
Gaussian plane. aspar
3
an you write down a formula for nding the modulus, | z|, of a
Factual
Wessel (1745–1818)
complex number z = x + iy?
also developed a similar
4
Factual
What do you notice about the modulus of a complex number?
5
Conceptual
approach using vectors.
What does the modulus of a complex number tell you
about that number?
Given the complex number z = x + iy,
x, y ∈ ℝ, the modulus of z is given by
2
2
z

Notice
x

iy
that

x
the
2
2

y

modulus
of
 Re  z  
a

complex
 Im  z  
number
is
always
a
positive
real
+
number,
|z | ∈
ℝ
.
163
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Example 12
Find
the
a
-
modulus
of
the
1
5
2
Check
your
answers
complex
i
20 i
-
b
12i
following
numbers.
- 21
c
4
29
with
a
GDC.
2
2
a
5 - 12i
=
5
+
( -12 )
=
2
1
i
b

2
4
20i
 21

2
1 



5






21 


the
formula
for
the
modulus.

Apply
the
formula
for
the
modulus.
Apply
the
formula
for
the
modulus.
Make
sure

4
16

4
2
 20 



29
Apply
5

2
c
= 13
2
 1 

169
29





29

841
=
= 1
841
Checking:
calculator
5–12×
i
1
nd
the
how
to
use
modulus
of
a
a
number.
You
can
use
this
to
0.559017
check
×
2
to
know
13
complex
1
you
the
results.
i
4
5
0.559017
4
20×
1
i–21
29
Operations with complex numbers
First
you
need
to
understand
Two complex numbers z
when
= a + bi and
1
two
complex
= c + di, a,
z
b,
numbers
c,
are
equal.
d ∈ ℝ
, are equal if
2
and only if their real par ts are equal and their imaginary par ts are equal,
a
=
c and b = d
Investigation 6
Use your calculator to complete the rst row of this table.
and z
z
1
are
2
complex numbers, and m and n are real constants.
Then add your own examples: for each blank row, dene any two complex
and z
numbers z
1
, and any two real constants m and n, and complete the
2
table.
164
/
z
z
1
= 2 + i
z
z
2
z
m
m
×
z
2
n
n
×
1
2
= 3 -2i
z
1
+
1
z
m
×
2
z
+ n
×
arbegla dna rebmuN
3.2
z
1
2
- 3
2
Use the table you just completed to answer the following questions.
1
Factual
How do you add two complex numbers?
2
Factual
How do you subtract two complex numbers?
3
Conceptual
Do we need two separate operations for adding and
subtracting complex numbers?
4
How do you multiply (or divide) a complex number by a real
Factual
number?
Conceptual
5
How are algebraic operations with complex numbers
similar to algebraic operations with polynomials?
InternationalWhen
adding
complex
numbers,
or
multiplying
them
by
a
constant,
mindedness
use
the
following
rules.
How do you use the
abylonian method of
+ z
z
1
= (a + bi)
+ (c + di) = (a + c)
+ (b + d)i, a, b, c, d
∈ℝ
2
multiplication?
lz = l(a + bi)
= (la) + (lb)i, l, a, b
∈ℝ
Try 36 × 14.
Example 13
If
z
=
4
-
3i
and
z
1
=
-2
+
5i ,
calculate
the
following
and
check
your
answers
with
a
2
calculator.
1
a
z
1
2
z
b
z
+
2
3
a
z
-
z
1
z
1
2
=
(4
-
3i)
-
(-2
=
(4
+
2)
+
(4
- 3i ) +
+
5
Use
5i)
the
properties
of
complex
2
numbers.
1
2
z
b
+
-
1
z
1
3
(-3
=
2
5)i
=
6
-
8i
2
( -2 + 5i )
Multiply
both
3
4
=
4
-
i
3
-
and
2i
=
5
+
(4–3×
i)
5
then
add.
sure
6–8×
you
to
know
add
numbers
and
numbers
by
to
and
how
to
subtract
multiply
use
a
complex
complex
i
a
constant.
You
can
use
8
×
i)+
given
i
Make
2
×
3
the
15
calculator
1
by
2
8
+
Checking:
i–(–2+5×
andz
5
constant,
4–3×
z
1
5
(–2+5×
i
i
15
this
to
check
the
results.
165
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Exercise 3F
1
Find
the
real
and
imaginary
parts
of
3
the
Given
the
complex
numbers
z
=
3
-
4i ,
1
following
complex
numbers.
z
=
5
+
i
and
z
2
a
z
=
b
-4i
z
=
following
5
=
-1
+
2i ,
calculate
the
3
and
check
your
answers
with
a
calculator.
5 - 12i
c
z
=
-24
+
d
7 i
z
=
a
13
z
-
z
1
2i
e
z
+
b
z
2
3
1
2
-
z
+
z
1
the
modulus
numbers
4z
1
3z
-
4z
2
- 5z
3
c
Find
+
1
3
=
5
2
2z
- 1
in
of
each
of
the
+
1
2z
2
d
z
2
2
complex
-
3
3
4
3
1
question
Investigation 7
Use your calculator and your algebra skills to ll in the following table. The rst
line has been done for you.
z
z
1
z
2
×
z
1
2
(2 + i)× (1 + 3i ) =
= 2 + i
z
= 1 + 3i
z
1
2
2
2 + i + 6i +3i
= -1 + 7i
1
= 1 - 2i
z
2
= 3 + 2i
z
2
= a + bi
2
= 3 + 2i
= c + di
z
1
2
= a + bi
z
= 3 + i
z
1
z
= 4 - 5i
z
1
z
= 3 + i
z
1
= c + di
z
1
2
1
Factual
Using words, express how the real par t of
×
z
the real and imaginary par ts of both z
Factual
2
and z
1
2
is related to
z
1
2
In a similar way, express how the imaginary par t of
×
z
related to the real and imaginary par ts of both
Conceptual
2
and z
z
1
3
is
z
1
2
How would you compare multiplication of two complex
numbers to multiplication of two algebraic linear factors?
To
summarize
the
ndings
from
the
investigation,
we
write
the
following.
×
z
= (a + bi) ×(c + di) = (ac - bd)
z
1
Although
for
the
formula
multiplying
simplify
to
the
multiply
nd
+ (ad + bc)i, a, b, c, d ∈ 
2
their
two
process
two
in
the
complex
of
key
point
numbers,
multiplication
complex
numbers,
it
above
in
reality
much.
is
gives
you
this
rule
Therefore,
often
simpler
a
to
general
does
when
use
rule
not
asked
algebra
to
product.
166
/
arbegla dna rebmuN
3.2
Example 14
Given
the
complex
numbers
z
=
1
-
3i ,
z
1
=
4
+
i
and
z
2
=
-2
+
3i ,
nd
c
z
the
following.
3
2
a
×
z
z
1
Use
-
b
z
2
z
 z
1

z
2
to
check
your

i
z
2
+
3
×
2z
1
z
2
3
answer.

1  3i   4

4

9  14 i

1  3i   4

 7  11i   2  3i 
3
×
z
1
technology
a
×
z
3

 2  3i 
Expand
the
brackets
and
simplify
the
expression.
z
b
 z
1
 z
2
3

i
 12i
 3 

2  3i
i   2  3i 
First
multiply
multiply
the
rst
two
the
product
with
the
expression.
factors,
the
then
third
factor.

14

 19 
21i

22i
 33
Simplify
43i
2
2
c
z

2z
1
 z
2

3
1  3i 
 2 4

i   2  3i 
Square
z
and
multiply
z
1
 1  6i
 9 

8  6 i

30  14 i
2  8  12i
 22 
 2i
 3
Expand
by
z
2
the
expressions
.
3
and
simplify.
20 i
Checking:
1–3×
i
→z1
1–3×
i
Make
4+i
→z2
–2+3×
i
→z3
4+
i
–2+3×
i
z1×
z2
z3
9–14×
i
z1×
z2×z3
19+43×
i
–30+14×
i
sure
you
calculator
to
numbers.
You
results.
z
,z
1
For
then
how
multiply
can
ease,
andz
2
know
in
two
use
the
the
GDC
use
a
complex
this
store
to
to
check
values
memory
the
of
and
3
work
with
the
variables.
2
z1
+2×
z2×
z3
Example 15
Find
Use
z
∈
ℂ
that
technology
Let
z
=
a
+
bi
bi)
×
satises
to
the
check
equation
your
z
×
(2
-
i)
=
z
+
3
+
2i
answer.
Rewrite
z
in
substitute
⇒
(a
+
⇒
(2a
(2
-
i)
=
(a
+
bi)
+
3
+
2i
3)
+
(b
Simplify
+
b)
+
(-a
+
2b)i
=
(a
+
+
 2a
 b

a
 3
a

b



a

2b

b

2
both
the
sides
and
equation.
of
the
equation,
collect
real
and
imaginary
parts.
3
If

into
form
2)i
and

it
Cartesian
two
complex
numbers
are
equal
then


a

b

2
both
the
real
imaginary
parts
parts
are
are
equal
and
the
equal.
Continued
on
next
page
167
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Simplify
 1
 a, b 

5 
,

1


z


2
solve
the
simultaneous
i

2

and
5
equations.
2

2
Use
your
calculator
to
check
the
Checking:
solution.
1
5
1
i
2
×
2–i
5
=
2
true
×
2
i+3+2×
i
2
Conjugate complex numbers
For any complex number z = a + bi, a, b ∈ , there is a conjugate complex
number of the form z* = a - bi . Their real par ts are equal, Re(z) = Re(z*),
and their imaginary par ts are opposite,
Notice
that
twice
the
Also,
the
when
real
when
adding
part ,
that
subtracting
imaginary
part,
two
two
is
z
Im(z) = -Im(z*)
conjugate
+
z*
=
conjugate
multiplied
by
i,
a
+
complex
bi
+
a
complex
that
is
z
-
-
numbers
bi
=
=
a
get
2a
numbers
z*
you
+
you
bi
-
get
(a
-
twice
bi)
=
2bi.
Investigation 8
opy and complete the following table. The rst row has been done for you.
Use a GD to check your answers.
|z |
z
z*
z
×
z*
2
2
z = -1 + i
z

 1
 1

2
z* = - 1 - i
z
×
z* = 2
z = 3 - 4i
z =

5

2i
z = a + bi
1
Factual
What is the relationship between the product and the
modulus?
2
Conceptual
Why do a complex number and its conjugate have equal
moduli?
*
3
Conceptual
Verify that z and z
are mutually conjugate.
168
/
3.2
In
Investigation
number
the
by
a
complex
6
real
and
Example
number.
conjugate
The
helps
13,
you
saw
following
us
to
how
to
example
divide
two
divide
a
complex
demonstrates
complex
arbegla dna rebmuN
Division of complex numbers
how
numbers.
Example 16
z
1
Given
two
complex
numbers
z
=
a
+
bi
and
z
1
=
c
+
di ,
a, b, c, d
∈
ℝ, z
2
≠
0,
nd
.
2
z
2
z
a
 bi
a
 bi
c
 di
c
 di
1


z
c

Multiply

di
c

di
by
2
the
numerator
complex
and
denominator
conjugate
of
the
denominator.
1

2
ac
- ad i
+ bci
- bd i
The
=
2
c
+
d
number,
a
ac
+ bd
=
bc
2
i
2
+
is
then
a
real
2
d
c
and
complex
you
know
number
by
a
how
real
to
divide
number.
- ad
+
c
denominator
2
Simplify
the
numerator,
real
imaginary
and
split
into
2
+
d
and
parts.
*
z
z
1
2
Notice that, since z
×
z* = |z|
 z
1
2

, we can write
2
z
2
z
2
Example 17
If
z
=
2
+
i ,
1
z
=
2
- 5i
and
z
2
=
-1
+
2i,
nd
the
following:
3
2
z
2z
z
1
a
z
z
2
your
answers
z
2 
with
i
a
z
 z
2
Check
2
c
b
3
 3z
1
1
- 3z
 z
1
3
3
calculator.
Multiply
2  5i
both
numerator
and
1
a
 3z

 3  1 

3
2i 
denominator
z
2  5i
by
the
conjugate
of
the
2  5i
2
denominator.
4
 10 i

2i

 5
 3  6i
Expand
4

and
simplify.
25
1  1 2i
 87  174 i

Write
the
expression
over
a
common
29
denominator,
and
simplify.
86  162i

29
Continued
on
next
page
169
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
2
2
2 
z
i
1
b

z
 z
2
 2  5i   1 
3
4
4i

2i 
 1
Expand

2 
3 
4i
 5i
 10
the
the
factors
numerator
in
the
and
multiply
denominator.
4i

8  9i
You
3 
could
multiply
top
and
bottom
by
4 i   8  9i 
the

conjugate
of
the
bottom.
Here,
we
2
*
8  9i
z
z
1
have
 z
1
used
2

2
z
z
2
2
24
 27 i
 32i
 36
Multiply
out
both
numerator
and

 81
64
denominator.
12 
60  5i
i
Simplify

145
2z
i
 3  2  5i 
2
Expand

z
 z
1
answer.
29
22 
 3z
1
c
the

3
4
2 
i   1 

 6  15i
2i
2i 
the
the
factors
numerator
in
the
and
multiply
denominator.

2 
4i

i

2
2  17 i
Simplify.

4
 3i
*
 2  17 i   4
 3i 
z
z
1
Using

 z
1
2

2
2
4
z
 3i
z
2
2
8  6i
 68 i
59  62i
 51

Multiply

16  9
and
simplify
Use
a
For
ease,
×
29
2
z3
2×z1–3×
z1×
z2
z3
numerator
store
and
expression.
to
all
check
three
then
the
results.
values
work
with
in
the
the
variables.
1
+
z2×
the
i
29
12
z1
in
162
z3
z2
the
calculator
memory
86
–3×
factors
25
Checking:
z1
the
29
29
59
62
25
25
×
i
×
i
Example 18
z
Find
z
∈
ℂ
that
satises
the
+
2
equation
1 -
Use
your
calculator
to
check
the
z
- 3i
=
i
2 +
i
solution.
170
/
z
+
2
z
- 3i
=
1 -
Multiply
i
2 +
(1
2)
×
(2
+
zi
+
4
-
z
⇒
(z
+
⇒
2z
⇒
2z
⇒
z
⇒
z(1
+
+
2zi
+
both
sides
of
the
equation
by
i
+
zi
=
+
2i)
2i
+
-7
=
i)
zi
-
-7
=
=
=
(z
z
-
-3i
-
zi
-
×
3i)
-
3
3i
-
(1
-
4
-
+
Multiply
Isolate
2i
i).
out.
the
terms
in
z
on
one
side
of
the
equation.
Simplify.
5i
-
i)(2
i)
3
-
-
arbegla dna rebmuN
3.2
Factorize.
5i
7  5i
z
Make

1 
z
the
subject.
2i
*
 7
z
 5i  1  2i 
z
1
Using
 z
1
2


2
2
1 
7  5i
z
z
2
2i
2
 14 i
 10

1 
4
17  9 i

5
Checking:
–17+9×
Use
i
–17
→z
5
z–3×
z+2
calculator
to
check
the
solution.
9
+
5
your
×
i
5
i
true
=
1–i
2+i
Exercise 3G
1
Given
the
complex
numbers
z
=
1
+
3
i ,
Find
the
real
numbers
a
and
b
that
satisfy
1
z
=
3
-
2i
and
z
2
=
-2
+
3i,
calculate
the
the
following:
3
following
and
check
your
answers
with
a
a
(1
+
3i)(a
+
bi)
=
5
+
5i
calculator.
a
z
a
×
z
z
1
+
+
bi
b
2
b
2z
2
z
1
*
=
-3 +
i
-
3
1 +
z
2i
5
3
z
c
z
-
3z
1
Find
the
complex
number
z
that
satises
3
d
z
2
4
z
1
2
the
*
3
following
equations.
z
2
*
2z
a
- 4z
1
2(z
+
i)
=
3i(z
-
1)
=
(z
-
+
2
2
e
*
z
z
3
z
-
2
b
2
Find
the
real
following
and
imaginary
complex
parts
of
1
b
i
2 +
i
(z
+
2i)(2
z
1 -
i
+ 1 +
d
i)
1)(1
-
i)
i
z
- 3i
=
1 - 4i
2i
+
2i
+
i
c
i
numbers.
2i
a
1 +
-
=
1 - 2i
the
c
1 +
z
2
2i
+ 5
1 - 2i
-
1 - 2i
1 +
2i
171
/
3
E X PA N D I N G
5
Given
the
b
nd
∈
,
such
THE
complex
the
NUMBER
number
relationship
z
=
S YS T E M :
a
+
C O M PL E X
8
bi , a,
between
a
and
NUMBERS
Prove
the
complex
b
that
following
properties
of
conjugate
numbers.
*
*
z
1 
a
(
b
(z
z
)
=
z
z
)
i
b
 
a
is
purely
imaginary.
*
2 
i
z *
+
1
*
=
z
2
*
6
Solve
a
|z|
for
+
z
z
=
∈
c
ℂ
×
(z
z
1
z
2
*
)
=
*
×
z
2
1
*
+
1
1
z
2
*
*


z
z
1
b
|z|
-
z*
=
1
d
i



*
z

2
c
z
+
z*
=
Prove
the
modulus
following
of
properties
complex
of
× z
1
|
=
|z
2
| × |z
1
+
z
1
|
≤
|z
2
|z*|
1
b
|
=
2
z
2
|z
=
z
z
c
|z|
numbers.
1
|z
2
the
z
a

2
e
7
z
2
|
+
|z
1
2
|
2
Powers and roots of complex numbers
In
order
to
raise
a
complex
number
to
a
power,
you
rst
need
to
TOK
investigate
what
happens
when
you
raise
the
imaginary
unit,
i,
to
a
power
ould we ever reach a
point where everything
impor tant in a mathe-
matical sense is known?
Investigation 9
Fill in the following table by calculating the powers of
n
0
1
1
i
2
3
4
5
6
i
7
8
9
n
i
1
Describe a pattern that your table shows.
2
Find a general rule for i
3
Use your general rule to nd i
4
Does your rule also apply for negative powers of
n
, n ∈ .
2019
5
Conceptual
i, that is, when n ∈ 
?
an you generalize the powers of imaginary numbers?
Now you will use your general rule to raise any complex number to an integer
power.
n
6
Let z = a + bi. In hapter 1 you learned the binomial expansion of (a + b)
n ∈
. Use the binomial expansion, and your results about powers of i from
2
question 2, to nd expressions for z
3
and z
4
7
,
Use the same method to nd (z
)* and (z*)
4
. What do you notice?
172
/
3.2
following
example
uses
some
higher
powers
of
a
complex
arbegla dna rebmuN
The
number.
Example 19
Given
the
answer
complex
with
a
number
z
=
2
+
i,
nd
5
3
a
expressions.
Check
your
(
5
z
c
)
(
z
)
3
z
=
(2
+
i)
3
=
these
*
*
b
z
of
calculator.
3
a
each
2
2
+
3
×
2
2
×
i
+
3
×
2
×
i
Use
the
binomial
Use
the
powers
theorem
for
n
=
3.
3
+
i
of
i.
Simplify
the
expression.
=
8
+
12i
-
6
5
-
i
=
2
+
11i
5
*
b
z
)

2
(
=
(2 -
i)
Use
the
binomial
Use
the
powers
theorem
for
n
=
5.
2
5
4
3
 5  2
 i 
 10  2
 i 
3
4
5
2

10  2
 i 
 5  2  i 

 i 
of
i.
Simplify
the
expression.
=
32
-
80i
-
80
+
40i
+
10
-
i
=
-38
-
41i
*
5
5
c
First
calculate
5
z
and
then
(z
)
5
z
=
(2
+
i)
5
4
=
2
+
=
32
5
×
2
i
3
+
10
×
2
2
2
i
+
10
+
10
×
2
3
i
4
+
5
×
2i
-38
+
41i
Use
the
binomial
Use
the
powers
theorem
for
n
=
5.
5
+
i
of
i.
Simplify
the
expression.
+
80i
-
80
-
40i
*
+
i
=
*
5

z


  38 
41i 

 38 
41i
Find
Use
the
a
conjugate
calculator
to
complex
check
number.
the
solutions.
Checking:
3
(2+i)
2+11×
i
–38–41×
i
–38–41×
i
5
i)
5
(2–i)
You can nd square roots of complex numbers algebraically by letting the
root be equal to a complex number a + ib. You then square this, and equate
real and imaginary par ts.
You can use a similar technique for nding higher roots of complex numbers,
such as cube roots or four th roots.
Example
20
will
show
you
how
to
do
this.
173
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Example 20
Evaluate
Let
z
=
7 - 24 i
a
+
bi , a, b
∈

Write
z
in
Cartesian
form.
2
z

7 
24 i

a
 bi 

7  24 i
Square
both
sides
and
expand
the
product.
2
⇒
2
a
+
2abi
-
b
=
7
-
144

2
a
2
a

2

7


7
2

a
Equate

2ab


 b


24i
parts
and
imaginary
parts.
12
24

b


Solve


4
a
2
2
 7a
 144
simultaneous
equations
by
using
a
the

method
of
substitution.
2

a
0


real

 9
 a
 16


0



12
b


Factorize


b
a

the
expression.
Since
a
is
12

2

real,
we
discard
(a
+
9)
=
0.
a

2
a
a
 16




A
4



For
-
calculator
can
will
=
-4
also
only
+
3i
be
values
of
a
nd
two
values
of
b
give
you
to
evaluate
one
this
square
root,
both
but
solutions.
the
solution.
4-3i
You
have
to
that
is
+
3i.
saw
in
-4
you
requires
or
Find
used
i
As
two
 3
4

3iorz
calculator


a
4
12
b


=

12
b

z


know
that
the
Example
algebraic
skill
20,
in
second
solution
nding
solving
square
is
the
roots
simultaneous
opposite
of
number,
complex
equations
of
numbers
degree
two
higher.
Exercise 3H
2018
1
Calculate
3i
2019
+
2i
HINT
d
7
a
i
17
+
i
27
+
i
2020
37
+
i
4i
2021
- 3i
Similar to ∑-notation
173
b
i
272
-
i
351
+
i
766
-
2019
1010
i
k

77
c
(3
+
i
)
2k
i

93
×
(1
-
2i
1
for a sum there is
i
a product notation
)
k 1
k 1
e
f
2019
1010
n
k

k 1
i
2k

k 1
i
k

i
1

i
2
 i
n
 ...  i
k 1
174
/
3.3
Calculate
check
the
your
following
answers
4
and
with
Show
a
that
n
n
a
|z
b
(z*)
|
=
arbegla dna rebmuN
2
+
|z|
,
n
=
(z
∈

calculator.
n
n
*
+
)
,
n
=
1
∈

3
a
(3
+
2i)
b
(1
-
3i)
5
Given
that
z
+
i,
nd
the
4
+
values
4
c
(1
-
of
n
∈

such
that
4
2i)
+
(1
+
2i)
n
5
d
(1
+
i)
a
z
b
z
is
real
is
purely
5
+
(1
-
i)
n
3
Evaluate
the
following
and
6
check
your
answers
imaginary.
with
Show
that
a
2n
calculator.
a
(1
-
i)
b
(1
-
i)
n
=
(-2i)
,
n
∈

2n+1
a
b
i
n
5
c
-21 +
d
20i
(1
-
i)(-2i)
,
∈

i
-
36
3.3
n
=
-i
3
Polynomial equations and
inequalities
Investigation 10
n
1
Draw the graphs of f
(x) = x
, n
∈
, up to at least n = 7
n
2
Conceptual
How could you classify these graphs by their shapes?
3
Conceptual
How could you classify these graphs by their symmetrical
proper ties?
Polynomials are functions which map a real variable, often called
real number. We write f :
→

.
n
Polynomials are functions of the form
f(x)
= a
x
n
+ a
n
∈ , i = 0,
where a
...,
n
x, to another
are called the
coecients.
x
n
1
+ ... + a
1
x
+ a
1
,
0
The highest power
i
(n)
of the variable
x
is called the
degree of the polynomial, and we write
= n
deg(f )
A linear combination of two functions f and g is an expression of the form
a
×
f(x) + b
×
g(x), where a and b are some real numbers.
n
A linear combination of n functions is an expression of the form

a
i

f
i
 x ,
i 1
where f
i
are functions and a
∈ .
i
175
/
3
E X PA N D I N G
In
general,
you
THE
can
say
NUMBER
that
2
power
functions
{1,
x,
x
polynomials
3
,
x
S YS T E M :
4
,
x
are
C O M PL E X
linear
NUMBERS
combinations
of
the
Did you know?
5
,
x
,
...}.
There is still a lot of
5
By
taking
a
linear
combination
of
such
powers,
for
example
3
×
x
-
2
8x
-
11.
×
discussion going on among
2
x
5
+
8
×
x
-
11
×
1,
you
obtain
a
polynomial
f(x)
=
3x
2
-
2x
+
mathematicians regarding
Polynomial
most
functions
frequently
different
so
degrees
are
far.
that
the
Here
you
functions
are
have
some
that
you
examples
learned
have
of
worked
the zero polynomial.
with
polynomials
Some consider that the
of
zero polynomial is more
about.
than just a special case
•
Zero
θ(x)
polynomial:
=
0.
Its
graph
is
a
horizontal
line
along
y
=
0
of the constant function.
(the
x-axis).
The degree of the zero
•
•
Constant
=
polynomial:
line
y
The
degree
f(x)
=
c ∈ ,
c,
c
≠
0.
Its
graph
is
the
horizontal
c
polynomial is sometimes
dened as -1 or even -∞
of
a
constant
polynomial
is
0.
y
HINT
3
f (x)
=
2
is
a
constant
polynomial
Notice that the notation
used for the zero
1
i(x)
=
0
is
the
zero
polynomial is θ(x) = 0 so
polynomial
x
that you can distinguish
0
–5
–4
–3
–2
–1
1
2
3
4
5
–1
it from other polynomials.
The zero polynomial has
•
Linear
polynomial:
f(x)
=
mx
+
c,
m
≠
0,
is
also
called
a
polynomial
an impor tant proper ty,
of
the
rst
degree.
The
graph
is
c
change
a
straight
line.
By
changing
the
being an additive identity
parameters
m
and
we
the
steepness
of
the
line
and
the
element for the set of
y-axis
intercept,
respectively.
polynomials, that is
y
f(x) + θ(x) = θ(x) + f(x) =
6
f(x) for all polynomials f
5
4
1
f1(x)
×
=
x
+
2
3
3
1
x
0
–5
–1
y
–2
6
–3
5
–4
4
–4
f2(x)
×
=
–5
x
+
1
2
f2(x)
3
=
–3
×
x
+
2
×
x
+
4
3
–6
•
2
Quadratic
of
the
function:
second
symmetry
is
f(x)
degree.
a
Its
vertical
=
ax
+
graph
line
bx
is
which
+
a
c, a
≠
0,
is
parabola
passes
a
polynomial
whose
through
axis
the
of
x
vertex
–1
of
the
graph.
By
changing
the
parameters
a,
b
and
c,
we
–2
change
–3
❍
the
shape
of
the
❍
the
concavity
❍
the
position
graph
(wide
or
narrow)
–4
(open
of
the
upwards
parabola
or
downwards)
relative
to
the
coordinate
–5
1
f1(x)
axes.
2
×
=
x
–
3
×
x
–
1
2
–6
176
/
3.3
3
Cubic
the
third
looks
The
function:
degree.
like
a
shape
up
and
=
+
are
shape
is
the
2
ax
There
“Flex“
second
concave
f(x)
a
bx
+
two
like
cx
is
d, a
shapes
those
combination
other
+
of
you
of
concave
≠
0,
is
a
cubic
met
two
in
polynomial
graphs.
One
shape
Investigation
U-shapes,
one
of
of
arbegla dna rebmuN
•
10.
which
is
down.
y
y
6
6
5
5
4
4
3
3
2
–1
f2(x)
3
×
=
x
2
×
x
2
1
2
+
×
+
2
x
–
2
2
1
1
x
x
0
0
3
f2(x)
2
f1(x)
3
×
=
x
=
–x
2
–
6
×
x
–
12
×
x
–
6
2
–
2
×
x
+
2
×
x
+
1
3
–3
–3
–4
–4
3
f1(x)
=
x
2
+
4
×
x
–5
–5
–6
–6
+
3
×
x
–
2
International-
Investigation 11
mindedness
2
In hapter 2, you learned that a quadratic equation
f(x) = ax
+ bx + c, a ≠ 0,
The Sulba Sutras in

b
has axis of symmetry
x
=
and ver tex
-

2a

b

b

,
.

f

2a

ancient India and the

2a


akhshali manuscript
In this investigation, you will look at how the parameters of a cubic function
contained an algebraic
aect the graph of the function.
formula for solving
3
onsider the cubic function f(x) = ax
1
2
+ bx
+ cx + d, a ≠ 0
quadratic equations.
Using graphing software, investigate the eect of changing
a. What can
you conclude about the eect that a has on the graph?
2
What eect does changing d have on the graph of the function?
3
Investigate the eect of changing the parameter
3
f(x) = x
b in the graph of
HINT
2
+ bx
. How does the shape of the graph change as
b changes? Do
With the help of graphing
all graphs of this nature have any common points?
technology, you might
4
Investigate
the
effect
of
changing
the
pa ramet er
c
in
the
graph
of
3
f(x) = x
+ cx. How does the shape of the graph change as
c changes?
Do all graphs of this nature have any common points?
want to investigate how
the parameters of a
four th-order polynomial
aect the shape,
5
Conceptual
How do the parameters in quadratic and cubic functions
symmetry, ver tices and
aect their graphs?
intercepts of the graph.
177
/
3
E X PA N D I N G
One
interesting
same
leading
different
THE
feature
of
coefcient
you
can
nd
NUMBER
a
S YS T E M :
polynomials
is
that
larger
even
of
the
though
window
in
C O M PL E X
same
locally
which
NUMBERS
degree
they
they
with
might
look
the
look
very
y
very
similar.
y
6
50
5
3
f2(x)
=
x
2
+
3
×
x
+
3
×
x
+
40
2
3
f2(x)
4
=
x
2
+
3
×
x
+
3
×
x
+
2
30
3
3
3
f1(x)
=
x
f1(x)
2
+
x
–
x
–
1
=
x
2
+
x
–
x
–
1
HINT
20
2
Polynomials are
10
continuous functions,
x
x
which means that
0
–1
you can draw their
graphs without lifting
–2
your pen from the
–3
paper. Notice that
–4
you always have
–5
to proceed in one
direction (usually you
–6
draw them from left
They
behave
terms
and
for
can
like
a
polynomial
extremely
be
large
neglected.
that
values
This
is
the
of
has
x
only
are
the
leading
insignicant
so-called
“end
to
term,
the
behaviour”
since
total
to right). Their graphs
other
are also smooth
value
property
therefore they have
of
no sharp points.
polynomials.
The factor and remainder theorems
In
order
to
divide
to
divide
two
two
polynomials
numbers
by
using
you
long
need
to
remind
yourself
how
division.
Example 21
4
Use
all
the
steps
in
the
long
division
of
two
numbers
to
divide
3
3x
+
2
4x
+
2x
+
3x
-
1
by
2
x
+
x
-
2.
2
3x
4
2
+ x - 2
x
3x
3
+7
+3x
-1
4
Dividing
3x
Multiply
x
+ 3x
Dividing
x
- 2x
Multiply
x
Dividing
7x
Multiply
x
2
by
x
2
gives
3x
+2x
2
4
.
2
+4x
3x
+x
3
2
+3x
-6x
3
+8x
3
2
x
x
-
3
2
x
2
+
2
by
3x
and
subtract.
2
by
x
gives
x
2
+x
+
x
-
2
by
x
and
subtract.
2
7x
2
+ 5x
-1
+ 7x
-14
2
by
x
gives
7.
2
7x
-2x
2
can
x
-
2
by
7
and
subtract.
+13
Stop
You
+
is
write:
when
smaller
the
than
degree
the
of
the
degree
of
remainder
the
divisor.
4
3x
3
+
4x
2
+
2x
2
+
3x
-
1
=
(x
+
2
+
x
-
2)(3x
(-2x
+
13)
+
x
+
7)
178
/
arbegla dna rebmuN
3.3
Theorem
For any two polynomials f and g, there are unique polynomials q and r such
that f(x)
=
g(x)
+
q(x)
×
r(x) for all real values of x.
The polynomial q is called the quotient and the polynomial r is called the
remainder. Notice that deg(g) > deg(r)
Exercise 3I
1
Use
long
division
polynomial
f(x)
=
2x
b
f(x)
=
3x
divide
polynomial
f
2
by
Use
g
long
division
remainder
3
a
to
5x
+
x
when
polynomial
2
+
-
11x
+
4andg(x)
=
x
+
4
3
+
2
6x
+
5x
+
=
+
quotient
polynomial
f
is
and
divided
by
g.
a
f(x)
=
3x
b
f(x)
=
2x
2
-
6x
-
3x
4
x
the
+
x
-
+
4x
1andg(x)
=
x
-
1
6and
2
g(x)
nd
4
3
5
to
2
3
2
+
x
-
5and
2
6
c
f(x)
=
5
x
-
x
4
+
3
2x
+
x
3
g(x)
2
x
-
5x
+
2x
-
=
x
2
andg(x)
=
+
x
5
x
-
+
+
2
6
c
f(x)
=
x
4
+
x
3
+
1andg(x)
=
x
+
1
Did you know?
Polynomial remainder theorem
The polynomial
n
Given a polynomial f(x)
=
a
n
x
+
a
n
i
=
1,
2,
...,
n,
a
≠
1
2
x
n
+
...
+
1
a
x
+
a
2
0, and a real number
x
+
a
1
,
a
0
∈
,
i
remainder theorem is
p, then the remainder when
f(x)
also known as “Little
n
is divided by a linear expression (x
p) is f(p)
-
ézout’s theorem”
as it was proposed
by Étienne ézout
Proof:
(1730–1837). ézout
In
the
unique
decomposition
of
the
polynomial
f(x)
=
(x
-
p)
×
q(x)
+
r,
was inspired by the
where
the
remainder
r
is
a
constant
(one
degree
less
than
the
divisor),
work of Euler and
we
input
x

p

f (p)  (p

p) q(p)  r

f (p) 
r .
decided to become



0
a mathematician. In
1763 he was appointed
examiner of the Gardes
Factor theorem
de la Marine (French
n
A polynomial
f(x)
=
a
x
n
+
a
n
1
x
n
2
+
...
+
a
1
x
+
2
a
x
+
a
1
,
0
a
∈ℝ,
Naval Academy)
k
with a special task to
k
=
0,
1,
2,
...,
n,
a
≠
0,
has a factor
(x
-
p),
p∈ℝ
,
if and only if
f(p)
=
0.
n
compose a textbook for
teaching mathematics
The
factor
and
the
theorem
proof
is
Mathematician
that
at
value
given
a
as
William
algorithm
a
left
is
direct
an
exercise
George
simplies
of
consequence
the
for
Horner
process
of
you
want
select
the
them
in
to
remainder
theorem
to the students.
(1786–1837)
nding
the
formalized
value
of
a
an
polynomial
x
nd
the
coefcients
tabular
the
you.
3
If
of
value
of
all
of
f(x)
terms,
=
3x
2
–
2x
including
–
5x
missing
–
1
when
terms,
x
and
=
2,
organize
form.
179
/
3
E X PA N D I N G
3
–2
–5
–1
+
+
+
6
8
6
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
HINT
f(x)
=
((3
×
x
–
2)
f(2)
=
((3
×
2
–
2)2
=
((6
–
2)
=
(8
=
3
×
2
=
6
–
l
=
5
×
(x
–
–
5)
5))
×
x
×
–
l
2
–
1
2×
3
4
Horner’s
3
polynomial
f(x)
also
as
works
Since
a
when
you
factor
searching
is
for
nds
the
divided
synthetic
synthetic
linear
2
–
5)
×
2
–
1
5
algorithm
known
×
divide
a
division
g,
it
is
an
factors
of
by
–
5)
–
quotient
a
linear
division .
us
the
effective
2
–
l
1
and
remainder
polynomial
Notice
polynomial
tells
×
by
a
that
and
quick
g(x).
This
synthetic
linear
quotient
when
a
algorithm
division
is
only
polynomial.
when
f
method
is
to
divided
use
by
when
f
Example 22
Use
synthetic
a
(x)
division
3
f
a
x

=
3x
2

to
nd
the
remainder
when
dividing:
2
x
3
+ 11x
- 7x

 2 

11
–7
–19
+
+
+
–6
–10
34
5
–17
15
3
r
- 19

f
by
 2 
g ( x)
=
x
+
2,
b
f
(x)
=
4x
- 3x
+ 8 by
Use
the
Use
synthetic
g ( x)
remainder
=
x
- 5
theorem.
division
to
nd
the
remainder.
–2×
3
r
x
b

f
 5
 2 
 15


r
4
f
5 
0
–3
8
Use
the
remainder
Use
synthetic
theorem.
division
to
nd
the
remainder.
+
+
+
20
100
485
20
97
493
5×
4
r
=
f
(5)
=
493
180
/
Reect
arbegla dna rebmuN
3.3
What do you notice about the quotient polynomial q(x) and the result
you obtained from synthetic division in the example above? an you use the
remainder theorem to explain why this is the case?
Example 23
3
The
polynomial
a
Use
b
Hence
f(x)
synthetic
fully
=
2
2x
+
division
factorize
x
to
-
15x
show
the
-
18
is
given.
(x
-
3)and(x
that
+
2)
are
Use
1
of
the
polynomial.
polynomial.
a
2
factors
–15
synthetic
division
twice:
rst
to
18
(x
3),
–18
3
divide
+
+
+
6
21
18
7
6
0
+
+
–4
–6
3
0
then
2x
to
2
+
x
-
divide
15x
the
-
by
quotient
by
-
(x
+
and
2).
3×
2
–2×
2
Since
and
the
(x
+
3
b
2x
remainder
2)
are
is
factors
zero
of
in
the
each
case,
(x
-
3)
Use
factor
theorem.
polynomial.
2
+
x
-
15x
-
18
In
=
the
(x
-
3)(x
+
2)(2x
+
the
synthetic
division
table,
you
can
3)
see
the
last
factor
is
the
second
quotient.
Exercise 3J
4
1
In
each
the
case,
quotient
polynomial
use
and
f
is
synthetic
the
division
remainder
divided
by
to
2
nd
when
The
f(x)
=
x
b
f(x)
=
2x
c
f(x)
=
+
2x
3
3x
2x
-
6x
x
4
-
-
f(x)
=
3x
andg(x)
3x
+
2
-
32x
+
31x
+
60
is
g
1andg(x)
-
10x
-
9x
3
5
d
x
Show
that
(x
+
1)and(x
of
polynomial.
-
3)
are
factors
=
x
-
2
the
2
+
4
=
3
-
2
a
g(x)
2x
given.
a
3
polynomial
-
=
20x
x
+
=
x
+
3
b
Hence
fully
factorize
the
polynomial.
2
+
4
+
2andg(x)
7x
-
11and
3
+
10x
2
-
13x
+
7x
-
3
6
181
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Corollary
n
Given a polynomial f(x) = a
n
+ a
x
n
k = 1,
1
2
+ ...
x
n
+ a
1
+ a
x
2
x + a
1
, a
0
∈,
k
≠ 0, and real numbers a and b, a ≠ 0, then the remainder when
2, ..., n, a
n
b
f(x) is divided by a linear expression (ax - b) is f
a
The
proof
is
left
to
you
as
an
exercise.
HINT
In
order
to
use
synthetic
division
when
dividing
by
a
linear
expression
Use the same
(ax
-
b)
we
need
to
modify
the
algorithm.
technique we
f(x)
=
(ax
-
b)
×
q(x)
+
used to prove
r
the remainder
b


f ( x) 
a
x




a

b


f ( x) 
x


a

So,
in
when
the
a  q  x   r

applying
quotient
theorem.



 q  x   r

Horner’s
will
have
a
algorithm,
common
notice
factor
of
that
all
the
coefcients
a
Example 24
Use
synthetic
f(x)
=
3
division

nd
the
quotient
and
remainder
-
7x
-
15x
3x
 1 
3

6
+
1
x

by

=
3x
+
dividing
1.
Rewrite
 

3

–7
g(x)
1  


the
1
the
divisor
remainder
so
that
you
can
use
theorem.
 
–15
1
Use
–
when
2
6x

g( x )
to
+
+
+
–2
3
4
–9
–12
5
Horner’s
algorithm.
×
3
6
2
So
the
quotient
remainder
3
6x
is
is
r(x)
q(x)
=
5.
1
=
=
2
-
7x
2x
-
3x
-
4
and
Divide
the
in
the
the
coefcients
table
by
of
the
remainder
3.
2
-
15x
+
(2x
-
3x
-
4)(3x
+
1)
+
5
182
/
arbegla dna rebmuN
3.3
Example 25
3
When
Find
x
+
polynomial
the
2
value
⇒
r
=
of
f(x)
=
x
2
+
3x
-
ax
-
5
is
divided
by
+
2)
the
remainder
is
3.
a
f(-2)
1
(x
3
–a
Use
the
remainder
Use
synthetic
+
–2
–2
division
to
nd
the
–5
remainder
+
theorem.
in
terms
of
a
+
2a
+
4
2a
–
1
–2×
1
2a
-
1
1
=
3
⇒
–a
a
=
–
2
Equate
2
the
the
table
remainder
with
3,
and
you
solve
obtained
for
in
a.
Example 26
2019
Find
the
remainder
when
polynomial
f(x)
=
3x
1019
+
5x
-
7x
+
4
is
divided
2
by
x
-
1.
2
f
 x 

Write
x
the
unique
the
polynomial
remainder
2019
=
1
⇒
x
=
-1
f(1)
=
3
×
1
5
×1
-
7
×
1
+
4
=
5
Calculate
at
f(1)
=
3
×
(-1)
+
4
=
is
a
and
notice
linear
that
the
polynomial.
1019
+
2019
⇒
of
 b
  q  x   ax

r  x 
x
decomposition
 1
the
the
zeros
value
of
the
of
the
polynomial
divisor.
1019
+
5
×
(-1)
-
7
×
(-1)
3
2

f
1

f
 1
1
 1
  q 1  a  1  b
Use
 5
the
unique
decomposition
of
the

polynomial

and
form
simultaneous
2



 1
 1

 q  1  a 
 1
 b
 3
equations.

a
 b

5
b



a

b

3
Therefore
r(x)
=
x
+
the

4

a

Solve
the
simultaneous
write
the
remainder.
equations
and
 1
remainder
is
4.
183
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Exercise 3K
5
1
In
each
the
case,
use
quotient
polynomial
and
f
is
f(x)
=
4x
b
f(x)
=
6x
c
f(x)
=
2x
the
5x
+
11x
6
nd
when
+
+
3x
13x
f(x)
is
g
2andg(x)
=
4x
-
1
+
4andg(x)
=
2x
+
3
+
4x
(x
2x
+
5x
3
-
=
g(x)
3x
=
7
When
6x
4
-
3x
the
1)
by
(2x
the
a
-
30x
-
1)
2
-
and
remainder
of
and
When
polynomial
45x
+
ax
when
is
-40.
+
b
divided
Find
the
b
f
is
divided
by
(x
+
2)
the
(x
-
2
-
+
3and
remainder
+
3
5x
+
x
is
4,
and
when
divided
by
5)
remainder
is
-3.
Find
the
remainder
when
2
-
2x
+
8x
+
2
2
and
polynomial
f
is
divided
by
(x
-
3x
-
10).
1
8
2
-
3
13x
1
5
f(x)
+
values
the
d
4
6x
divisible
by
-
=
2
4
=
by
to
2
-
3
g(x)
division
remainder
divided
3
a
synthetic
polynomial
f
is
divided
Find
the
remainder
2019
by
f(x)
=
x
when
2018
+
x
+
...
+
x
+
1
is
divided
by
2
the
polynomial
g(x)
=
2x
-
3x
+
1
you
2
and
(x
+
1).
2
obtain
the
quotient
q(x)
=
x
+
2n
9
the
remainder
r(x)
=
x
-
3.
Find
Show
that
the
polynomial
f(x)
n
(x
polynomial
=
(x
+
2)
+
the
+
3)
2
-
1
is
divisible
by
(x
+
5x
+
6)
for
all
f
+
5
3
f(x)
=
6x
4
+
divisible
3
17x
by
(x
-
+
20x
2).
n ∈ 
2
-
Find
35x
the
+
44x
value
+
of
a
.
is
n
a
10
Given
a
polynomial
f(x)
=
a
x
n
+
a
n
1
x
n
+
...
1
2
4
4
f(x)
=
2x
3
+
5x
2
-
4x
+
bx
+
1
is
divisible
+
by
a
x
+
2
(2x
+
1).
Find
the
value
of
b
a
≠
a
x
+
a
1
0,
and
,
0
real
∈ ,
a
k
=
1,
2,
...,
n,
k
numbers
aandb,
a
≠
0,
show
n
4
5
f(x)
(x
-
=
x
1)
3
+
5x
and
(x
2
+
5x
+
2).
+
ax
Find
+
b
the
is
divisible
values
of
that
by
a
and
the
remainder
when
f(x)
is
divided
is
f
by
b
b
the
linear
expression
( ax
-
b)
a
Developing inquiry skills
Return to the opening problem.
an you use a polynomial to model the roller coaster?
3.4
The fundamental theorem of algebra
Investigation 12
1
Factorize the following numbers into prime factors. 30, 504, 1155, 35 200
2
Fully factorize the following polynomials.
2
x
3
3
+ 10x + 25, x
Conceptual
2
- 2x
3
- 15x, x
2
- 3x
4
+ 3x - 1, x
2
- 5x
+ 4 .
What are the similarities and dierences between factorizing
a number into prime factors, and factorizing a polynomial expression?
184
/
3.4
fundamental
theorems
theorem
or
is
The
the
in
of
theorem
mathematics.
arithmetic
product
of
fundamental
zeros
of
a
a
of
It
algebra
is
(which
unique
theorem
an
is
one
algebraic
states
that
algebra
the
of
refers
most
version
every
combination
of
of
of
the
number
prime
to
important
the
arbegla dna rebmuN
The
fundamental
is
either
prime
factors.)
existence
of
complex
polynomial.
The fundamental theorem of algebra (F TA)
n
A polynomial
f ( x)

a
n
x
 a
n
1
x
n
2
 ...  a
1
x
 a
2
x
 a
1
or complex coecients has at least one zero. There is an
f
( )
Gauss
=
,
a
0
0 , with real

n
 
such that
0
proved
this
theorem,
but
the
proof
goes
beyond
the
level
of
this
textbook.
There
from
are
the
a
lot
of
theorems,
fundamental
manipulation
of
lemmas
theorem
equations
and
of
and
corollaries
algebra
that
polynomial
can
that
help
are
derived
with
algebraic
functions.
Lemma
n
Each polynomial
f ( x)

a
x
n
 a
n
x
n
1
2
 ...  a
1
x
 a
2
x
 a
1
, a
0

0,
n
with real coecients can be written in a factor form,
f
( x )
=
a
n
( x
- 
1
)( x
- 
) ... ( x
- 
f(x)
x
2
n
), such that

 , k
 1, ..., n
24x
a
k
Example 27
4
Express
check
the
your
polynomial
answer
4
f(x)
by
3
=
x
-
=
x(x
6x
=
using
your
2
-
19x
+
3
-
24x
6x
2
-
19x
+
as
Use
linear
factors,
and
the
the
lemma
terms
to
factorize
have
the
the
polynomial.
common
factor
x,
so
2
-
6x
-
19x
+
24)
you
1
of
calculator.
All
3
product
–6
–19
24
+
+
+
can
Notice
of
the
factorize
that
–5
x
–24
=
1
sum
remaining
therefore
1
the
to
apply
this
rst.
of
the
cubic
expression
synthetic
factorize
it
coefcients
division
further.
(x
-
is
0,
for
1)
is
a
1×
factor
1
–5
–24
2
x
since
the
remainder
is
0.
0
2
- 5x
- 24
=
x
-8 x
+ 3x




- 24
Factorize
the
remaining
quadratic
5x
expression
by
splitting
the
middle
Continued
term.
on
next
page
185
/
3
E X PA N D I N G
=
x(x
f(x)
-
=
8)
x(x
+
-
THE
NUMBER
(x
-
-
8)
3(x
-
8)
=
1)(x
+
3)(x
8)(x
S YS T E M :
+
C O M PL E X
Use
3)
NUMBERS
all
Make
the
sure
calculator
factors
you
to
in
the
know
check
factor
how
the
to
form.
use
your
zeros.
Checking:
4
x
3
–6×
x
2
–19×
x
+24×
x, x
–3, 0, 1, 8
Notice
write
If
a
that
the
a
calculator
polynomial
certain
factor
multiplicity.
n
different
gives
in
the
appears
So
if
zeros,
we
then
zeros
factor
more
have
the
a
and
you
than
of
to
use
the
FTA
to
form.
once
polynomial
sum
need
their
we
f
say
of
that
nth
the
degree
multiplicities
will
factor
with
add
has
less
up
than
to
n
17x
-
k
p
p
1
f
 x 

a
 x
n
 
1

p
2
x
 
2

k
...  x
 
k

,
k

n,

p

n
k
i 1
Example 28
5
-1
is
a
zero
Express
Check
of
f(x)
your
2
as
the
the
polynomial
product
answers
by
of
using
f(x)
=
linear
a
2x
4
+
7x
3
+
3x
2
-
13x
-
3
–13
–17
–6
+
+
+
+
+
2
11
6
5
–2
–11
–6
0
+
+
+
+
–2
–3
5
6
3
–5
–6
0
+
+
+
–2
–1
6
1
–6
0
multiplicity
3.
factors.
Successively
apply
with
to
respect
given
–5
with
calculator.
7
–2
6
zero,
in
synthetic
the
this
division
multiplicity
case
three
of
the
times.
–1×
2
–1×
2
–1×
2
186
/
2
2
2x
+
x
- 6
=
2x
-3 x
+
4 x
- 6
Factorize




the
remaining
quadratic
x
expression
by
splitting
the
middle
term.
=
x(2x
-
3)
+
2(2x
-
3)
=
(2x
-
3)(x
+
2)
3
f(x)
=
(x
+
1)
(2x
-
3)(x
+
Use
all
Checking:
Use
your
Notice
5
x
4
+7×
x
3
+3×
x
the
factors
in
the
factor
form.
2)
2
–13×
x
–17×
form,
x
that
so
obtain
1
calculator
the
you
the
last
need
linear
to
check
zero
to
is
the
in
zeros.
arbegla dna rebmuN
3.4
fraction
multiply
it
by
2
to
factor:
–2, –1, –1, –1,
2
3 

2
x




2

 2x
 3

Did you know?
Investigation 13
Due to the imperfection
1
Use the remainder theorem and synthetic division to show that the
of the calculator ’s
complex number z is a zero of the polynomial f
algorithm when nding
3
2
a
z = i, f(x) = 2x
b
z = -4i, f(x) = 2x
c
z = -2 + 3i, f(x) = 3x
- 3x
zeros, you obtain app-
+ 2x - 3
3
roximations of the
2
- x
+ 32x - 16
multiple zeros, without
3
2
+ 10x
+ 31x - 26
their multiplicity. This
4
3
d
z = 1 + i, f(x) = x
+ x
e
z = -1 + 3i, f(x) = x
2
- 2x
5
+ 2x + 4
4
+ x
3
+ 9x
will occur even more
2
- 9x
times when complex
+ 8x - 10
zeros occur.
2
Use the remainder theorem and synthetic division to nd the values of the
polynomial f at z* for all the complex numbers given in question 1
3
Conceptual
If z is a zero of a polynomial f, what can you say about the
TOK
complex conjugate z*?
Aliens might not be
4
y considering the fundamental theorem of algebra and your
Factual
able to speak an Ear th
answer to question 3, what can you say about the zeros of an odd-degree
language but would
polynomial?
they still describe the
equation of a straight line
in similar terms?
Is mathematics a formal
Your
ndings
from
Investigation
13
can
be
summarized
by
the
language?
following
theorem.
Conjugate root theorem
Given a polynomial
n
f ( x)

a
x
n
n
 a
x
n
1
1
2
 ...  a
x
2
 a
1
x
 a
,
0
a

, k

1
,
2, ..., n, a
k

0,
n
*
that has a complex zero z, then its conjugate z
is also a zero of the
polynomial f
187
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Example 29
3
Given
the
that
other
Check
=
your
1
+
+
3i
zeros
Method
x
1
is
of
a
complex
zero
of
the
polynomial
answers
using
a
⇒
x
x
2
-
5x
+
16x
-
30,
nd
all
calculator.
Use
1
=
f
1
3i
f(x)
=
1
-
3i
is
(1
-
3i))
+
3ix
a
zero
of
the
conjugate
zero
theorem.
f
2
(x
-
=
x
(1
+
3i))(x
-
2
2
-
x
-
3ix
-
x
+
2
1
+
3
2
=
x
-
2x
+
10
Expand
Use
x
and
long
nd
division
the
to
quadratic
nd
the
factor.
remaining
-3
factor.
3
2
x
- 2x + 10
x
2
16x
-5x
3
-30
2
-x
-2x
+ 10x
2
0
6x
-3x
-30
2
f(x)
x
-
=
3
(x
=
-
0
Method
(1
⇒
x
+
3i))(x
=
-3x
+ 6x
-30
0
0
0
-
(1
-
3i))(x
-
3)
Fully
factorize
the
remaining
Use
synthetic
2
numbers
–5
16
–30
+
+
+
1
+
+
3i
–13
–4
+
3i
3
–
–
+
1
–
1
nd
zero.
9i
division
for
complex
twice.
30
–
9i
0
+
3i
–3
+
9i
3i)×
1
f(x)
=
(x
-
–3
(1
+
0
3i))(x
-
(1
-
3i))(x
-
Fully
3)
the
x
and
3i)×
1
(1
polynomial
3
1
(1
the
-
3
=
0
⇒
x
=
factorize
remaining
the
polynomial
and
nd
zero.
3
Make
sure
that
you
know
how
to
use
Checking:
your
3
x
calculator
to
check
your
answers.
2
–5×
x
+16×
x–30, x
1–3×
i, 1+3×
Notice
i, 3
roots
that
of
you
need
to
use
complex
polynomials.
188
/
Example 30
4
The
polynomial
a
Find
b
Hence
the
f(x)
value
nd
all
=
of
a
the
3
x
-
2
x
+
given
2x
that
remaining
+
2i
ax
is
-
a
zeros
24, a∈ℝ,
complex
of
f,
and
is
given.
zero
check
a
of
f.
your
answers
using
Apply
2
–1
2
a
+
4i
+
–2i
–
8
–6
–
2i
0
⇒
a
is
a
–12i
calculator.
synthetic
division
for
–24
the
+
a
complex
number
arbegla dna rebmuN
3.4
2i.
+
+4
(8
+
2a)i
+
24
×2i
2
f(2i)
b
x
=
–1
=
(8
+
2i
⇒
x
1
+
4i
2a)i
=
=
-2i
(4
=
+
a)
–
12i
(8
+
2a)i
-4
root
of
Use
the
remainder
Use
the
conjugate
theorem.
f
2
zero
theorem.
2
–1
+
4i
–6
–
2i
–12i
Use
+
+
a
=
-4
and
apply
synthetic
+
–4i
2i
12i
–1
–6
0
division
a
second
Identify
the
time.
×(–2i)
2
2
q(x)
=
2x
-
x
-
6
=
0
⇒
0orx
-
2
=
(2x
+
3)(x
x
=
-
2)
=
0
its
3
⇒
2x
+
3
=
0
⇒
-
orx
=
quotient
and
nd
zeros.
2
2
Checking:
Use
your
your
4
x
3
x
calculator
to
check
answers.
2
+2×
x
–4×
x–24, x
–3
, –2×
i, 2×
i, 2
2
189
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Exercise 3L
1
Find
a
polynomial
of
integer
coefcients
a
and
the
smallest
whose
zeros
degree
5
with
are
In
each
zero
z.
case,
Find
f
is
the
3
0,
2
b
7
-3,
-2,
1
and
3
a
f(z)
=
z
b
f(z)
=
2z
a
remaining
c
1,

,
and
2
with
zeros.
2z
+
4z
-
8,
z
=
2i
2
-
13z
+
32z
-
13,
z
a
polynomial
integer
3
-
1
3
c
Find
=
2i
5
2
2
complex
2
-
3
1
polynomial
of
coefcients
the
smallest
whose
zeros
degree
f (z )

3z
3
2

4z
 4z
 7,
z


i
2
with
include
4
d
f(z)
=
z
e
f(z)
=
z
3
-
2z
+
5z
2
2
+
6z
-
+
11z
2z
+
5,
z
=
-i
1
4
3
a
2
and
b
2
1,
and
3
3
2
+
13z
+
10,
2
z
=
-2
-
i
3
c
1 -
3
and
2
4
3
Factorize
check
the
your
following
answers
polynomials
using
a
f (z )
f
and
calculator.
f(x)
=
x
b
f(x)
=
3x
-
3x
3
+
4x
-
2
4
In
each
of
the

 19z
x
+
2x
+
6
3
2x
-
case,
i
3
5x
k
is
Given
that
z
is
a
zero
of
+
a
11x
zero
-
3x
with
-
5
nd
multiplicity
all
n
the
the
a
fully
and
check
your
answers
using
calculator.
using
a
zeros
k
=
-2,
n
a
z
=
2,f(x)
b
z
=
-5,f(x)
c
z
=
of
=
2,
polynomial
f,
f(x)
=
and
nd
check
your
2
x
=
f
Hence
calculator.
+
ax
+
x
3
3
a
the
coefcients.
remaining
answers
f.
missing
3
f
 6,
2
polynomial
Factorize
 12z
2

3
6
=

2
2z
2
-
4
f(x)
3

2
a
c
3z
1
z
3

2x
-
2,
a
∈

2
+
10x
+
ax
+
15,

bx
a
∈

2
3x
+
7x
-
8x
-
20
4

5 i,
f ( x) 
x
3
 2x
2
 ax
 85,
2
3
b
k

,
n

2,
f ( x) 
9x
2

24 x

44 x
 16
a, b  
3
4
4
c
k
=
1,
n
=
2,
f(x)
=
3
x
-
2x
d
2
-
3x
+
8x
-
z
=
1
-
2i,f(x)
=
3x
3
-
7x
2
+
18x
+
ax
+
b,
4
a,
b
∈

1
4
d
k


,
n

3,
=
4,
f
 x 

8x
3

20 x
2
 18 x
 7x
 1
2
5
e
k
=
1,
n
f(x)
=
4
5x
-
-
13x
23x
+
3
+
2
2x
+
22x
7
Sum and product of polynomial roots
Just
as
you
have
already
shown
for
quadratic
polynomials,
François
InternationalViète
a
developed
polynomial.
did
it
just
for
formulas
Viète
was
positive
that
the
real
connect
rst
zeros,
to
the
zeros
investigate
whilst
Albert
and
this
the
coefcients
connection
Girard
was
the
but
rst
of
he
mindedness
to
François Viète (1540–
extend
it
to
complex
zeros.
1603) was a French
amateur mathematician
and astronomer, whilst
Investigation 14
Alber t Girard was a
1
Use your calculator to nd the zeros of the following polynomials.
French mathematician
and musician.
3
(x) = x
f
2
+ x
3
- 4x - 4,
1
3
(x) = x
2
- 6x
+ 11x - 6,
2
3
(x) = x
f
f
2
- 4x
3
- 7x + 10,
f
(x) = x
2
+ 27x
+ 71x - 1155
4
190
/
2
arbegla dna rebmuN
3.4
Find the sums and the products of all the zeros for each polynomial in
question 1
3
How do the sum and product of the zeros you found in
Factual
question 2 relate to the coecients of each polynomial?
4
Use your calculator to nd the zeros of the following polynomials.
3
2
(x) = 2x
f
3
- 11x
+ 17x - 6, f
5
3
2
(x) = 15x
f
- 16x
3
+ 19x
- 4, f
7
5
2
(x) = 3x
+ 3x + 10,
6
(x) = 14x
2
+ 291x
+ 1402x - 1155
8
Find the sums and the products of all the zeros for each polynomial in
question 4
Conceptual
6
How do the sum and product of the zeros of a cubic relate
to their coecients?
Your
results
from
Investigation
3
Given a cubic equation
ax
14
form
part
of
the
following
key
point.
2
 bx

cx


x
d

0, a , b , c , d

,

x
a
0, with

b

x

1

x
2


3
a

c

x
,
x
1
and
x
2
as solutions, then

x

x
1

x
2

1
x

x
3
2

3
3
a


d
x


x
1

x
2


3
a

Example 31
3
The
roots
of
a
cubic
equation
3x
2
+
5x
+
2x
-
4
=
0
are
x
,
1
Without
solving
the
equation,
x
+
x
1
+
b
x
2
x
3
×
x
1
×
2
3,
b
=
5,
c
=
2,
d
=
-4
3
1
+
+
3
1
=
1
c
x
x
a
andx
2
nd
1
a
x
x
x
2
1
x
x
3
x
2
3
Identify
the
coefcients
of
the
cubic
polynomial.
5
a
x
+
x
1
+
x
2
=
-
Use
the
theorem
for
the
sum.
Use
the
theorem
for
the
product.
3
3
 4
b
x
1

x

2
x


4

3
3
3
Continued
on
next
page
191
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
5
1
1
x
1
+
x
3
c
+
x
x
1
+
x
2
x
1
+
x
2
x
3
=
x
2
x
3
x
1
=
Rewrite
-
4
x
2
5
3
1
=
the
expression
in
terms
of
the
4
3
sum
and
product
and
then
use
the
3
results
Make
from
sure
calculator
Use
all
the
the
for
you
to
previous
know
check
complex
possible
sum
Notice
the
and
that
decimal
your
roots
zeros.
product
the
form
how
are
the
results
in
fraction
degree
of
accuracy.
to
use
nder
to
the
nd
to
same
the
to
as
a
obtain
features
obtained
form
a
results.
Use
results
parts.
results.
in
the
certain
Investigation 15
1
Use your calculator to nd the zeros of the following polynomials.
4
(x) = x
f
3
2
+ x
- 7x
4
- x + 6, f
1
4
(x) = 6x
f
(x) = x
3
2
+ 7x
- 13x
(x) = 35x
- 86x + 280,
3
+ 381x
2
- 73x
- 309x + 110,
4
4
(x) = 4x
3
- 4x
2
+ 65x
4
- 64x + 16, f
5
2
2
- 39x
4
- 4x + 4, f
3
f
3
+ 4x
2
(x) = 6x
3
2
- 11x
+ 117x
+ 397x - 145
6
Find the sums and the products of all the zeros for each polynomial in
question 1
Conceptual
3
How do the sum and product of the zeros of a four th-degree
polynomial relate to its coecients?
n
Given a polynomial
f ( x)

a
x
n
 a
n
real or complex coecients
a

n
1
x
n
2
 ...  a
1
x
 a
2
0  and zeros
x
,
x
x
1
, ...,
2
with
 a
1
0
of f,
x
then
n
a
n
x

x
1

x
2
 ... 
x

3
1

n
a
n
n
a
0
and
x
x
1
x
2
 x

3
n
 1
a
n
The
proof
you
saw
Since
comes
when
every
as
a
direct
proving
polynomial
the
consequence
equivalent
can
be
written
of
the
result
in
for
the
factor
a
theorem,
cubic
form
f(x)
polynomial.
=
a
(x
n
(x
-
x
)...(x
2
-
x
),
by
expanding
this
expression
we
as
obtain
the
-
x
)
1
given
n
formulas.
192
/
arbegla dna rebmuN
3.4
Example 32
Find
the
sum
and
the
product
of
the
zeros
of
the
following
polynomials.
4
3
a
f(x)
=
x
b
f(x)
=
2x
c
f(x)
=
-5x
+
x
2
-
5
2x
+
4
-
5x
n
=
4,
a
2
2
15x
-
511x
1457
+
=
-
3
+
2019
a
3x
1,
7x
a
4
=
1,
-
4x
250
-
4x
a
=
3
15x
-2
x
+
x
2
+
x
3
+
21
Use
the
formulas
and
product
for
sum
0
=
from
the
key
1
3
+
1
8
47
-
a
x
+
point.
-
=
-
=
-1
4
a
1
4
a
4
-2
0
x
x
1
x
2
x
3
=
( -1)
4
=
=
a
Notice
-2
1
formulas
4
only
b
n
=
5,
a
=
2,
a
5
=
-5,
a
4
that
=
the
for
you
those
need
three
to
use
coefcients.
8
0
5
a
5
5
4
x






k
a
k 1
2
2
5
5
5
a
8
0

x

k
 1
n
=

a
k 1
c


 4
2
5
2019,
a
=
-5,
a
2019
=
Use
0,
your
check
a
=
calculator
to
2018
the
results.
21
0
2019
a
0
2018

x





Notice
that
due
the
someti me s ,
0
k
a
k 1
5
to
accura cy
of
2019
2019
2019
a
21
the
calculator’s
the
results
are
al g o r i thm,
not
e x a ct
21
0

k 1
x

k
 1

a


5
(the
complex
part
is
5
2019
almost
In
this
0).
example
calculator
nd
high
to
the
zeros
an
technique
sum
cannot
order,
have
and
the
so
even
due
to
it
helpful
is
the
algebraic
to
nd
their
product.
193
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Exercise 3M
1
Find
the
sum
and
product
of
the
zeros
of
these
polynomials.
4
3
a
f(x)
=
x
b
f(x)
=
2x
c
f ( x)

23 x
-
2
3x
+
6
2x
-
5
-
5x
+
4
4
6x
+
11x
3
+
13x
-
5x

5
17
9

4
x

13 
x

x
11
2020
d
f(x)
=
3x
2019
-
The
cubic
46
573
4x
+
17x
-
2x
3
2

3
equation
+
115x
-
8
2
4x
+
x
-
7
=
10
has
roots
x
,
x
1
Without
a
x
+
solving
x
1
b
x
×
×
x
2
3
+
10x
1
+
3
3
3

The
x

are
x
results
of
the
x
-
x
+
x
×
+
x
×
+
using
x
+
x
3
13x
-
x
3x
.
Without
+
3x
+

x
your
3
equation,
nd
3x
3
4
6
2
the
4
6

Check
solving
x
2
x
0
x
6

=
4
×
d
1
calculator.
+
3
3
1
x
a
equation
and
2
3x
3
4
x
2
1
x
quartic
3
,
1
x
11x
2
,

2
2
2x
1
x
3
3
+
x
1
your
4
c
x
roots
6x
b

2
Check
a
3

1
3
10x
2
d
x
nd
3
x
10x
equation,
.
3
x
2
1
c
+
the
andx
2
6

x

1
results
x
2

x
x
4
using

1
a

x
3

x
4
x
2

x
3

x
4
calculator.
194
/
3.5
Since
in
Solving equations and inequalities
you
have
Section
solving
3.1,
already
in
this
polynomial
Historically,
a
extensively
section
you
equations
group
of
Italian
of
covered
are
solving
mainly
degree
3
or
going
quadratic
to
be
equations
focused
on
higher.
mathematicians,
dal
Ferro,
Tartaglia,
Did you know?
and
Cardano
(13th–14th
century),
developed
a
formula
for
solving
a
abylonians were able
general
cubic
arbegla dna rebmuN
3.5
equation.
to solve some cases
That
formula
is
very
complicated
and
very
difcult
to
use,
so
you
of cubic equations
are
going
to
be
studying
rational
solutions
that
can
be
found
by
(2000–1600 BC).
alternative,
The
most
simpler
common
methods.
method
They had tables with
that
you
are
going
to
use
order
to
be
for
perfect squares and
solving
perfect cubes and their
polynomial
equations
is
factorization.
In
able
to
factorize
sums. y using those
the
polynomial
you
are
going
to
use
theorems
that
are
valid
for
tables, they were
polynomials
of
all
degrees.
solving equations of
Before
you
start
on
factorization
you
are
going
to
look
at
how
3
many
the form
real
zeros
you
may
expect
for
a
given
ax
+ bx
=
c .
polynomial.
Solving polynomial equations
In
this
section
algebraic
you
methods
are
going
and
then
to
solve
use
a
polynomial
calculator
to
equations
verify
the
by
using
solutions.
Example 33
3
Solve
the
equation
3
x
- 7x
+
6
=
0
by
using
algebra.
3
x
- 7x
+ 6
=
x
-
x -6 x
+ 6
Split


 




in
the
linear
term
so
that
you
can
factorize
pairs.
2

x
x

1
  6x
=
( x
- 1)
(x (x
=
( x
- 1)
(x
 1
+ 1)
- 6

x
x
 1  x
 1
 6  x
)
 1
Factorize
the
and
factorize
then
difference
Factorize
the
Use
product
the
quadratic
of
two
squares
expression.
factor
by
inspection.
2
+

 x
 1  x

x
 1,
1
x
- 6
 2  x

2
x
2,
x
)
=
(x
 3


- 1) ( x
0
- 2) ( x
+ 3)
zero
theorem.
3
3
Make
sure
result
by
mial
you
know
nding
using
a
the
how
zeros
to
of
check
the
the
polyno-
calculator.
195
/
3
E X PA N D I N G
On
of
a
calculator
the
you
THE
can
NUMBER
also
use
S YS T E M :
the
C O M PL E X
graphical
method
NUMBERS
for
nding
zeros
polynomial.
y
3
f1(x)
=
x
–
7x
+
6
6
4
2
(2, 0)
(1, 0)
(–3, 0)
x
0
–2
–4
–6
Apart
from
quadratics,
polynomials.
You
are
you
haven’t
going
to
drawn
graph
more
many
graphs
polynomials
of
higher-order
once
calculus
is
introduced.
Exercise 3N
3
1
Solve
real
the
solutions,
using
a
x
+
x
check
your
Give
all
3
the
The
answers
equation
repeated
2x
root
x
-
equal
4x
-
4
=
a
Find
the
values
b
Find
the
remaining
2x
- 9x
- 18
=
f
 x 
- 3x
+ 3x
3
2
- 2
=
x
+
x
- 3x
-
x
+
3
equation
root
has
one
3.
of
a
and
b.
root.

2
x
 ax
 bx

c , a , b, c

.
0
Two
The
0
2
4
2
=
0
3
d
to
+ 9
0
4
x
+ bx
2
+
3
c
2
+ ax
2
3
b
and
equations.
calculator.
3
a
following
equal
to
2
=
0
+
x
+ ax
-
4
=
0
has
Find
the
value
b
Find
the
remaining
of
this
polynomial
are
opposite
a
Show
b
Find
that
ab
=
c
one
-2.
a
of
numbers.
2
x
zeros
the
third
zero.
a.
two
roots.
Solving polynomial inequalities
When
you
the
solving
need
signs
structing
to
of
a
polynomial
fully
the
sign
polynomial
is
inequalities
factorize
factors
table.
either
on
Yo u
the
the
an
polynomial
whole
should
positive
by
or
set
and
of
identify
negative,
algebraic
real
for
and
then
method,
investigate
numbers
which
use
by
values
this
to
con-
of
x
solve
the
the
i n e q u a l i t y.
When
solving
calculator
of
x
the
the
to
polynomial
graph
the
polynomial
is
inequalities
polynomial
either
by
and
positive
or
a
graphical
again
method,
identify
negative,
and
for
use
use
which
this
to
a
values
solve
inequality.
196
/
Example 34
3
Use
an
algebraic
answer
using
a
method
to
solve
the
inequality
x
2
+
4 x
+
x
- 6
3
3
4x
+
x
and
check
Factorize
2
+
0
your
calculator.
Factorization:
x
>
- 6
=
x
splitting
2
+
4 x
+
4 x
- 3x
the
the
expression
linear
by
term.
- 6
arbegla dna rebmuN
3.5
2
2
=
x
(x
+
4x
+
4
) - 3( x
+
2)
=
x
(x
+ 2)
- 3 ( x
Use
+ 2)
to
the
distributive
factorize
the
property
expression.
2
=
( x
+
2)
(x (x
+
2)
- 3
)
=
(x
+ 2)
(x
+ 2x
- 3
)
Factorize
the
quadratic
=
( x
+
2) ( x
- 1
)( x
+
3
–3
]–3, –2[
–2
]–2,
1[
1
]1,
∞[
Construct
–
x
+
0
+
+
+
+
3
x
4x

+
sign
table
3
x
+
2
x
–
1
x
–
6
–
–
–
0
+
+
+
–
–
–
–
–
0
+
0
+
0
–
0
+
2
+
the
+
and
x
factor.
)
]–∞, –2[
x
remaining
3,  2

4 ,
investigate
Identify

where
the
the
the
sign.
intervals
sign
is
positive.
y
6
Draw
4
the
graph
polynomial
and
of
the
nd
all
the
2
zeros.
(–3, 0)
Identify
the
values
the
graph
(1, 0)
(–2, 0)
x
of
x
for
which
is
0
3
4
above
the
x-axis.
–2
–4
3
f(x)
=
x
2
+
4x
+
x
–
6
–8
Notice
that
boundaries
when
the
in
solution.
the
inequality
is
not
strict,
you
need
to
include
the
197
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Example 35
3
Given
the
values
of
polynomials
x
Method
such
f
that
f
( x )
 x 

=
2
3x
+
g  x 
by
2x
3
- 3
using
and
a
g ( x )
=
graphical
2
x
-
x
+ 3x
method
on
- 1,
a
nd
all
the
calculator.
1
Graph
y
of
both
polynomials
and
nd
the
points
intersection.
8
3
f(x)
=
3x
2
+
2x
–
3
4
(1, 2)
(–2, 0)
x
3
(–0.5, –2.88)
–8
–12
3
g(x)
=
x
2
–
x
+
3x
–
1
–16
(–2, –19)
–20
–24
x

 2,
 0 .5

1,
Identify
 
the
Method
3
3x
2x
+
3x
3
2x
graph
intervals
of
f
is
on
above
the
the
x-axis
graph
of
where
g
2
2
+
the
3
-
3
≥
-
3x
x
2
-
x
≥
0
+
3x
-
Rewrite
1
and
2
-
2
the
nd
where
its
the
inequality
zeros.
as
Then
polynomial
one
nd
is
polynomial
the
above
intervals
or
on
the
y
x-axis.
15
3
f(x)
=
2x
2
+
3x
–
3x
–
2
10
5
(–2, 0)
(1, 0)
x
3
–5
–10
–15
x

 2,
 0 .5

1,
 
HINT
In the case of Method 1 you may have needed to consider changing the
scale on your GDC to nd all the points of intersection between the graphs
that might not be visible in the original window. Also, it can be dicult to
read from the graph which function is the upper one and which the lower
one. This is even more dicult on calculators with lower resolution.
Method 2 is more suitable because you don’t need to nd the points of
intersection but only nd zeros, which are always on the x-axis. This saves
time as you don’t need to change scales in the window, etc.
Sometimes
polynomial
polynomials
that
you
equations
can
sketch
or
inequalities
and
nd
the
can
be
solution
easily
by
broken
to
simpler
inspection.
198
/
Example 36
3
Use
a
simple
3
x
polynomial
2

3
x

4

0

graph
to
solve
the
inequality
x
2
-
x
-
4
£
0
2
x


x

4
Rewrite



f (x)
g( x )
the
inequality
by
using
two
simple
polynomials.
Or
y
arbegla dna rebmuN
3.5
14
Simply
enter
the
original
inequality
into
the
12
GDC.
10
(2, 8)
8
6
3
f(x)
2
g(x)
=
x
+
=
x
4
Graph
2
by
x
those
two
inspection
simple
the
point
polynomials
of
and
nd
intersection.
0
–2
–4
–6
–8
–10
–12
–14
x


, x

Identify
2
below
3
Notice
that
example
of
an
equivalent
simple
Algebraic
methods
degree
this
in
equations
polynomials
for
course.
and
inequality
solving
A
GDC
inequalities
to
graph
a
be
4
£
and
used
higher
the
interval
graph
of
where
the
graph
of
f
is
g
2
-
polynomials
may
of
x
the
x
nd
are
to
would
the
also
an
solution.
restricted
solve
be
to
the
third
polynomial
degree.
Example 37
2
Use
a
GDC
7
-2 x
to
solve
the
inequality
4 x
7
<
2x
+ 1
2
+
4 x
- 1 <
0
Rewrite
the
inequality
as
one
and
nd
polynomial.
y
14
12
7
f1(x)
=
–2x
2
+4x
–1
10
8
6
4
2
(0.504, 0)
(1.1, 0)
(–0.496, 0)
x
–2
–4
–6
–8
–10
–12
–14
-0.496
<
x
<
0.504
or
x
> 1 .10
Graph
the
polynomial
which
part
of
the
graph
is
the
below
zeros.
the
Identify
x-axis.
199
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
7
Notice
that
the
equivalent
inequality
NUMBERS
2
2x
- 4 x
+ 1 >
0
gives
the
same
result.
Exercise 3O
1
Solve
the
following
inequalities
in
the
set
2
of
Given
the
polynomials
3
real
numbers
3
a
the
method
4
0
of
your
choice.
f
( x )
=
3x
g ( x )
=
x
2
+ 5x
- 6x
- 9
and
2
x
+
x
-
3
b
by
4x
-
<
3
2
+
2x
-
4x
-
4,
nd
+
2x
- 9x
- 18
>
3
x
d
4 x
such
that
f
( x )
£
g ( x )
- 3x
+ 3x
3
2
- 2
£
3
0
Use
simple
solutions
+ 8x
+
x
- 3
polynomial
of
the
graphs
following
3
+
4 x
+ 7x
3
12 x
+
2
>
a
x
c
x
0
the
inequalities.
3
+
x
+
x
4
+
2
>
b
0
2 x
2

x
 1 
0
- 16 x
4
- 81x
3
+
2
x
- 35
<
2
£
0
0
2
Use
a
calculator
-
x
+
2
-
x
2
+
x
solve
the
following
0
5
3
to
inequalities.
- 3x
4
2x
-
2
4
-
x
- 1 <
a
x
b
x
c
x
3
 4 x

2x
 1 
0
0
13
8
+
4
<
+
2x
15
3.6
nd
2
3x
h
to
0
3
x
of
2
c
g
values
0
x
f
the
2
x
e
all
3x
+ 5x
14
8
+ 5x
2
>
4 x
- 1
Solving systems of linear
equations
Systems of t wo linear equations with t wo unknowns
Up
to
this
systems
point
of
linear
where
there
satisfy
both
In
this
can
is
a
your
studies,
equations
unique
you
(such
as
solution:
a
have
a
only
pair
single
of
been
asked
to
simultaneous
pair
of
real
solve
equations)
numbers
x
and
y
that
equations.
section,
occur
in
you
when
will
solving
learn
about
systems
of
the
possible
linear
types
of
solutions
that
equations.
Investigation 16
1
Without using a calculator, solve the following simultaneous equations.



a

4 x
 3y

 18
b


2x
 5y

10 x
 4 y

3
4
c


4
7 x

 4 y
 13
 6 x

 15 y

3
 2 x


2
3
y


5
5
For each set of equations a–c, sketch both lines on the same axes. Where
possible, show on your sketch the solution to the pair of equations.
3
Factual
What are the dierent types of solution to a system of two linear
equations with two unknowns?
4
Conceptual
Explain the geometrical signicance of each type.
200
/
When solving systems of two linear equations with two unknowns there are
TOK
three possible types of solutions that can occur. Each type has a dierent
If we can nd solutions
geometrical interpretation.
in higher dimensions
1
When there is a unique pair of numbers that satisfy both equations, the
can we reason that
lines intersect at one point.
these spaces exist
beyond our sense of
2
When there are no real numbers that satisfy both equations, the lines are
perception?
arbegla dna rebmuN
3.6
parallel and distinct.
3
When there are innitely many pairs of real numbers that satisfy both
equations, the lines coincide.
All
the
results
checked
this
on
using
a
of
so l v i ng
calcul a tor.
your
s ys te ms
Ma ke
of
s ur e
l in e ar
y ou
eq ua t i on s
know
how
can
to
be
do
GD C .
HINT
In type 3, innitely many pairs of real numbers (x,
y) satisfy this system
where y can be any real value, whilst x is a value dependent on y
Instead
of expressing y in terms of x as usual, a calculator may provide an answer
where x is expressed in terms of y
Example 38
Find
the
value
of
a
real
parameter
m
such
that
the
system


4 x

y
 1
has

3x



my

no
solution.
2
Apply

4 x

 1 
y



3x


m 4 x
 1

2
method
4 x
 1 
y
of
substitution
by

3 


4m  x

2 
m
expressing
y
as
the
in

2 +
x
the

m
subject
rst
the
equation,
and
=
3 +
substituting
4m
second
it
in
the
equation.
3
3 
4m

0

m


The
system
has
no
4
solution
when
3
0,
+
4m
then
=
you
dividing
since
would
by
be
zero
201
/
3
E X PA N D I N G
The
system
of
THE
NUMBER
equations
in
S YS T E M :
Example
38
can
C O M PL E X
either
have
NUMBERS
no
solutions
TOK
3
(in
the
case
where
m
=
-
),
or
a
unique
solution
(in
the
case
where
How accurate is a visual
4
3
representation of a
m


).
4
It
is
mathematical concept?
not
possible
solutions
since
for
the
this
particular
coefcients
of
system
the
are
not
proportional,
have
variable
4
coefcients
to
x
innitely
and
the
many
constant
1
ie

3
2
Example 39

2 x
Consider
the
system
of
equations

p
 1 y
 1

px

 3y

2

a
Find
the
values
b
Find
the
unique
of
the
real
solution
parameter
in
terms
of
p
such
that
the
system
has
a
unique
solution.
p

2 x
a

p
 1 y
 1
 p, p

0

 3y
px


Use
the
the
rst
method
of
elimination;
multiply
 2
2

by
equation
by
p
and
the
second
2.


 2px
p  p
 1 y

2
p


p

p
 6
 y

p
 4
Subtract
the
two
equations
to

eliminate
2px

 6y

x.
4

2
⇒
p
-
p
-
6
≠
0
⇒
(p
-
3)(p
+
2)
≠
In
0
order
to
coefcient
p
≠
3andp
≠
have
next
a
to
unique
y
must
solution
not
be
the
equal
-2
to
p
0.
 4
2
b
p

p
 6
 y

p

4

y

Use
the

p
3  p

3 y

2
in
part
a
to
 6
express
px
elimination
2
p

px
y
in
terms
of
p
 4


Substitute
2
your
value
of
y
into
the
2
p

p
 6
second
equation.
2
3p

 12 
2p
 2p
 12
Solve

px
to
nd
x
in
terms
of
p.
2
p

p
 6
2
2p

x
 5p
2p


x
2
2
p
p

 x, y


Notice
that
 5
 x, y

the

2p
p
 6
p

 5
p

2
p

2

case
p
 6
when
p
p
p
 6

 4
,

 5


p
 6
=
0
is
,
p

2, 3

covered
by
the
unique
solution
2
,


6
3
202
/
Exercise 3P
1
In
each
case,
nd
parameter
m
equations
has
the
such
no
value
that
the
of
the
3
real
system
In
of
each
parameter
solution.
equations
the
 mx
 6y


m
2
7x
 3y

 1 x

2y

has
solution
a
in
values
that
the
unique
terms
of
the
system
solution.
of
real
of
Hence
nd
s
5


2 x
3x


my
a
 11

sy

 1
b

s

2 x

y

5  2s  x

sy
s




In
such
the
1

b



2
s
nd


a
case,
arbegla dna rebmuN
3.6
each
case,
parameter
equations
p
nd
the
such
has
values
that
the
innitely
of
the
system
many
3x

y


2

4

real
4
of
The
system
solutions.

Find
the
a( x

solutions.

y)  b( x

y)
 1

y)  b( x

y)
 1
, a,

a( x

6 x
a

py

b 
, a ,
b

0
of
a


8

3x

y

4
is
given.
a
Find
b
State



px

b

y

p

for
p
 4 x

p
 1 y

the
solution
and
the
(x,
explain
system
to
y)
in
terms
whether
have
no
it
is
and
b
possible
solution.
2


Now
let’s
method
take
of
two
linear
elimination
to
equations
nd
a
in
a
general
general
form
and
use
the
solution.
Example 40

 ax
Find
the
general
formula
for
a
unique
solution
of
 by

, a,


cx

dy

Multiply



ax
 by

e

b,
c,
d

0.
f
the
rst
equation
by
d
and
the

adx
 bdy



e

ed


second
equation
by
b
to
obtain
equal

cx


dy
⇒
 bdy
bcx
f



fb 

adx
-
bcx
⇒
coefcients
of
ed
-
x(ad
-
bc)
⇒
Subtract
equations
ed
-
x
ed

ad
 bc
y.
the
to
eliminate
the
y
fb
Use

variable
fb
variable
⇒
the


distribution
to
make
x
the
subject.
fb

ed

fb
a 
 by
ad
 bc

e
Substitute
eg
the
rst
x
in
one
of
the
equations,
equation.
Continued
on
next
page
203
/
3
E X PA N D I N G
e(ad

by
THE
NUMBER
 bc )  a(ed

S YS T E M :
C O M PL E X
NUMBERS
fb)

 bc
ad
ead
- ebc
- aed
+ afb
=
ad
afb
- bc
- ebc
=
ad
b(af

y
y
 ec )

 bc
af
 ec
ad
 bc
Make
y
 x, y

ed
af
 ec
ad
 bc
fb

, ad

 bc

State
0
the
have
40
that
the
denominator

the
Example
condition

 bc
ad
cannot
would
subject.

,

In
the
b



1

ad

- bc
you
given
could
the
have
same
rst
end
eliminated
x
be
equal
system
instead
of
y,
to
to
have
0,
which
a
unique
is
required
for
solution.
which
result.
Did you know?
Gabriel ramer (1704–1752) has developed this formula within his work on
a
b
c
d
determinants. The form
a
is called a determinant. It is a numerical value such
b
=
that
c
ad
- bc
d
All the expressions in the values of x and y in Example 40 can be written as
determinants in the following way.
a
e
b
=
b
D,
=
D
a
e
c
f
,
=
x
c
d
f
d
D
y
Hence, to calculate the values of x and y we need to calculate those three
D
D
y
x
determinants and then we use the formulas
x

,
y
D
Using
determinants
when
the
complex
to
coefcients
calculate
of
the
unique
linear
,
D

0
D
solutions
simultaneous

works
very
equations
efciently
are
numbers.
204
/
arbegla dna rebmuN
3.6
Example 41
Use
the
formula
from
Example
40
to
solve
these
simultaneous
equations
with
complex
coefcients.

3 x

2iy

5  7i
1 
i x

y



i

Check
a
=
your
3,
b
3×(-1)
=
-
answers
2i,
c
=
2i×(1
1
-
using
a
i,
-1,
-
i)
=
d
=
-3
-
calculator.
2i
e
=
5
+
7i,
-
2
=
-5
-
f
=
Identify
i
Using
2i
the
(5
+
3×i
7i)×(-1)
-
(5
+
-
×
i
-
i)
=
7i)×(1
3  7i
x
2i
5 

=
-5
3i
2i
-
-
5
7i
+
5i
15  6i

+
2
-
=
7i
-3
-
7
-
=
5 
-12

2i
-
i
1 
your
denominator
12 
i
5 
5  2i
A
GDC
5 
can
60  24 i
 5i

also

2i
be
2 
to
conrm
the
-
-
fb.
The
numerator
for
y
is
af
-
ec.
x
i
ed
-
ad
- bc
i
y
40,
bc.
ed
fb
=
.
Divide
multiplying
of
af
- ec
ad
- bc
top
the
complex
and
numbers
bottom
by
the
denominator.
=
29
used
ad
is
2

is
Example
x
29
2i

from
for
conjugate

result
numerator
by
y
coefcients.
The
 35i + 14

5  2i
7i
the
result.
Exercise 3Q
Solve
the
following
simultaneous

equations
1 

and
check
your
answers
with
a
2
GDC.
4i  x
 3iy

2 
4i

 3  5i  x


5 
4i  y

21  27 i


 3 x
1

1 
i y

4
 5i

4


i x

1 
i y

7

Systems of three linear equations with three unknowns
When
you
are
unknowns,
with
you
or
two
will
presented
you
have
unknowns,
focus
on
to
with
a
reduce
which
solving
system
them
you
to
know
systems
of
a
three
equations
system
how
to
algebraically,
of
two
solve.
either
In
with
three
equations
this
by
section,
substitution
elimination.
205
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
Example 42
2x


y
C O M PL E X

z

NUMBERS
5

Solve
the
simultaneous
equations

x
 2y
 3z

0

 1
by

5x

y
2z

a
the
method
of
substitution
b
the
method
of
elimination.
Isolate
one
of
the
variables
from
one
of
the

2  2y

a
x

2y
 3z
 3z


y

z
 5
equations

and
substitute
it
in
the
remaining

5  2y

 3z


y

2z
 1
two
equations.
Here
we
nd
x
in
the
second

equation.
Try
to
target
the
simplest
possible
expression.




5y

5

=
1
 17 z
y


9y


y
 5z

z
 1

9y
 1
 17 z
 1


-
z
⇒
9(1
-
z)
+
17z
=
1
⇒
8z
=
-8
You
can
again,
⇒
z
=
-1
⇒
y
=
2
⇒
x
=
apply
by
the
method
expressing
y
in
of
substitution
terms
of
z,
or
vice
1
versa.
⇒
(x,
y,
z)
=
(1,
2,
-1)
Find
z
and
then
the
values
of
y
and
x
respectively.
b
Add
equations
1
Add
2
This
eliminates
two
unknowns.
and
3.


2x

y

z


5


×
equation
1
to
equation
2.


5x

y

2z
 1






4 x

2y
 2z
 2y
 3z
0


7x

z

6
 5z
 10

to
give
two
equations
in

When


y

5 x

 10

x



eliminating
eliminate
use
a
the
a
same
combination
variable
variable,
of
all
you
and
three
must
you
must
equations
in
the
process.



7x



z

6



x

z

2


8x

8

x
 1
Add
the
two
equations
to
eliminate
z.
Find

the

value
of
x.
to
nd


⇒
7
+
z
=
6
⇒
⇒
2
+
y
-
(-1)
⇒
(x,
y,
z)
z
=
-1
Use
nd
You
can
your
=
=
(1,
again
5
2,
use
⇒
y
=
this
x
z,
and
then
use
x
and
z
to
y.
2
-1)
a
calculator
to
check
solution.
206
/
3.6
that
once
you
you
don’t
unknowns,
this
system.
with
vice
two
you
used
unknowns,
solving
variables’
variables).
you
can

you

just
of
organize
2
1
 1
5
1
 2
 3
0
to
two
same
nd
use
a
equations
method
system
with
again
of
substitution
2
1

two
to
two
to
solve
equations
solve
them,
or
1
 2
 3
0
0
5
5
5
0
9
17
1

operating
coefcients
(independent
rewriting
everything
with
obtain
the
 3
0
2
1
 1
5
5
 1
2
1
1
 2
 3
0
in
rows
and
the
of
on
the
variables
calculate
the
result.
Swap
the
rst
two
rows
to
simplify
the
calculations.
Multiply
the
rst
from
second
row
by
2
and
subtract
it
0
1
1
1
0

9
17
1

the
by
5
and
subtract
it
row.

1
 2
 3
0
0
1
1
1
 1

 2
 3
0
1
1
1

Multiply
the
from
third
the
Divide



0
the
rst
row
row.
second
row
by
5
to
simplify
the
calculations.

0
0
8
0
 8

x

 2y
 3z

0
y
 1
 1
x
1
 1

2y
 3z

Multiply
the
from
third
the
second
row
by
9
and
subtract
it
row.
0



y


Divide
2

the
third
row
by
8
to
simplify
the
into
the
equa-
from
the
third
calculations.


z
1
z

x
0



1

 2  2
 3  1

0


x
 1
y


y



2


Now
rewrite
tions
and
equation
the
nd
coefcients
the
solutions
up.
2

z

1
z

1


This
just
free
 2

are



you



that

 1


the
coefcients
to
 1
notice

5

of
the
then
can
and
the
operations


you
always


apply
could
coefcients


to
system
elimination
equations
Instead
elementary

have
the
versa.
When
the
If
obtain
arbegla dna rebmuN
Notice
method
generally
is
used
called
for
row
reduction
or
the
Gaussian
method
and
it
is
matrices.
Investigation 17
1
Use your calculator to solve the following systems of simultaneous
equations.

2x

y

z

5


a


y
 2z
5x

y


4

x
 2y
 3z


 1
0

b

z

3

5x

3x

y
2z
6x


2y
 4z

5
Continued
on
next
page
207
/
3
E X PA N D I N G

x

2y
THE
 3z

NUMBER
S YS T E M :
4
 6z

8
d
11x


2y


z
3
 10 y

2z
 1

3 x
 6y
 9z

12
23 x

 18 y
 3z

3

x


y

2z

 12
6 x
 15 y
 21z

4


3x

 3y

2z

22
f


8x

2y
 11z
 15

7x
 2y
 2z

3
2 x

 5y
 7z
 17

Conceptual
2

NUMBERS

 4 y
2x

e
7x


c
C O M PL E X
How can you classify the types of solutions to systems of
three linear equations with three unknowns?
3
What is the dierence between the sets of innitely many
Factual
solutions to the equations in par ts c and e?
4
What is the relationship between the coecients of the three
Factual
equations in par t c?
When
solving
unknowns
systems
there
are
of
three
a ga i n
simultaneous
three
possible
linear
types
of
equations
solution
with
that
three
ca n
o c c u r.
1
There is a unique triplet of numbers that satisfy all three equations.
2
There is no triplet of real numbers that satisfy all the equations.
3
There are innitely many triplets of real numbers that satisfy all the
equations.
Example 43
x

y

mz

2

y

z

3
 2z

5

Discuss
all
possible
types
of
solution
of
the
system

2x
with
respect
to
the

x

y

real
x
parameter

y

mz

m.
Eliminate
2
the
variable
y
by
adding
the



2x

2x

y
z

3
2z

5



m
 2 z

7
rst
the


x

y

and
third
equations
and
subtracting

x


z

second
and
third
equations.
2


Eliminate
x
by
multiplying
the
second


2 x


m
 2 z

7
equation
 


2x

2z

4


m

4 z
by
2
and
subtracting
it
from
 11
the
rst
equation.



208
/
11

z
If

, m
m

a
real
by
 4

2 
 4
2m


y
m
≠
4
4
there
2m

 x , y, z 


m

11



m
the
4
⇒
m
–
4)
which
cannot
be
0.


a
 4
m
unique
 3
x
in
terms
of
m
Find
y
in
terms
of
m
When
m
=
 5
solution
 5
7m
,
 4
m

11

,
,
m

4

m
 4
m
 4
m

4

×
z
=
11
⇒
0
=
11,
which
is
false
=
4
the
system
has
no
4
from
the
statement,
equation
therefore
the
you
obtain
system
has
false.
no
When
to
(m


0
expression
Find
a
=
divide


is
7m

2



m
to
 4
 3 
5 
When
need

m

you
2m  3
11
x
exists,
4
equal

solution
arbegla dna rebmuN
3.6
solution.
solution.
Determining the equation of a polynomial from a system
of linear equations
Suppose
you
graph
the
to
of
are
linear
formulate
this
system
to l d
two
of
the
co or di na tes
functi on
f(x)
e q ua tions
equa ti o ns,
in
you
=
ax
two
can
of
+
t wo
b.
p oi nt s
You
can
u n kn own s,
de t e r m i ne
a
which
use
and
th e
l ie
thes e
b.
By
e qu at i on
on
the
p o i nts
solving
of
the
function.
In
a
similar
way,
of
a
quadratic
when
you
know
three
points
through
which
the
graph
2
equations
gives
the
in
function
three
f(x)
=
ax
unknowns
equation
of
the
a,
+
bx
+
b
and
c
c.
passes,
you
Solving
can
this
formulate
system
of
three
equations
quadratic.
Example 44
2
Find
the
points
equation
(-1,
10),
of
(2,
the
-2)
quadratic
and
(4,
function
f(x)
=
ax
+
bx
+
c
that
passes
through
the
0).
2
Input
 1, 10 
:
a  1
 b  1
 c
the

a
 b

c
the
coordinates
of
each
point
on
 10
 10
parabola
into
the
equation.
(1)
2
 2,
 2
:
a  2
 b  2 


4a
2b

c
c


2
2
(2)
2
 4, 0 
:
a  4
 16a

 b  4

c
4b

0

c

0
(3)
Continued
on
next
page
209
/
3
E X PA N D I N G
(2
)  (
1
)
3a

THE
 3b

NUMBER
12
a


(3
)  (
1
)
15a

 5b

S YS T E M :

b

C O M PL E X
Eliminate
4

10
rst
3a


b

NUMBERS
variable
equation
c
from
by
the
subtracting
the
second
third
and
2
equations.
2a
=
2
⇒
a
=
1
⇒
1
+
b
=
-4
⇒
b
=
-5
Subtract
to
1
+
5
+
c
=
10
⇒
c
=
rst
eliminate
Find
4
the
a,
b
equation
from
the
second
b
and
c
2
⇒
f(x)
=
x
-
5x
+
Write
4
your
the
equation
values
of
a,
b
of
the
and
function
using
c
HINT
You can use your GDC in two dierent ways to solve this type of problem.
You can formulate the three linear equations and use your GDC to solve
•
them.
You can use the statistical feature for nding the regression equation of
•
a quadratic cur ve which passes through three specied points.
Exercise 3R
1
Find
that
the
the
value(s)
system
of
has
the
no
real
parameter
unique
m
so
3
Find
the
values
of
the
solution.
real
2x

parameter

y

z

m

 2x

y

z


mz
2y

4

x


y

2
b
my

z
m
 1
so
that
the
system

3x

Find

 1
z
x

my

z

m
 1

3x


y

mz

2


3
x
has
x

2y
z

y

mz

3



2



a
x
a
unique
solution.

the
value(s)
of
the
real
parameter
4
Find
the
equation
of
a
quadratic
function
2
k
so
that
the
solutions.

2x
system
Find
 3y

the
z
has
innitely
solutions.
 1
many
f(x)
=
ax
+
bx
+
c
that
passes
through
the
points
a
(-3,
1),
(-2,
b
(-1,
1),
(1,
-5)
and
(1, 4)

a
kx
 9y
 3z

3
3x

 7y
 5z

2

-9)
and
(2, 8)

x

 ky


z
0

b

4 x
 6y

k

2 z

6

2x
 3y

2z

3

Developing your toolkit
Now do the Modelling and investigation activity on page 2
16.
210
/
arbegla dna rebmuN
3
Chapter summary
•
The quadratic formula
2
+ bx + c = 0, a, b, c ∈ , a
For a quadratic equation in the form ax
≠ 0, the solutions or roots are
2
-b
x
±
b
-
4ac
=
2a
2
•
Given a quadratic equation of the form ax
+ bx + c = 0, a, b, c ∈ , a
≠ 0, the discriminant is the
2
expression in the formula that is under the square root and is denoted by Greek letter Δ, Δ = b
- 4ac
Case 1: Δ > 0
❍
2
-b
If the discriminant is positive, then
x
±
b
-
4ac
and there are t wo distinct real roots
=
2a
Case 2: Δ = 0
❍
-b
If the discriminant is equal to zero, then
x
. This is regarded as one repeated root
=
2a
Case 3: Δ < 0
❍
2
If the discriminant is less than zero, then
•
b
-
4ac
is not real. In this case, there is no real solution
Complex numbers are numbers of the form z = a + ib, where a, b ∈ .
a is called the real par t of z, and we write Re(z) = a.
b is called the imaginary par t
of z, and we write Im(z) = b.
The set of complex numbers is denoted by 
•
Given the complex number z = x + i y, x, y ∈ , the modulus of z is given by
2
2
z
•

x

iy

x
2
2

y

Two complex numbers z

Re  z


z

= c + di, a, b, c, d ∈ , are equal if and only if their real
= a + bi and z
1

Im
2
par ts are equal and their imaginary par ts are equal, a = c and b = d
•
Addition and multiplication by a scalar:
+ z
z
1
= (a + bi)
lz = l(a + bi)
•
+ (b + d)i,
a , b , c, d ∈

l, a, b ∈ 
= (la) + (lb)i,
Multiplication of complex numbers:
× z
z
1
•
+ (c + di) = (a + c)
2
= (a + bi) ×(c + di) = (ac - bd)
+ (ad + bc)i,
a , b , c, d ∈

2
For any complex number z = a + bi, a, b ∈ , there is a conjugate complex number of the form
*
z* = a - bi . Their real par ts are equal,
Re(z) = Re(z
), and their imaginary par ts are opposite,
*
Im(z) = - Im(z
•
).
Division of complex numbers:
*
z
z
1
 z
1
2

2
z
2
z
2

1, n

4k
i,
n

4k
 1
1,
n

4k

 i,
n

4k
 3


n
•
Powers of i:
i

,

k
 
2



Continued
on
next
page
211
/
3
E X PA N D I N G
•
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Polynomials are functions which map a real variable, often called
write f:

→
.
n
•
x, to another real number. We
Polynomials are functions of the form f(x) = a
n
+ a
x
i = 0, ..., n are called the coecients.
1
+ ... + a
x
n
n
1
x + a
1
∈ ,
, where a
0
i
The highest power ( n) of the variable x is called the degree of
the polynomial, and we write
deg(f) = n
•
A linear combination of two functions f and g is an expression of the form a
×
f(x) + b
×
g(x),
where a and b are real numbers.
n
A linear combination of n functions is an expression of the form
a

f
i
i
 x  , where
are functions
f
i
i 1
∈ .
and a
i
n
•
Two polynomials f(x) = a
n
+ a
x
1
m
2
+ ... + a
x
n
n
1
+ a
x
2
x + a
1
, a
0
≠ 0, and g(x) = b
n
x
m
+ b
x
m
m
1
+
1
2
... + b
x
+ b
2
x + b
1
, b
0
≠ 0, are said to be equal if and only if:
m
❍
they have the same degree, n = m
❍
all the corresponding coecients are equal: a
= b
i
1,
•
...,
for all i = 0,
i
n
For any two polynomials f and g there are unique polynomials q and r such that f(x) = g(x) × q(x) + r(x)
for all real values of x
The polynomial q is called the quotient and the polynomial r is called the remainder. Notice that
deg(g) > deg(r)
•
Polynomial remainder theorem
n
Given a polynomial f(x) = a
n
+ a
x
n
1
2
+ ... + a
x
n
1
+ a
x
2
x + a
1
∈, i = 0, 1,
, a
0
2,
..., n, a
i
≠
0, and
n
a real number p, then the remainder when f(x) is divided by the linear expression (x - p) is f(p)
•
Factor theorem
n
A polynomial f(x) = a
n
+ a
x
1
2
+ ... + a
x
n
n
1
x
+ a
2
x + a
1
∈ , k = 0, 1, 2, ..., n, a
, a
0
k
≠ 0, has a
n
factor (x - p), p ∈  if and only if f(p) = 0
❍
Corollary
n
Given a polynomial f(x) = a
x
n
+ a
real numbers a and b, a
1
2
+ ... + a
x
n
n
1
+ a
x
2
x + a
1
∈ , k = 1,
, a
0
k
2, ..., n, a
≠
0, and
n
≠ 0, then the remainder when f(x) is divided by the linear expression
b
(ax - b) is
f
a
•
The fundamental theorem of algebra (F TA)
n
A polynomial f(x) = a
x
n
+
a
n
1
2
+ ... + a
x
n
1
x
+ a
2
x + a
1
, a
0
≠
0, with real or complex coecients
n
has at least one zero. There is an w ∈ ℂ such that f(w) = 0
•
Lemma
n
Each polynomial f(x) = a
x
n
+
a
written in a factor form f(x) = a
1
1
(x - w
n
2
+ ... + a
x
n
n
x
2
)(x - w
1
+ a
x + a
1
)...(x - w
2
, a
0
≠
0, with real coecients can be
n
∈ , k = 1, ..., n
) such that w
n
k
212
/
•
Conjugate root theorem
n
Given a polynomial f(x) = a
n
+ a
x
1
2
+ ... + a
x
n
n
1
+ a
x
2
x + a
1
∈ , k = 1,
, a
0
2, ..., n, a
k
≠
0,
n
*
that has a complex zero z, then its conjugate z
n
•
If a polynomial f(x) = a
x
n
+ a
1
2
+ ... + a
x
n
n
is also a zero of the polynomial f
1
+ a
x
2
x + a
1
with real or complex coecients (a
k
, x
x
1
, ..., x
then

n
2
x
x
i
1
1 i
i
1
2
 ...  i
... x
i
2
i


0)
a
n
has zeros
≠
n
0
k
 1
, 1 
k

arbegla dna rebmuN
3
n; specically, for the sum
k
n
k
a
n
and the product,
a

n

x


x
1
 ... 
x
2

x
3
1

n
a

n

n
a

0
x
x
1
x
2
... x
3

n
 1

a

•
n
When solving systems of two linear equations with two unknowns there are three possible types
of solutions that can occur. Each type has a dierent geometrical interpretation.
❍
When there is a unique pair of numbers that satisfy both equations, the lines intersect at one
point.
❍
When there are
no real numbers
that satisfy both equations, the lines are parallel and
distinct.
❍
When there are innitely many pairs of real numbers that satisfy both equations, the lines
coincide.
•
When solving systems of three simultaneous linear equations with three unknowns there are
again three possible types of solutions that can occur.
•
❍
There is a unique triplet of numbers that satisfy all three equations.
❍
There is no triplet of real numbers that satisfy all the equations.
❍
There are innitely many triplets of real numbers that satisfy all the equations.
Finding polynomials by points
2
When you know three points through which a quadratic function f(x) = ax
+ bx + c passes, you can
formulate three equations in three unknowns a, b and c. Solving these gives the equation of the
quadratic.
Developing inquiry skills
Return to the opening problem. Your city council has asked you to prepare
a plan of a roller coaster for a new amusement park.
How would you go about this task using what you have learned in this
chapter?
213
/
3
E X PA N D I N G
THE
NUMBER
S YS T E M :
C O M PL E X
NUMBERS
Click here for a mixed
Chapter review
1
Show
that
the
equation
review exercise
x(x
-
a
-
b)
=
1
-
ab
7
The
polynomial
2
always
b
has
real
solutions
for
all
of
a,
(f(x))

∈
3
2
values
The
polynomial
f(x)
=
f
is
4
=
9x
a
What
b
Given
+
is
such
that
3
2
12x
the
-
26x
degree
of
aandb
that
-
the
are
d
∈
,
has
two
zeros,
+
z
bx
+
cx
and
w.
+
d,
a,
+
+
f,
that
z
=
(1
64
Find

the
zeros
of
f
?
the
1

3
i)
63
i
b
-
polynomial
+
nd

Show
25.
b,
polynomial
3
a
+
2
ax
1
c,
20x
(1
i)
=
-4.
18
 i
8
 ...  i

in
the
Find
the
complex
numbers
z
w
and
such
form
*
1 
2i 
z

that
i

3

iz

6 
that
z


*
a
c
+
bi,
a,
Hence
b
nd
integer


.
∈
the
polynomial
coefcients
which
f
that
are
has
i
2
9
Given
-
z
+
1
=
0,
show
that
2019
relatively
z
=
-1.
prime.
Exam-style questions
3
The
system
ax

y

of
simultaneous
equations
2
10
, a  ,

is
P1: a
Show
that
the
solutions
to
the
given.
2

x
a
 ay

Find
2
the
system
b
equation
values
has
a
the
solution
For
which
of
a
for
unique
in
solution
terms
value
of
which
a
of
written
the
and
where
nd
x
in
-
6x
the
-
43
form
pandq
are
=
0
=
p
x
may
± q
positive
13 ,
integers.
(4
a.
does
the
b
system
Hence,
or
otherwise,
be
solve
marks)
the
2
inequality
have:
x
-
6x
-
43
≤
0.
(2
i
no
ii
marks)
solution
innitely
many
solutions.
equation
of
line.
the
Write
the
2
11
P1: a
Solve
by
the
equation
8x
+
6x
factorization.
-
(4
5
=
0
marks)
2019
(1 +
4
a
Prove
i)
b
that
is
an
integer
its
k
What
for
which
8x
must
n
∈

satisfy
so
+
6x
-
5
=
k
has
(3
marks)
P1: Find
the
range
of
values
of
k
for
which
=
has
2
is
(1 -
equation
3kx
+ k
3x
+ 3
0
an
imaginary
number?
two
real
roots.
(6
marks)
i)
13
your
GDC
to
solve
the
P1: a
Solve
the
inequality
(x
+
4)(3
-
x
4
+
3x
2
-
2x
4
+
2x
+
4
≥
x
x)
>
0.
inequality
(3
5
no
i)
n
Use
of
that
the
5
values
solutions.
n+ 2
(1 +
of
value.
12
b
range
i)
real
nd
the
2
2017
(1 -
Determine
and
3
+
3x
marks)
2
+
2x
+
3.
2
b
6
Solve
the
simultaneous
Solve
the
inequality
2x
-
11x
+
(4
3 x

z

2y
c
 8y

<
0.
marks)
 1

6x

9
equations.
3z
Hence,
nd
the
range
of
values
of
x
 6
which
satisfy
both
the
inequalities

4 y

 7z
 12 x

4
2
(x
+
4)(3
-
x)
>
0
and
2x
-
11x
+
(1
9
<
0.
mark)
214
/
3
P1: Find
the
possible
values
w ∈ ℂ
of
which
18
P1: Prove
that
wz
w,
∈
*
-
zw
is
purely
imaginary
2
w
satisfy
=
77
-
36i.
(9
3
15
P1: The
polynomial
marks)
for
all
z
ℂ.
(5
marks)
2
2x
+
ax
-
10x
+
b
has
19
P2: Find
the
equation
of
a
+
bx
quadratic
2
a
factor
of
15
of
(x
when
values
of
P1: f(z)
=
z
Given
aandb
P1: Find
of
(3
the
each
and
+
i)
b
(x
∈
is
48z
a
-
f(x)
+
remainder
2).
).
zero
176z
of
Find
(7
f,
+
sum
(9
and
f(x)
=
through
the
and
-71).
(10,
ax
points
+
( -1,
c
passing
-5),
(3,
-1),
(7
marks)
product
of
z ∈ ℂ
marks)
the
roots
function.
=
5x4
-
5
g(x)
function
the
4x
2
+
3x
+
2x
-
1
(4
b
the
marks)
260,
nd
zeros.
3
a
a
2
8z
-
has
by
(a,
3
-
remaining
17
1),
divided
4
16
-
arbegla dna rebmuN
*
14
=
5x
4
-
4x
3
+
3x
marks)
2
+
2x
+
-7x
(4
+
10
marks)
215
/
Approaches to learning: ritical thinking,
Making a Mandelbrot
ommunication
Exploration criteria: Mathematical
communication (), Personal engagement (),
Use of mathematics (E)
ytivitca noitagitsevni dna gnilledoM
IB topic: omplex numbers
Fractals
You may have heard about fractals.
The image above is from the Mandelbrot set, one of the most famous examples of a fractal.
This is not only a beautiful image in its own right. The Mandelbrot set as a whole is an object of great
interest to mathematicians. However, as yet, no practical applications have been found.
This image appears to be very complicated, but is in fact created using a remarkably simple rule.
Exploring an iterative equation
Consider
this
iterative
What
do
you
think
is
happening
found
in
this
calculation?
to
the
values
equation:
2
z
=
(z
n+1
)
+
A dierent iterative equation
c
1
Consider
a
value
of
c
=
0.5,
so
you
have
Now
consider
a
value
of
c
=
-0.5,
so
you
have
2
z
2
z
=
(z
n+1
)
+
0.5.
(z
)
-
0.5.
1
1
Given
that
z
=
0,
nd
the
value
of
1
Now
=
n+1
nd
the
for
a
.
Again
start
with
2
value
of
z
few
more
z
=
0.
1
.
3
Repeat
z
What
happens
this
time?
iterations.
216
/
3
The Mandelbrot set
The
of
Mandelbrot
c
for
which
set
the
consists
sequence
of
all
those
starting
at
values
z
=
0
1
not
escape
where
the
That
only
The
is
innity
calculations
value
need
to
part
of
c
of
in
to
be
real.
It
could
be
complex
If
you
pick
a
Mandelbrot
Consider
off
story
of
the
not?
starting
with
a
to
of
c
innity).
does
form
number
or
values
however.
function
complex
set
zoom
the
the
(those
ytivitca noitagitsevni dna gnilledoM
does
a
for
value
+
c,
of
not
bi.
is
z
it
in
the
=
0
+
0i,
of
1
+
i,
1
rather
than
Consider,
just
for
“0”.
example,
a
value
of
c
2
so
you
have
z
=
(z
n+1
z
=
0
+
)
+
(1
+
i).
1
0i
1
Find
z
2
Repeat
to
nd
z
,
z
3
Does
this
(Does
it
diverge
zoom
,
etc.
4
off
or
converge?
to
innity
or
not?)
HINT
One way to check is to plot the dierent values of z
, z
1
, z
2
, z
3
, etc,
4
on an Argand diagram and to see whether the points remain within a
boundary square. Or you could calculate the modulus of each value
and see whether this is increasing.
Repeat
for
Try,
example:
a
for
c
=
What
0.2
a
–
few
some
You
may
as
can
values
b
0.7i
happens
Try
more
if
values
c
=
of
c
=
of
c
-0.25
+
0.5i
i?
your
own.
Ex tension
it
number
nd
be
a
GDC
used
gets
for
very
will
help
complex
big,
the
with
these
numbers.
calculator
calculations
When
the
will
not
be
that
some
•
Try to construct your own spreadsheet or write a
able
code that could do this calculation a number of
to
give
a
value.
You
may
also
notice
times for dierent chosen values of c
numbers
get
“big”
a
lot
quicker
than
others.
•
It
is
clearly
a
time-consuming
process
to
try
all
Explore
edges
There
are
programmes
that
can
be
used
to
output
after
several
iterations
when
happens
as
you
zoom
in
to
the
of
the
Mandelbrot
set.
calculate
•
the
what
values!
you
What is the relationship between Julia sets and
input
the Mandelbrot set?
a
number
belong
The
for
to
the
c.
This
those
points
values
indicate
Mandelbrot
Mandelbrot
colouring
will
of
c
set
set
d i a gr a m
on
an
that
and
is
values
which
cre a t e d
Ar g a nd
escape
which
to
don’t.
by
di ag r a m
inni ty
of
of
in
c
•
What is the connection to Chaos Theory?
•
What are Multibrot sets?
•
How could you nd the area or perimeter of the
a ll
one
Mandelbrot set?
colour
(say
bounded
in
black)
a nd
anothe r
a ll
thos e
co lo ur
( s ay
that
r em ai n
r ed ) .
217
/
ANS WERS
Exercise 3A
1
x
=
2
−5orx
=
a
34
1

−3
Exercise 3G
If
m


1

two
distinct
real
a
1
5i
1
b

roots

i

2
x
=
−7orx
=
65
5
2
65

x
1

1
3

or
x


2
m


c

one
repeated
real
5
d
90

3
37i

4
x
5

or
x
1
m



122
or
x
no
real
169
a
Re(z)
=
2,
b
Re(z)
=
−1,
c
Re  z
roots
2
b


i
169
5

2
6
x
i


2
5
+
e

5
1
root
t



two
distinct
real
Im(z)
=


0 , Im  z 

16
5

73

1
a
1,
0
8
roots

Exercise 3B
−1
73

t
=
If
2
5
Im(z)
7
b
3, 10
t



one
repeated
real
ro
o ot
3
a
a
=
2,
b
b
a
=
−5,
=
1
16

1 
1
5

c
2
t

3


no
real
15
4
5
15
1
f
,-
2
−5
10
a
c
Two
distinct
real
1
1
2
2
b
i
13
-
=
roots
16

e
b
73
,2
d
i
13
roots:
1 
s
5
2
<
0
or
s
>
40
4
c
1
–
i
d
-
3
+
i
13
2
a
b
2.41,
One
repeated
s
or
root:
0.4141
0.382,
=
0
s
=
4
5
a
a
=
6
a
x
2b
b
a
=
−b
2.62
No
c
real
26
1.00,
real
roots:
0
<
s
<
1
4
=
, y
=
b
0
(x,
y)
∈
Ø
1.50
2
d
2.26,
Exercise 3E
0.736
c
1
x
∈
]
4,
2
x
∈

−2, y
0
3
1 ±
5
b
5,
3
c
-
d
a
0
d

b
0
x
∈
]
∞,
2] ∪ [15,
+∞[
6
1, y
=
0
 3
4
x


i
2
5
2
x

 , 



5
13
9
+
c
14
46i
b
28
d
2
a
+
96i
6
x
2


i
2
2



3

2
1
b
Exercise 3F
b
8i
2

 2
f
3
1
2, 
2
4
a

3

f
4 

e
1
6


e
25

e
2
5i

4

121 - 4
5
, 2

11 ±
c
,2
2
2
a,
=
17

25
a
x
6
1
2
or
Exercise 3H
6,
e
=
{4}
1
a
=
2[
Exercise 3C
1
x
2


i
b,
2
2
2
1
3
c
k,
2k
-
d
Re(z)
=
0,
Im(z)
=
−4
b
Re(z)
=
5,
Im(z)
=
0
1
1
,
p
a
c
Re(z)
=
−24,
Im(z)
=
Re ( z
)
=
12
,
Im
(z )
=
37
>
0
⇒
two
e
roots
4
a
Use
3G
13
Re ( z
)
=
-
result
from
Question
induction
2
,
Im
(z )
=
−7
c
Δ
=
0
<
0
⇒
no
real
5
b
P(n):
P(1)
one
(and
as
below).
(z∗)
n
=
(z
+
)∗,
n∈

roots
2
⇒
Exercise
7a
n
Δ
i
3
=
5
b

distinct
1
real
1

d
-
13
=
5i)
7
5
d
Δ
+
2
Exercise 3D
a
(2
p
c
1
±
repeated
a
4
b
5
d
1
e
1
c
is
true
25
real
Assume
root
P(k)
is
true
for
+
some
d
Δ
=
−37
e
Δ
=
π
<
0
⇒
no
real
roots
3
a
3
3i
b
25
k∈

13i
k
(z∗)k
+
1
=
(z∗)
k
z∗
=
(z
)∗z∗
2
8
>
0
⇒
two
25
c
distinct
real
13

roots
12
i
d
3
+
10i.
k
=
(z
z)∗
using
Question
f
Δ
=
0
root
⇒
one
repeated
Exercise
3G
6
8c
real
k+1
=
(z
)∗
774
/
2
4
5
a
n
=
4k,
k
a
f(x)
=
(x
+
2)
(3x
5)

∈
Then,
x

x
1

x
1
2
2
b
b
n
=
4k
+
2,
k
f(x)
=
(3x
2)
(x
+
4)

∈
a

2
c
f(x)
=
(x
1)
(x
+
2)(x


x

a
2
2)
1
Exercise 3I
2
3
d
f(x)
=
(2x
+
(x
1)
+
1)
and

x

x
1
x
1

x
2
x
1
2
2
1
a
q(x)
=
2x
3x
+
1
4
e
3
b
q(x)
=
f(x)
=
(x
1)
(5x
+
b
7)

2
3x
+
x
+

=
i
b
1
5
4
q(x)

x
1
1
c

b
2
a
2i,
2
b
3 
2i ,
2
x
x
So
2
the
three
zeros
are
2
2
2
a
q(x)
=
3x
3x
2,
r
=
1
−3
3
c
q(x)
=
2x
=
6x
i, -
2
2
b
5x
+
5,
a ,  i
7
+
d
2
i,
1
±
b
4i
and
3
the
product
  x
  x
of
the
roots
is
r(x)
x

15
1 
e
2 
7
1
1


2

i
i
b
i

b

a
i,
2
c
q(x)
=
x
+
x,
r(x)
c
2

ab



c
2
=
−x
x
+
1
1 
1
2 i
,
f

6
i

ab
Exercise 3J
b
a
q(x)
=
x
+
4x
+
5,
r
=
11
6
a
a
=
−2,
±
i
a
b

3,
Exercise 3O
2
b
q(x)
=
2x
c
q(x)
=
2x
d
q(x) = 3x
3x
3
1,
r
=
1
c
2
+
2x
4
x
3
+
3,
r
=
a

22,
b

10,

5 i, 1  4 i
1
1
a
1
<
x
<
2
2x
x + 13,
1 
a

7, b
b
(x
−81
+
11i
6
4)(x
+
b
x
>
3,
c
x
≤
2
1)(2x
5)(x
x

Exercise 3M
3
<
a
sum
=
3,product
=
x
4
>
=
x
b
q(x)
=
3x
x
+
3,r
=
7
b
+
x
+
1,r
=
c
q(x)
=
x
d
q(x)
=
x
c
2
+
2x
sum
=
3,product
=
0
f
x
+
1,r
=
sum
=
0,product
=
4
3
2x
2
+
x
x
+
d
3,
sum
=
=
x
b
=
−1
g
x
>
≥
product
=
h
-
<
x
<
-
5
,
3
2
1,
1
≤
x
≤
1,
x
≤
−2
<
x
< 1
3
2
3
2x
1
-
1
-
−1
4
4
1
8
,
3
−12

2
4
4
=
x
1
2
3
a

2
3
1
2
3
−2
-
2
q(x)
=
<
1
a
f(x)
x

2
e
1
2
−2
3
,
Exercise 3K
r
<
3)
d
1
x
 5,1  2 i ,
1
2
,
2
+ 2x
d
=
a
i

2
r
c
6
2
1

3
3x
2
+
3x
2x
2
2
a
1
17
b
2
4
2
3
x
a
<
1.17227...
x
>
−1
b
x
≤
1
6
c
5
d
c
1
≤
x
≤
1
x
>
1.78897...,
<
x
<
17
5
a
=
−5
b
=
−6
4
6
a
=
4
b
=
a
a
b
2
b
x
+
1
12
3
7
1.8947...
1
1
3
2
c
1
d
0.746571...
<
x
<
4
1.27299...,
8
0
x
Exercise 3N
Exercise 3L
c
1
3
1
a
9x
+
+
−2,
c
x
=
2
1
b
x
=
−3,
d
x
=
1,
2,
0.505312
<
x
<
0.50533....
3
2
Exercise 3P
2
x
11x
9x
+
18
2
a
4
b
1,
1
a
14
b
2
a
2
b
3
a
2,
3
3
11x
2x
3
2
=
14x
3
x
4
c
x
−1.09526
2
x
4
b
a
<
+
23x
+
10
2
2
2
a
x
2x
b
2x
2x
+
4
2s
5
4
+
3
x
3
2
x
6x
3x
+
a
a
=
−11,
b
=
12
3
x
1
=
1
, y
3s
=
2
3s
2
1
5
c
4
x
3
2x
2
2x
2x
+
4x
+
4
b
b
2
2
2
3
a
(x
2x
1)(x
+
s
2)
x
4
a
Let
the
three
(x
+
1)(3x
roots
=
4x
+
2s
, y
be
-
s
+ 8
+ 4s
- 5
=
2
s
2
b
2
- 4
2
+ 4s
- 5
s
6)
x
1
,
x
2
and
x
2
2
c
(2x
+
1)(x
1)(x
2x
+
5)
775
/
ANS WERS
2
5
1
4
a
x
=
, y
a
b
The
=
x
∈
2.45,
1.26]
(
[
does
not
0.339, 0.715]
solution
when
a
+
∪
[1.34,
k
3
b
=
0
6
 x , y, z 
1

>
0
(1
mark)
(1
mark)
2

, 
3k
36k
7
3
2
a
2

2
b
12k
8
z
=
1
3i, w
>
0
0.4
k(k
2  i
0

k
Exercise 3Q
>
, 2


x
4 ( 3k ) ( 3)
)
∞[
have
 2
a
∪
0
+ b
system
[
=
3
+
12)
>
0
2i
1
Critical
values
are
k
=
0
and

y
2i
x
9
Using
of
a
geometric
k
12
(1
Solution
2

mark)
is
k
<
0
or
k
>
12
3
y
103
26  12i
 z 
56i
 1
2
1 
z

(2
3
z


z
0

z

 1
13
673
2019
m
≠
−2
b
m
≠
2
and
m
≠
a
Using
sketch
of
y
=
(x
+
4)
673
3
z
a
marks)
1
Exercise 3R
1
=
sequence,
87  35i
1
sum
z
1


1
(3
x)
(1
Correct
3
sketch
or
mark)
table
Exam-style questions
Solution
2
a
k
≠
6
b
k
=
is
4
<
x
<
3
2
2
-b
10
1
3
m
=
0
or
m
= 1
or
m
a
x
±
b
(2
- 4 ac
2
2a
=
b
2
9
4
a
a
(1
21
=
,
b
=
c
=
6 ±
x
4
4
2x
11x
(2x
9)(x
+
9
<
0
mark)
7
,
marks)
=
1)
<
0
208
=
(1
mark)
(1
mark)
(1
2
mark)
2
22
34
Using
b
a
=
,
b
=
-5,
c
=
x
3
=
3 ±
52
sketch
of
y
=
(2x
9)
3
(x
x
=
3 ± 2
(1
13
1)
(1
mark)
Correct
Chapter review
b
Using
sketch
or
mark)
sketch
or
table
table
9
1
x(x
a
b)
=
1
ab
Solution
3 - 2
13
£
x
£
3 +
2
is
1 <
x
<
13
2
2
⇒
x
(a
+
b)x
+
ab
1
=
0
(2
marks)
(2
marks)
2
Δ
=
(a
+
b)
4(ab
2
1)
11
a
8x
+
6x
5
=
0
c
2
=
a
=
=
2
+
2ab
+
b
a
2ab
+
b
(a
b)
4ab
+
(4x
4
+
5)(2x
1)
=
answers
a
and
b
gives
1
<
x
<
3
marks)
2
+
from
0
(2
2
Comparing
(1
4
mark)
5
2
+
4
>
0
4 x
 5

0

x

14

Let
w
=
a
+
bi
where
a,
b ∈

4
2
So
there
are
two
distinct
(a
real
(1
solutions
for
b
∈
bi)
2
all
 1 
0

x
=
77
36i
(1
b
+
2abi
=
77
36i


(1
2
(1
a
(1
i)
3
+
(1
+
mark)
mark)
2
3
2
mark)
2
a
1
2x
a,
+
mark)
Equating
i)
reals:
2
a
b
=
77
(1)
2
b
2
=
(1
3i
+
3(
i)
+
3i
+
3(
i)
+
6x
5
k
=
0
3
i
2
(1
8x
)
+
No
3
+
i
real
solutions
(1
mark)
(1
mark)
(1
mark)
⇒
)
Equating
imaginary
2
4ac
b
=
(1
3i
3
3i
3
+
i)
+
0
(1
2ab
4×8×(
5
k)
<
(
2
=
−4
2i)
=
−36
(2)
0
i)
18
(1
=
mark)
+
36
(1
<
+
(2i
mark)
(2)
gives
b
=
-
a
2)
36
+
32(5
+
k)
<
0
Substitute
in
(1):
36
2
2
b
1
5 + k
<
-



i
32
18 

2

a



5
77

5
a


36
3
c
f(x)
=
5x
2
+
16x
15x
+
4
k
<
-
- 5
32
324
2
a
2

a
a

1,  x , y 

40
k
a
 1
<
-
-
1  a
8
8
4
2
a
i
a
=
1
a
=
−1,
−
77a
y
=
x
+
<
(1
-
mark)
Two
real
roots
0
b
n
is
to
factorise,
or
the
quadratic
formula:
implies
2
–2
=
8
2
12
a
324
Attempting
using
4
−
49
k
ii
77
a
,


b
=
2
9
3
-
2
(a
even
2
+
4)(a
−
81)
=
0
(1
mark)
2
b
4ac
>
0
(1
mark)
776
/
2
Since
a ∈
,
a
=
81
(1
mark)
4

17
a
Sum
of
roots

=
±9
a
=
9
a
=
−9
b
⇒
=
b
(2
−2
=
Product
2
4
So
w
=
±(9
2i)
(2
marks)

3
15
Let

5

⇒
p(x)
=
 1
+
10x
+
1


=
0
⇒
2
+
a
10
+
b
(2
=
5
=
8
(1)
4

marks)
(1
b
2)
=
=
15
⇒
−16
+
4a
of
roots

b
=
11
(2)
+
(2
20
(1)
limit


5
5

1
Continuous
2
Not
3
Continuous
4
Continuous
5
Not
continuous
6
k

marks)
continuous
of
roots
marks)
5
equations
0
+

Solving
No
6
4


15
+
7
3
mark)
Product
4a
2
4
Exercise 4B
Sum
(2
p(
5
limit
0

b
No
6
marks)

+
3
2
5

b
b
a
4

(2
p(1)
Exercise 4A


ax
marks)
1
1 

5

roots

2
2x
of
Chapter 4


a
4


and
 1
 10 


(2)
2

5


1
1
simultaneously:
(1
mark)
(2

k
7

marks)
2
3
a
=
1
(1
mark)
b
=
7
(1
mark)
18
Suppose
w
=
a
+
bi
and
z
=
c
+
di
8
for
16
Since
3
i
is
a
zero,
a,
b,
is
also
a
d ∈

(1
a
Not
continuous
at
b
Not
continuous
at
±1
c
continuous
d
not
continuous
at
x
=
−5,
e
not
continuous
at
x
=
1
f
continuous
±3
mark)
its
Then
conjugate
c,
zero,
w∗
=
a
bi
and
z∗
=
c
di
i.e.
(1
mark)
∗
(3
i)
=
3
+
i
is
a
zero.
So
(1
wz∗
the
factor
(3
i)]
=
(a
+
bi)(c
1
di)
+
di)(a
bi)
theorem,
=
[z
zw∗
mark)
(c
By
−
is
a
factor
of
f,
[ac
+
bd
+
i(bc
ad)]
[ac
+
bd
+
i(ad
bc)]
and
Exercise 4C
[z
(3
+
i)]
is
a
factor
of
f.
(1
(1
1
mark)
=
Therefore
[z
(3
mark)
i)][z
i(bc
ad)
i(ad
a
5
b
3
c
6
d
2
e
none
f
0
h
a
bc)
(3
(1
mark)
2
+
i)]
=
z
6z
+
10
is
also
a
2
=
factor
of
f.
(2
3
z
=
6z
+
+
i(bc
48z
176z
2(bc
+
kz
+
+
b
ad)i
1
2
+
which
10][z
2
g
ad)
3
2
[z
+
2
8z
2
260
ad)
marks)
=
4
Writing
i(bc
is
purely
a
3
b
2
c

3
imaginary
26]
(1
mark)
d
none
e
0
g
none
h
4
a
x
f
0
2
Equating
coefcients
of
z
20
gives
48
=
26
6k
+
At
(
1,
5):
a
b
+
c
=
−5
10
(1
(2
1
mark)
marks)
3
=
1
;
y
=
6
At
So
k
=
−2
(1
(3,
1):
9a
+
3b
+
c
=
(1
mark)
2
z
2z
+
26
=
0
At
(10,
71):
100a
+
10b
+
z
 2

4  1  26
=
−71
(1
b
x
c
x
d
x
e
x


3;
y
 1
c
2
  2  
2
−1
mark)
=
1;
y
=
−1
2;
y
mark)

2
2 
100

Solving
2  10i



0
using
GDC:
(1
mark)
a
(1
mark)
=
0;
no
horizontal
 1  5i

2
simultaneously
asymptote
2
=
−1
1
(2
marks)
f
b
=
3
(1
x
=
1
;
x
=
zeros
and
(1
are
therefore
(3
±
y
=
mark)
2
The
1;
2
i)
c
=
−1
(1
mark)
Exercise 4D
±
5i).
1
a
Diverges
b
Diverges
777
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