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Complex
Numbers :
Natural ✗ + 2=
I
imaginary
numbers
Integers
x
2 =\
it
= - I
④ Rational 2x
i = Fl
IR
Real ✗ 2=
2
c >
= ±F
x
=
IF ✗ FT
x
=
I2i
← 2 distinct
imaginary
solutions
2x
X
2x
10
= 0
b
49C
§R
(x
2
= 0 C. T. S
C- 2)
2
2 = -
'
. doesn't
cross oc axis x
= I
3i
= I ±3i
Solutions
...
= I
and ✗
= I + 3i
Real
no.
an imaginary
no.
'
. they're
complex
numbers ,
Every
complex number is
= ✗ +
iy
X
, YER
a pair of
real numbers
Re (2)
=
x Im (2)
=y
where 2 = 0C + iy
T
meaning
x meaning
y
is the real is the
part
of
. imaginary
part
of 2 .
Quadratic
equation
...
AZZ + bz + C
= 0 eg
←
if
this is
negative
then
I
✓ 42-4× 1 × 13
=
there are
2A complex numbers 2 =
2 × 1
involved.
=
I
-36T
✓
4+-6 i
z
=
=
I
3i ✓
Complex
Conjugates
:
The
complex
conjugate of
2 ,
written 2.* or 2-
,
is the
complex
number with the same real
part,
and opposite imaginary part
.
eg
. 12 + 7- i)* = 2 - 7- i This works for any polynomials
if
0C +
iy
then 2
= x
iy
as long
as the coefficients
are
real.
Arithmetic with
complex numbers :
Addition : Subtraction :
Multiplication
:
if
,
= 2 + Si
if
,
= 2
si iz = - I ← most important part
And
Zz
= I
Zz
= I
= ( 2+ i
1-
3i )
=
2i
= I + 8i
= 2
Si
' ←
multiply
the brackets
= 2 + (5- 6) i
15 ×1-1 )
← factor out i
=
Ii + 15 ← i 2=-
and simplify
= 17
expansion
:
(I + i )
"
I + 4i + 6i2 + 4i3 +
i
"
if
i 2=-
= I + 4i
-6 +
4i + 1 then i
= iz ✗ i = - i
=
4
= IZ
✗ i
2
i
'
is
key facts
iO
= 1
i
'
i
[
2 = - I
i.
= - i
i.
i.
4
1
is = i
i.
°
...
Ji
Argand diagrams
:
2- x
iy
iy
A method of plotting
complex
numbers on a 2D
plane
...
±
Cartesian
form
→ 0C
iy
Q1)
= 2 + i
and Zz
= I + 3i
Show 2 , ,
Zz and their
complex
conjugates
on an Argand diagram
.
imaginary. Zz
The
operation of
conjugation
corresponds
to a reflection i 1 I ' l l real
in the real axis.
z
2
① 1- b) Now show 2 , , Zz and their sum on an Argand diagram
in
Z sum
+ i
I
+ 3i
z ,
i →
Zsum
real
The
points
can be joined
with a parallelogram
which suggests
they're
behaving
like vectors .
odulus and argument
:
2- x
iy
we can think of
a complex
number as the
position
vector of
a
iy
iz
!.
.
'
'
point
in the
plane
where...
.
.
'
'
)
Arg
2
,
The magnitude , from the origin,
is called the modulus
✗
And the angle,
measured anticlockwise between the positive
real
axis and the vector is called the argument Arg
2
im
2 = ✓ ✗
2
yz
121 tan ( arg
= ¥
y
L / Arg
Re
x
let
r and let O =
Arg
2
SO..
.
Which means ✗ = rcoso and
y=
rsino so
...
= r ( Cosa + isinc)
I
r This is the modulus
rsino
argument form
RCOSO
eg
. express
53 + i in modulus argument form
in
i
0C
y
= i
y
r
TÉy
→ (FsY → %
= 54 = 2
re
0C
Angle
= tan
= to e-
In modulus argument
form
...
2 = r ( cos
I +
isino )
(cos
76 + isin
Transposing
a matrix:
Here
you
're trading
rows for
columns by flipping
the matrix
about the
leading diagonal
.
the
leading diagonal
is point
and
2,2 and
the line that
joins
them .
eg
. Original Transposed
Notation
T=
transposed
I
I -3 I -
=
Leading
← diagonal
still in 2
Position .
leading
diagonal
Multiplying different
dimensions :
- I 2 2 0 -3 Two matrices can
only
be
=
multiplied if
the number of
Columns
of
the first
( → )
matches
the
number
of
rows
Of
the second (
t ) .
2 ×
-1 = -2 2 × 4 = 8 2 × 0
=
I
× 2 = 8
4 × 0
= 0 4 ×- =
3 ×-
=
÷
-3×4=-12 -
× 0
= 0
2 × 2
=
4 2 × 0
= 0 2 ×- = -
I ✗
4-
I ✗ 0 =
D-
= 11 -
4 - 6
The
identity
matrix :
I 0
Iz
Therefore
, if
A
=
any
2in matrix then
IA
= A
if
I
=
' O
o ,
and A
=
3 o
then.. . And
if
B=
any
mx2 matrix then
BI
=
B
IA
=
3 o
= A
1 0 0 This works
in
a
Ig
=
° I 0
similar
way
to
Iz
O O l
1
Q ) Given
A
=
I 1 3 and B
= -
l s calculate AB
2
- I ✗ 3 = - - I ✗ I = - l l ✗ 3 = 3
I ✗ I = I
- 4 × - = 4 - 4 × 5 = - I ✗ - I = - I 1 × 5 = 5
6 × 0
= 0 6 * 2 = 12 3 ✗ 0
= 0 3 × 2
= 6
=
I
= I
A
transformation to a matrix :
a b A b
c d
and
points
1
and
are
mapped
to and
O l
C d
eg
. find
a matrix that
represents
a rotation of
'
about the
origin
.
(9)
Now rotate These
form
the
.
(f)
matrix.
..
c- (f)
T.ie.
Answer =
For
any angle
:
After
Before rotation length of original
vector
✗
i.S.in . ?
.. -
since
Cosa
:
COSO
so after
the
transformation,
I
is mapped
to
COSI
sun
Cosa
since
These
form
=
Sino Cosa
°
the matrix...
is
mapped
to
I COSQ
Invariant
points
:
0
During
the multiplication of
any
points
in the form
-0C will be
mapped
onto
itself
because the points lie on the line y=
>
these are invariant
points because their location stays
the same.
All matrix
transformations
have the origin
as an invariant point.
To
find
the
line
of
invariant points
...
eg
. if
the transformation
is
12 5
2 I X
=
3C
y y
which when
multiplied gives
...
2x
y
= x
5y =y
and then
simplifies
to
...
30C
y
= 0
= 0
and rearranges
to Ayy points along
this line will be invariant
for
this transformation.
y
=
30C
Invariant lines :
Moreover , any
line
at
right angles
to the line
of
reflection
can be called an
invariant line
and here ,
the
points
may
not
map
onto themselves but
they
will be somewhere
along
the
line
.
5
leg . find
the two invariant lines of
the transformation given by
2
I 0C
'
say
that
y
is mapped to y
' and the invariant line has
gradient
m and intercept
c.
so y=mx
y
' = Msc
'
c
matrices
summary
:
Reflection
in the line 0 1
Reflection
in the line
= -0C .
I 0 y=x
I 0
Reflection
in the
in the
axis
Cosa
since
Anticlockwise k O
enlargement
with Centre (0,0 )
rotation of
O
Sind Cosa about origin
.
0 U
scale
factor
K
.
K o stretch parallel
I 0
stretch parallel
1 to x axis 0 k to
y
axis
A b 0C
=
0C
'
go
0C
'
ax +
by
C d y y
'
y
'
Cx
dy
Algebra
and functions
:
roots of
polynomials
Quadratic
:
5=0 You will
get
the solutions to be the
roots : 2=
complex
conjugates if
the polynomial
= 1 - Zi contains oxy
real terms .
If
the
polynomial
doesn't contain
only
real values
eg
.
5iZ -4=0 then the solutions
arerit
the
complex
conjugates
roots : 2=0 And 2
4i
Cubic :
Because a cubic
produces
terms , they
cannot all be non
This is due to the fact
that if
one of
the roots is non
real then its
conjugate
is
also a solution ( non
come in pairs
Therefore,
one root is always
real
3 real roots
one real
root
I
i
Quartic :
They
could all be real ,
real and 2 non
they might
all be non real.
Q1 ) 23
Find
the other roots .
i ←
because one solution is non
the other is its conjugate .
is a root if
and
only if
( 2- a) is a factor
so...
( z
= (z
2
iz
→
( z
2
172
- 15 = ( 22 - 42 + 5)( 2- ✗
= - ✗ .
'
Roots =
,
2 + i ,
2 - i
① 3) Two
of
the roots
of
the
quartic
equation
924+623 + CZZ + dz + e=o are
given by
Li
and 3-
i.
Given all
of
the
coefficients
are real , find
their values .
quartic
= 4 roots → 2
pairs
of
non
Roots
: Zi
,
2i
,
i.
=
i use the reverse 2 = 2i ← Repeat for
the other root .
= i
of completing
the 22
= (2i)
(2-3)
' = i
2
square
to find
the 22
= 4i
(2- 3)
Original quadratic
.
=
(Z
= 0 I
1=
22-62+10=
Now your
solutions are (22-62+10)
(22+4) but
this
isn't
in the correct form
...
Expand
out the brackets
...
22
24 - 623
422 -242 40
= 24 - 623
←
simplify
'
. A
= I ← state all
of
the
values
of
the
coefficients.
= 14
D= -
f-
40
Product and
sum of
the roots
:
bz
C =O
Which has roots 2 = ✗ and 2--
22 + ¥ 2
§
=o Divide through by
a
Any
quadratic
can be written in the
form
...
(Z
expands
to 22
so we can
compare
the two.
..
C
22
¥ 2
-a
= 22
Which means we can say
sum
of
Roots : Product of
Roots :
✗ +
=
a
13
=
c-
a
Q )
calculate the sum and
product of
the roots
of
the
quadratic equation
222
32
A
= 2
B
=3 C
SO...
b- C
a
= 3- and I
=
sum of
roots
✗
=
-3-
Product
of
roots
xp
- Given that 323
622+42+12=0 has
roots ✗ , B ,
V
.
Find
£
¥
É
and
2
132
part
- a =3 b. =
C-
4 D= 12
← define
the
coefficients
B+
r=
equation
✗ + B
+ V
=
Bt
✗ r+Br= Ea
=
¥
←
equation
Br
=
%
=
13=-
←
equation
I 1
138-1×8+
← make over a common denominator 1-
g-
=
✗ Br
=
I ,
=
← sub
' in the values
part
- (✗
B
2
(✗ + Btr )( ✗
Btr)
= ✗ 2+132+
( ✗
✗ r t
Br
)
2
132
'
( ✗
B
+ f)
2 (✗
B
✗ r t Br
=
Ex )
B
= 22
=
§
Q2) Given
that
223-
PZ
+9=
has a root
of
3i and
p
and
q
are real
numbers , find
...
i) the
remaining
roots
ii
p
iii
q
i)
Roots
:
2 + 3i
,
2- 3i so ii
B
= (
(2-30)
= 2+3 i
4- 9iZ
=
13=
so ✗
T
= 2( 2. + 3i )
= 4 +
6i
Then ✗ +
Btr
=
Ia
=
¥
=
so if
then 8--
Br
= 2 (
2- 3i)
= 4-
6i
Meaning
the
roots are :
sum
,
3i ,
2- 3i
Sum
of
Products
of
Roots
xp
artpr
= Ea
so
Pla
21
P
.
.
iii) ✗ 138=212+
i )(
3i)
=2(
4- 9i2)
=
Product
of
Roots
pr=
da
so
=
912=
meaning 9=-
