Post on 11-Apr-2023
133
COMPLEX NUMBERS A complex number is represented by an expression
of the form π + ππ where a and b are real numbers
and i is a symbol with a property ππ = β1.
π = ββ1 was introduced by a Swiss mathematician
Euler. Traditionally the letters Z and W are used to
stand for complex numbers.
Given a complex numbers π§ = π + ππ.
The real part of a complex number z is π π(π§) = π
and the imaginary part of z is πΌπ(π§) = π.
Both π π(π§) and πΌπ(π§) real numbers.
Thus the real part of Z = 4 β 3π is π π(π€) = 4 and
imaginary part of Z is Im(π) = β3
By identifying the real number a with a complex
number π + ππ we consider β (real numbers) to be
subset of β (complex numbers).
Consider the equation π₯2 + 9 = 0, this can be written
as π₯2 = β9 and we can see that the equation has no
real roots since we cannot find the real root of a
negative number, But with π2 = β1 (Euler) we are
able to find the square root of complex numbers.
π₯2 = β9
π₯2 = 9ππ
βπ₯π = β9 π2
π₯ = Β±3π
π₯ = 3π π = β3π
Example
Solve the following equations
(a) 4π₯2 + 49 = 0
(b) π₯2 + 2π₯ + 6 = 0
Solution
4π₯2 + 49 = 0
4π₯2 = β49
π₯2 = β49
4
π₯2 =49
4π2
π₯ = β49
4π2
π₯ = Β±7
2π
(b) π₯2 + 2π₯ + 6 = 0
From, 2 4
2
b b acx
a
π₯ =(β2 Β± β(2)2 β 4(1)(6))
2 Γ 1
π₯ =β2 Β± ββ20
2
π₯ =β2 Β± β4π2 Γ 5
2
π₯ =β2 Β± 2πβ5
2
π₯ = β1 + πβ5
π₯ = β1 β πβπ
With this new concept we are in position to find the
roots of any quadratic equation.
When the imaginary part of a complex number is
zero, the complex number becomes a real number.
Thus, all real numbers are complex numbers.
Definition
Given a complex number π = π₯ + ππ¦, the complex
conjugate of Z denoted by οΏ½Μ οΏ½ ππ π§β is a complex
number given by οΏ½Μ οΏ½ = x β iy. Therefore if π§ = 4 + 3π,
π€ = β2+ 4π
Then οΏ½Μ οΏ½ = 4 β 3i, οΏ½Μ οΏ½ -2 β 4i
Algebra of complex numbers 1. Addition
Given that two complex numbers
π§1 = π₯1 + ππ¦1, π§2 = π₯2 + ππ¦2. Then
π§1 + π§2 = π₯1 + ππ¦1 + π₯2 + ππ¦2
= π₯1 + π₯2 + π(π¦1 + π¦2)
Therefore if π§1 = 3 + 5π and π§2 = 2 β 7π
π§1 + π§2 = 3 + 5π + 2 β 7π
= (3 + 2) + 5π β 7π
= 5 β 2π
134
Example
1. Subtraction:
1 1 1
2 2 2
1 2 1 1 2 2
1 2 1 2
1 2 1 2
( ) ( )
( ) ( )
z x y
z x y
z z x y x y
x x y iy
x x y y
i
i
i i
i
i
π§1 = 4 β 3π
π§2 = 6 β 14π
Find (π§1 β π§2)
(π§1 β π§2) = (4 β 3π) β (6 β 14π)
= 4 β 6 β 3π + 14π
= β2+ 11π
2. Multiplication
π§1 = π₯π + ππ¦π, π§2 = π₯2 + ππ¦2
π§1π§2 = (π₯1 + ππ¦1)(π₯2 + ππ¦2)
= π₯1π₯2 + π₯1π¦2π + π¦1π₯2π + πππ¦1π¦2
= π₯1π₯2 β π¦1π¦2 + (π¦1π₯2 + π₯1π¦2)π
Example
π§1 = 3 + 5π, π§2 = 2 β 7π
Find π§1π§2
Solution
Find π§1π§2 = (3 + 5π)(2 β 7π)
= 3(2 β 7π) + 5π(2 β 7π)
= 6 β 21π + 10π β 35π2
= 6 + 35 + (10 β 21)π
= 41 β 11π
3. Division
π§1 = π₯1 + ππ¦1 πππ π§2 = π₯2 + ππ¦2
π§1π§2=π₯1 + ππ¦1π₯2 + ππ¦2
π§1π§2=π₯1 + ππ¦1(π₯2 β ππ¦2)
π₯2 + ππ¦2(π₯2 β ππ¦2)
π§1π§2=π₯1π₯2 β π₯1π¦2π + π₯2π¦1π β π
2π¦1π¦2(π₯2)
π β πππ¦2π
=π₯1π₯2 + π¦1π¦2 + (π₯2π¦1 β π₯1π¦2)π
π₯22 + π¦2
2
=π₯1π₯2 + π¦1π¦2π₯22 + π¦2
2+(π₯2π¦1 β π₯1π¦2)π
π₯22 + π¦2
2
Example I
Simplify π§ =2+6π
3βπ
π§ =2 + 6π
3 β π=2 + 6π(3 + π)
(3 β π)(3 + π)
=2(3 + π) + 6π(3 + π)
32 β π2
=6 + 2π + 18π + 6π2
10
=6 + 20π β 6
10
=0 + 20π
10
= 2π
Example II
Express β1+2π
1+3π in the form a + bi
Solution
β1 + 2π
1 + 3π=β1 + 2π(1 β 3π)
1 + 3π(1 β 3π)
=β1+ 3π + 2π β 6π2
(1)2 β (3π)2
=5π + 5
10
=5 + 5π
10
=1
2+1
2π
The Argand Diagram Complex numbers can be represented graphically on
a graph of Real (Re) and Imaginary (Im) axes called
a complex plane. The complex plane is similar to the
Cartesian plane where the imaginary axis
corresponds to the y-axis and the real axis
corresponds to the x-axis. The diagram representing
the complex number in complex plane is called an
argand diagram named after JR argand 1806.
On the argand diagram a complex number is
represented by a line with an arrow on the head to
show direction
If π§ = π₯ + ππ¦ we can represent z on argand diagram
as shown below.
π₯ π π (ππ₯ππ )
(π₯, π¦) Im Axis
y
135
If π§1 = π₯1 + ππ¦1 and π§2 = π₯2 + ππ¦2 then
π§1 + π§2 = π₯1 + π₯2 + π(π¦1 + π¦2)
π§1, π§2 πππ π§1 + π§2 is represented by vectors ππ1ββ ββ ββ β,
ππ2ββ ββ ββ β πππ ππββββββ respectively. The diagram shows that
π1πββββββ β is equal ππ2ββ ββ ββ β in magnitude and direction
ππββββββ = ππ1ββ ββ ββ β + π1πββββββ β = ππ1ββ ββ ββ β + ππ2ββ ββ ββ β
Thus the sum of two complex numbers π§1and π§2 is
represented in the argand diagram by the sum of the
corresponding vectors ππ1ββ ββ ββ β πππ ππ2ββ ββ ββ β
Representing ππ β ππ on the argand diagram.
(z1 β z2) = OP1 β OP2
= π1π2ββ ββ ββ ββ
Since ππ1ββ ββ ββ β β ππ2ββ ββ ββ β = π1π2ββ ββ ββ ββ
π§1 β π§2 can be represented by π1π2
Example
Represent the following complex numbers on the
argand diagram.
π§1 = 3 + 4π, π§2 = β2+ π, π§3 = β5 β 4π,
π§4 = 2 β 3π, π§5 = β4β 2π,
Solution
Modulus of a complex number Given a complex number π§ = π₯ + ππ¦, the
magnitude or length of z is denoted be |π§| is
defined by
|π§| = βπ₯2 + π¦2
Example I
Given π§ = 1 + β3π find |π§|
Solution
π§ = 1 + (β3)π
|π§| = β(1)2 + (β3)2
= β4
= 2
Example II
Find |π§| ππ π§ = β1
2ββ3
2π
Solution
π§ = β1
2ββ3
2π
2 2312 2
( ) ( )z
= β1
4+3
4
= β1
= 1
βΉ |π§| = 1
Example III
π§ = β3 + 4π find |π§|
Solution
π§1
π1 | β π₯2 β | π¦1
π₯1 π
Im
axis
Real axis
y2
0 1 2 3 4
5 Real axis
Im
axis
β5 β4 β3 β2
β1 5
1
2
3
4
β1
β2
β3
β4
z1
z4
z2
z3
z5
136
π§ = β3 + 4π
|π§| = β(β3)2 + (4)2
= β9 + 16
= β25
= 5
Properties of modulus
If π§1 πππ π§2 are complex numbers then
(π) |π§1π§2| = |π§1||π§2|
(ππ) |π§1
π§1| 1
2
| |
| |
z
z
Example I
π§1 = 5 β 12π and π§2 = 3 β 4π
Find |π§1π§2| πππ |π§1
π§2|
Solution
π§1 = 5 β 12π , π§2 = 3 β 4π
|π§1π§2| = |π§1||π§2|
|(5 β 12i)(3 β 4i)|
= |5 β 12i||3 β 4i|
= β(5)2 + (β12)2β(3)2 + (β4)2
= β169 Γ β25
= 13 Γ 5
= 65
Alternatively
π§1π§2 = (5 β 12π)(3 β 4π)
= 15 β 20π β 36π + 48π2
= 15 β 48 β 56π
= β33 β 56π
|π§1π§2| = β(β33)2 + (β56)2
= 65 π’πππ‘π
π§1 = 5 β 12π, π§2 = 3 β 4π
|π§1π§2| =
|π§1|
|π§2|=β(5)2 + (β12)2
β(3)2 + (β4)2
=13
5
Alternatively, 1
2
5 12
3 4
z i
z i
(5 β 12π)(3 + 4π)
(3 β 4π)(3 + 4π)
π§1π§2=15 + 20π β 36π β 48π2
(3)2 β (4π)2
=63 β 16i
9 + 16
=63
25β16
25π
|π§1π§2| = β(
63
25)2
+ (β16
25)2
|π§1π§2| = β
(63)2 + (16)2
252
=65
25=13
5
Argument of a complex number Z (arg Z)
The argument of a complex number z is defined to be
the angle (ΞΈ) which the complex number z makes
with the positive x-axis.
From the diagram above,
tan π =π¦
π₯βΉ π = π‘ππβ1 (
π¦
π₯)
Note: For a given complex number, there will be
infinitely many possible values of the
argument, any two of which will differ by a
whole multiple of 360Β°.
To avoid confusion we usually work with the value
of π for which βπ < π < π or β180 < π < 180.
This is called the principle argument of z denoted by
arg z.
In practice the formula tan π =π¦
π₯
π = tanβ1 (π¦
π₯)
Which is often used to find the principal argument of
a complex number z, despite the fact that it tends to
two possible values for π in the permitted range. The
formula is necessary but not sufficient to help us
π₯
π¦
π
x
y
137
obtain the arg z. The correct value of arg z is chosen
with the aid of a sketch.
Example
Find the principal argument of the following
complex number
(π) 1 + π (π) β 1 β πβ3 (π) β 5
(π) β β3 + π (π)β3 β π
Solution
Consider z1 = 1 + i
π = π‘ππβ1 (1
1) = 45Β°
Since 180Β° = π radians
π =45π
180=π
4
βΉ arg π§1 = 45Β°
1
45arg
180z
=
π
4
(π) Let π§2 = β1β πβ3
tan π =β3
1
π = π‘ππβ1 (β3
1)
π = 60Β°
arg π§2 = π1
βΉ arg π§2 = β120Β°
OR arg π§2 = β2
3π
π§3 = β5+ 0π
arg π§3 = 180Β° or arg π§3 = π
(d) Let π§4 = ββ3 + π
π = π‘ππβ1 (1
β3) = 30Β°
π§4 = ββ3 + π
arg π§4 = 150Β°, from the sketch above
(e) β3 β π
π§5 = β3 β π
tan π =1
β3
π = π‘ππβ1 (1
β3) = 30Β°
arg π§5 = β30Β° from the above diagram
Properties of Arguments
Given the two complex numbers π§1 and π§2 then
ππ«π ( ππππ) = ππ«π ππ + ππ«π ππ
ππ«π (ππππ) = ππ«π ππ β ππ«π ππ
1
π
1 π πππ ππ₯ππ
πΌπ ππ₯ππ
π2
π
πΌπ ππ₯ππ
π πππ ππ₯ππ π1
ββ3
z2
-1
πΌπ ππ₯ππ
π πππ ππ₯ππ
β5
150Β°
πΌπ ππ₯ππ
π
1
π πππ ππ₯ππ ββ3
πΌπ ππ₯ππ
π π π ππ₯ππ
β1
β3
138
Example I
Given that π§1 = β1 β πβ3 and π§2 = 1 + π. Find the
arg(π§1π§2) πππ arg (π§1
π§2)
Solution
π§1 = β1β πβ3
1 3tan
1
= 60Β°
arg π§1 = π1 = β120Β°
Z2 = 1 + i
arg π§2 = π‘ππβ1 (
1
1) = 45Β°
arg(π§1π§2) = arg π§1 + arg π§2
= β120 + 45Β°
= β75Β°
arg (π§1π§2) = arg π§1 β arg π§2
= β120 β 45
= β165
Modulusβargument form of a complex
number
(Polar form of a complex number)
Consider a complex number π§ = π₯ + ππ¦ making an
angle π with the positive π₯ β axis
arg π§ = π
From the diagram above cosπ =π₯
π sinπ =
π¦
π
π₯ = π cos π π sinπ = π¦
π§ = π₯ + ππ¦
π§ = π cos π + ππ sin π
π§ = π (cos π + π sin π)
(modulus argument form a complex number)
Where π = |π§| = βπ₯2 + π¦2
Example
Express the following complex numbers in modulus
βargument
a) 5 + 5πβ3
b) β2 + π
c) ββ3
2+1
2π
d) β3β2 + 3β2π
e) β5π
f) β5 β 12π
Solutions
π§1 = 5 + 5πβ3
π = β(5)2 + (5β3)2
= β25 + 75
= 10
arg π§1 = π‘ππβ1 (
5β3
5) = 60Β°
π§1 = 5 + 5πβ3 = 10(cos 60 + π sin60)
(b) π§2 = β2 + π
|π§2| = π = β(β2)2+ (1)2
= β3
π
πΌπ ππ₯ππ
π πππ ππ₯ππ β1
π1
ββ3
1
1
π
π¦
π
π₯ π πππ ππ₯ππ
πΌπ ππ₯ππ
π
(x, π¦)
139
arg π§2 = π‘ππβ1 (
1
β2) = 35.3Β°
π§2 = π(cosπ + π sinπ
β3[cos 35.3 + π sin 35.3]
(c) ββ3
2β1
2π
π§3 = ββ3
2β1
2π
|π§3| = β((ββ3
2)
2
+ (β1
2)2
)
= β3
4+1
4 = 1
π = π‘ππβ1(12β
β32β) = 30Β°
3
3 1
2 2z
arg π§3 = β150Β°
π§3 = π(cosπ + π sinπ)
π§3 = 1(cosβ150 + π sinβ150)
(d) π§4 = β3β2 + (3β2)i
|π§4| = β(β3β2)2+ (3β2)
2
= β36
= 6
π = π‘ππβ1 (3β2
3β2)
π = 45Β°
arg(π§4) = +135Β°
π§4 = 6(cos135 + π sin135)
(e) π§5 = β5π = 0 + β5π
|π§5| = β02 + (β5)2
|π§5| = 5
arg π§5 = β90
π§5 = 5(cosβ90 + π sinβ90)
π§6 = 3 + 4π
π = |π§6| = β(3)2 + (4)2
= β25
= 5
π = π‘ππβ1 (4
5) = 53.1
z6 = 5(cos 53.1Β° + i sin 53.1Β°)
(g) π§6 = β5β 12π
|π§6| = β(β5)2 + (β12)2
= β169
= 13
π = π‘ππβ1 (12
5)
π = 67.4
z7 = -5 β 12i
arg π§7 = β112.6Β°
|π§7| = β(β5)2 + (β12)2
= β25 + 144
= β169
= 13
πΌπ ππ₯ππ
π
3β2
π πππ ππ₯ππ
πΌπ ππ₯ππ
π πππ ππ₯ππ
β5
π
π
πΌπ ππ₯ππ
π πππ ππ₯ππ β5 π1
β12
π
πΌπ ππ₯ππ
π πππ ππ₯ππ ββ3
2 π1
-0.5
140
13(cosβ112 . 6 + π sinβ122.6)
Example II
π§1 = π1(cosπ1 + π sin π1)
π§2 = π2(cos π2 + π sinπ2)
Show that
π§1π§2 = π1π2(cos(π1 + π2) + π sin(π1 + π2))
And π§1
π§2=π1
π2(cos(π1 β π2) + π sin(π1 β π2))
Solution
π§1π§2 = π1(cos π1 + π sinπ1)π2(cosπ2 + π sin π2)
= π1π2[(cos π1 cos π2 + cos π1 (π sinπ2)
+ π sinπ1 cos π2 + π2 sin π1 sin π2)]
= π1π2[(cos π1 cos π2 β sinπ1 sin π2+ π(sinπ1 cos π2 + cosπ1 sin π2)]
= π1π2[(cos(π1 + π2) + π sin(π1 + π2))]
1 1 1 1
2 2 2 2
1 1 1 2 2
1 2 2 2 2
(cos cos )
(cos sin )
(cos sin )(cos sin )
(cos sin )(cos sin )
z r i
z r i
r i i
r i i
1 1 2 1 2 1 2 1 2
2 2
2 2 2
cos cos cos sin sin cos sin sin
cos sin
r i i
r
1 2 1 2 1 2 1 21
2 2
2 2 2
cos cos sin sin sin cos cos sinir
r cos sin
1 1 1 2 1 2
2 2
1 11 2 1 2
2 2
(cos( sin(
1
cos( ) sin( )
) )z r i
z r
z ri
z r
(as required)
Example III
Given that π§1 = 1 + π
π§2 = β3 β π
Find in polar form π§1π§2 ππππ§1
π§2
Solution
π§1 = 1 + π
|π§1| = β12 + 12 = β2
arg π§1 = π‘ππβ1 (
1
1) =
π
4
π§1 = π1(cos π1 + π sinπ1)
π§1 = β2(cosπ
4+ π sin
π
4)
π§2 = β3 β π
|π§1| = π2 = β(β3)2+ (β1)2
= β4
= 2
arg π§2 = β30Β°
= βπ
6 πππππππ
arg π§2 = βπ
6
π§2 = 2(cosβπ
6+ π sinβ
π
6)
π§1π§2 = π1π2(cos(π1 + π2) + π sin(π1 + π2)
= 2β2(cos (π
4+ β
π
6) + π sin (
π
4+βπ
6)
= 2β2(cosπ
12+ π sin
π
12)
π§1π§2=π1π2(cos(π1 β π2) + π sin(π1 β π2))
=β2
2[cos (
π
4ββπ
6) + π sin (
π
4ββπ
6)]
β2
2[cos (
5π
12) + π sin (
5π
12)]
Demoivreβs Theorem Demoivres theorem states that for real values of n
(cos π + π sin π)π = (cosππ + π sinππ)
Proving Demoivreβs theorem by mathematical
induction
(cos π + π sin π)π = (cosππ + π sinππ)
For n =1, (cos π + π sinπ)1 = (cos π + π sinπ)
Itβs true for n =1
Assume the results holds for the general value of n=k
(cos π + π sin π)π = (cosππ + π sinππ)
It must be true for the next integer π = π + 1
1
cos sin cos sin cos sink k
i i i
= (cos ππ + π sinππ)(cos π + π sin π)
= cosππ cosπ + π cos ππ sinπ + π sinππ cos π
+ π2 sinπ sinππ
= [(cos ππ cos π β sinππ sin π) +
sin cos cos sin ]i k k
= cos(ππ + π) + π sin(ππ + π)
= cos(π + 1)π + π sin(π + 1)π
141
βΉ (cosπ + π sinπ)π+1
= cos(π + 1)π + π sin(π + 1)π
For the next integer , π = π + 1 = 2
βΉ π = 1
(cos π + π sin π)2 = (cos2π + π sin2π)
Since itβs true for n=1, n = 2 and so on itβs true for all
positive integral values of n.
Example I
Find the value of (cos1
4π + π sin
1
4π)12
Solution
(cos sin ) cos sinni n i n
(cos1
4π + π sin
1
4π)12
=
(cos1
4π Γ 12 + π sin
1
4π Γ 12)
= cos 3π + π sin 3π
= β1
Example II
Express (1 β πβ3)4 in the form π + ππ
Solution
(1 β πβ3)4
Let π§ = 1 β πβ3
|π§| = β(1)2 + (ββ3)2= 2
arg π§ = β60
= βπ
3
arg π§ = βπ
3
z = r(cos ΞΈ + isinΞΈ)
π§ = 2 (cosβπ
3+ π sinβ
π
3) )
π§4 = 24 (cosβπ
3+ π sinβ
π
3)4
4 43 3
16 cos sini
= 16(β1
2+β3
2π)
= β8 + 8β3π
Example III
Evaluate 1
(1βπβ3)3
Solution:
1
1 β πβ3= (1 β πβ3)
β3
Let π§ = (1 β πβ3)
|π§| = β12 + (ββ3)2
= 2
arg π§ = βπ
3
(1 β πβ3) = 2 (cosβπ
3+ π sinβ
π
3)
(1 β πβ3)β3= 2β3 (cosβ
π
3+ π sinβ
π
3)β3
=1
8(cosβ
π
3Γ β3 + π sinβ
π
3Γ β3)
=1
8(cos+π + π sin+π)
= β1
8
Example IV
Express β3 + π in modulus βargument form. Hence
find
(β3 + π)10 and1
(β3 + π)7 in the form of π + ππ
Solution
Let π§ = β3 + π
|π§| = β(β3)2+ 1 = 2
arg π§ =π
6
π§ = 2 (cosπ
6+ π sin
π
6 )
(β3 + π)10= 210 (cos (
10π
6) + π sin (
10π
6))
= 210 (1
2ββ3
2π)
=1024
2βπ(1024)β3
2
= 512 β 512β3π
1
(β3 + π)7 = (β3 + π)
β7
= (2(cosπ
6+ π sin
π
6))
β7
= 2β7 (cosβ7 Γπ
6+ π sinβ7 Γ
π
6)
142
7 76 6
1cos sin
128i
=1
128(ββ3
2+ 1
2π)
= ββ3
256+
1
256π
= ββ3
256+1
256π
Example V
Express (-1+i) in modulus β argument form. Hence
show that (β1 + π)16 is real and that
1
(β1 + π)6 is pure imaginary.
Solution
π§ = β1 + π
|π§| = β(β1)2 + 12
= β2
arg π§ = 135Β°
π§ = β2(cos 135 + π sin135)
π§16 = (β2)16(cos135 Γ 16 + π sin135 Γ 16)
= 256(cos 2160 + π sin 2160)
= 256(1)
= 256
βΉ (β1+ π)16 = 256 So it is purely real
As required
6
6
1( 1 )
( 1 )i
i
π§β6 = (β2)β6(cos 135 Γ β6 + π sin 135 Γ β6)
=1
8(0 + π) =
1
8π
βΉ π§β6 is purely imaginary.
Example VI
a) (cos π + π sinπ)2(cosπ + π sinπ)3
b) 1
(cosπ+π sinπ)2
c) cosπ+π sinπ
(cosπ+π sinπ)4
d) (cos
π
17+π sin
π
17)8
(cosπ
17βπ sin
π
17)9
e) (cosπ+π sinπ)(cos2π+π sin2π)
(cosπ
2+π sin
π
2)
f) (cos
2π
5+π sin
2π
5)8
(cos3π
5βπ sin
3π
5)3
Solutions
(a) (cos π + π sin π)2(cos π + π sinπ)3
= 2 3(cos sin )i
= (cos π + π sinπ)5
= (cos 5π + π sin 5π)
(b) 1
(cosπ+π sinπ)2
= (cos π + π sinπ)β2
= cosβ2π + π sinβ2π
= cos 2π β π sin 2π
(c) cosπ+π sinπ
(cosπ+π sinπ)4
= (cos π + π sinπ)1(cos π + π sin π)β4
= (cos π + π sinπ)1+β4
= (cos π + π sinπ)β3
= cosβ3π + π sinβ3π
= (cos 3π β π sin3π)
(d) 8
17 17
9
17 17
(cos sin )
(cos sin )
i
i
817
917
(cos sin )
(cos sin )
i
i
8 917 17
1
(cos sin )
(cos sin )
i
i
(e) 2 2
(cos sin )(cos2 sin 2 )
(cos sin )
i i
i
12
2(cos sin )(cos sin )
(cos sin )
i i
i
12
3(cos sin )
(cos sin )
i
i
52(cos sin )i
5 52 2
cos sini
(f) (cos
2π
5+π sin
2π
5)8
(cos3π
5βπ sin
3π
5)3
((cosπ5+ π sin
π5)2)8
((cosπ5+ π sin
π5)β3)3
143
(cosπ5+ π sin
π5)16
(cosπ5+ π sin
π5)β9
(cosπ
5+ π sin
π
5)16ββ9
cos (25 Γπ
5) + π sin (25 Γ
π
5)
= (cos5π + π sin5π)
Example VII
Use De-moivreβs theorem to show that
tan 3 π =3 tan ΞΈ β 3tan3ΞΈ
1 β 3 tan2ΞΈ
Solution
(cos 3π + π sin3π) = (cos π + π sin π)3
ππ’π‘ (cosπ + π sinπ)3 =
= πππ 3π + 3(π sinπ)πππ 2π + 3(π sinπ)2 cos π
+ (π sinπ)3
= (cos3 π β 3 sin2 π cosπ) +
π(3 sinπ cos2 π β sin3 π)
= cos 3π + π sin3π
Equating real to real and imaginary to imaginary;
β sin3π = 3 sin π πππ 2π β π ππ3πβ¦ . . (1)
cos 3π = cos3 π β 3π ππ2π cos πβ¦β¦β¦β¦ . . β¦ (2)
Eqn (1) Γ· Eqn (2)
β tan3π =3 sin π πππ 2π β π ππ3π
πππ 3π β 3π ππ2π cosπ
tan 3π =
3 sinπ πππ 2πcos3 π
βsin3 ππππ 3π
cos3 πcos3 π
β3 sin2 π cos πcos3 π
tan 3π =3 tanπ β tan2 π
1 β 3 tan2 π
Example VIII
Use Demovreβs theorem to show that
tan 4π =4 tanπ β 4 tan3 π
1 β 6 tan2 π + tan4 π
Solution
(cos π + π sin π)4 = (cos π + π sin π)4
= cos4 π + 4 cos3 π (π sin π) + 6 cos2 π (π sinπ)2 +
4cos π (π sin π)3 + (π sinπ)4
= cos4 π + (4 cos3 π sinπ)π β 6 πππ 2π π ππ2π
β (4 cos π sin3 π) π + sin4 π
= (cos 4 π + π sin 4π)
Equating real to real and imaginary to imaginary
cos 4π = cos4 π β 6 πππ 2ππ ππ2π + sin4 πβ¦ . (π)
sin4π = 4 cos3 π sin π β 4 cos π sin3 π β¦ . . β¦ (ππ)
Eqn (ii) Γ· Eqn (i)
tan 4π =4 cos3 π sin π β 4 cos π sin3 π
cos4 π β 6 πππ 2ππ ππ2π + sin4 π
tan 4π =
4 cos3 π sin πcos4 π
β4 cos π sin3 πcos4 π
cos4 πcos4 π
β6πππ 2π π ππ2π
cos4 π+sin4 πcos4 π
tan 4π =4 tan π β 4 tan3 π
1 β 6π‘ππ2π + tan4 π
Example IX
Show that
π§π +1
π§π= 2cos ππ
π§π β1
π§π= 2π sinππ
Hence show that 4 1
cos (cos 4 4cos 2 3)8
Solution
π§ = cos π + π sin π
π§π = (cosπ + π sinπ)π
= (cosππ + π sin ππ)
π§βπ = (cosπ + π sinπ)βπ
= cosβππ + π sinβππ
= cosππ β π sinππ
π§π +1
π§π= cosππ + π sinππ + cos ππ β π sin ππ
= 2cos ππ
π§π β1
π§π
= (cosππ + π sin ππ) β (cos ππ β π sinππ)
= 2π sin ππ
from π§π +1
π§π= 2cos ππ
π§ +1
π§= 2 cos π
π§π β1
π§π= 2π sinππ
π§ β1
π§= 2π sin π
144
π§ +1
π§= 2 cos π
(π§ +1
π§)4
= (2 cos π)4
But 41( )
zz
π§4 + 4π§3 (1
π§) + 6π§2 (
1
π§)2
+ 4π§ (1
π§)3
+ (1
π§)4
(π§4 +1
π§4) + 4 (π§2 +
1
π§2) + 6 = (π§ +
1
π§)4
2 cos 4 π + 4(2 cos 2π) + 6 = (2cosπ)4
16cos4ΞΈ = 2cos4ΞΈ + 4(2cos2ΞΈ) + 6
cos4 π =1
16(2 cos 4π + 8 cos 2π + 6)
cos4 π =1
8(cos 4π + 4 cos 2π + 3)
Example XI
Given that π§ = cosπ + π sinπ show that
π§π β1
π§π= 2π sin ππ
Hence or otherwise show that
sin5 π =1
16(sin5π β 5 sin 3π + 10 sinπ)
Solution
π§π β1
π§π= 2π sinπ
π§ β1
π§= 2π sin π
(π§ β1
π§)5
= (2π sin π)5
(π§ β1
π§)5
= π5(32) sin5 π
(π§ β1
π§)5
= (32π Γ π4sin5 π)
(π§ β1
π§)5
= 32π sin5 π
but (z β1
z)5
= π§5 + 5π§4 (β1
π§) + 10π§3 (β
1
π§)2
+10π§2 (β1
π§)3
+ 5π§ (β1
π§)4
+ (β1
π§)5
= π§5 β1
π§5β 5 (π§3 β
1
π§3) + 10 (π§ β
1
π§)
5
5
12 sin
12 sin5
12 sin
n
nz i n
z
z iz
z iz
(π§ β1
π§)5
= (2πsinπ)5
2π sin5π β 5(2π sin 3π) + 10(2π sin π) = 32π sin5 π
sin5 π =1
32(2 sin 5π β 10 sin3π + 20 sinπ)
sin5 π =1
16(sin5π β 5 sin 3π + 10 sinπ)
Example XII
Prove that cos6 π + sin6 π =1
8(3 cos 4π + 5)
Solution
π§π +1
π§π= 2 cosππ
π§ +1
π§= 2 cos π
(π§ +1
π§)6
= (2 cos π)6
(π§ +1
π§)6
= 64cos6 π
But (π§ +1
π§)6= π§6 + 6π§5 (
1
π§) + 15π§4 (
1
π§)2+
20π§3 (1
π§)3+ 15π§2 (
1
π§)4+ 6π§ (
1
π§)5+ (
1
π§)6
= (π§6 +1
π§6) + (6π§4 +
6
π§4) + (15π§2 +
15
π§2) + 20
= 2cos 6π + 6(2 cos 4π) + 15(2 cos 2π) + 20
β 64 cos6 π = 2 cos 6π + 12 cos 4 π + 30 cos 2π + 20
64 cos6 π = 2 cos6π + 12 cos 4π + 30 cos 2π
+ 20β¦β¦β¦β¦β¦β¦β¦β¦β¦ . (1)
(π§ β1
π§) = 2π sinπ
(π§ β1
π§)6
= 64π6 sin6 π
(π§ β1
π§)6
= β64 sin6 π
But
(π§ β1
π§)6= π§6 + 6π§5 (β
1
π§) + 15π§4 (β
1
π§)2+
20π§3 (β1
π§)3+ 15π§2 (β
1
π§)4+ 6π§ (β
1
π§)5+ (β
1
π§)6
= (π§6 +1
π§6) β 6 (π§4 +
1
π§4) + 15 (π§2 +
1
π§2) β 20
= 2cos 6π β 6(2 cos 4π) + 15(2 cos 2π) β 20
β 2cos 6π β 12 cos 4π + 30 cos 2π β 20
= β64 sin6 π
145
β64sin6 π = 2 cos6π β 12 cos 4π + 30 cos 2π
β 20β¦β¦β¦β¦β¦β¦β¦β¦β¦ . (2)
Eqn (2) β Eqn (1)
β 64 cos6 π β β64sin6 π = 24cos 4π + 40
cos6 π + sin6 π =8
64(3 cos 4π + 5)
cos6 π + sin6 π =1
8(3 cos 4π + 5)
Solving Complex Equations Given that x and y are real numbers. Find the values
of x and y which satisfy the equation.
2π¦ + 4π
2π₯ + π¦β
π¦
π₯ β π= 0
Solution
2π¦ + 4π
2π₯ + π¦β
π¦
π₯ β π= 0
2π¦ + 4π
2π₯ + π¦=
π¦
π₯ β π
2π¦+4π
2π₯+π¦=
π¦
π₯βπΓ
x i
x i
2π¦ + 4π
2π₯ + π¦=π₯π¦ + ππ¦
π₯2 + 1
2π¦
2π₯ + π¦+
4π
2π₯ + π¦=
π₯π¦
π₯2 + 1+
π¦π
π₯2 + 1
Equating real to real and imaginary to imaginary
β2π¦
2π₯ + π¦=
π₯π¦
π₯2 + 1β¦β¦β¦β¦β¦β¦(1)
4
2π₯ + π¦=
π¦
π₯2 + 1β¦β¦β¦β¦β¦β¦ . . (2)
From equation (1)
2π¦(π₯2 + 1) = π₯π¦(2π₯ + π¦)
2π₯2π¦ + 2π¦ = 2π₯2π¦ + π₯π¦2
2π¦ β π₯π¦2 = 0
π¦(2 β π₯π¦) = 0
π¦ = 0 ππ π₯π¦ = 2
From Eqn (2), 2
4
2 1
y
x y x
4x2 + 4 = 2xy + y2
For y = 0, 4x2 + 4 = 0
x2 + 1 = 0
x2 = -1
x2 = i2
x = Β±i
For xy = 2, 2
yx
2
2
44 4 4x
x
2
2
44 0x
x
Let x2 = m
44 0m
m
4m2 β 4 = 0
m2 β 1 = 0
(m + 1)(m β 1) = 0
m = 1, m = -1
When m = 1, x2 = 1 x = Β±1
When m = -1, x2 = i2 x = Β±i
xy = 2
If x = 1, y = 2
If x = -1, y = -2
If x = i, y = -2i
If x = -i , y = 2i
Example II
Find the values of x and y in
π₯
2 + 3πβ
π¦
3 β 2π=6 + 2π
1 + 8π
Solution
π₯
2 + 3πβ
π¦
3 β 2π=6 + 2π
1 + 8π
π₯(2 β 3π)
2 + 3π(2 β 3π)β
π¦(3 + 2π)
(3 β 2π)(3 + 2π)
=(6 + 2π)(1 β 8π)
(1 + 8π)(1 β 8π)
(2π₯ β 3π₯π)
13β(3π¦ + 2π¦π)
13=(6 β 48π + 2π + 16)
65
2π₯ β 3π¦
13β3π₯ + 2π¦
13π =
22
65β46
65π
β2π₯ β 3π¦
13=22
65
5(2π₯ β 3π¦) = 22
10π₯ β 15π¦ = 22β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(1)
Similarly, 3π₯+2π¦
13=46
65
5(3π₯ + 2π¦) = 46
15π₯ + 10π¦ = 46β¦β¦β¦β¦β¦β¦β¦β¦β¦ . . (2)
Solving eqn (1) and eqn (2) simultaneously
βΉ π₯ = 2.8 π¦ = 0.4
Example III
Find the values of x and y if π₯
1+π+
π¦
2βπ= 2 + 4π
146
Solution
π₯
1 + π+
π¦
2 β π= 2 + 4π
π₯(1 β π)
(1 + π)(1 β π)+
π¦(2 + π)
(2 β π)(2 + π)= 2 + 4π
π₯ β π₯π
2+2π¦ + π¦π
5= 2 + 4π
5(π₯ β π₯π) + 2(2π¦ + π¦π) = 2 + 4π
5π₯ β 5π₯π + 4π¦ + 2π¦π = 20 + 40π
Equating real to real and imaginary to imaginary;
5π₯ + 4π¦ = 20 β¦β¦β¦β¦β¦ (1)
2π¦ β 5π₯ = 40 β¦β¦β¦β¦β¦ (2)
Solving Eqn (1) and Eqn (2) simultaneously;
π¦ = 10
π₯ = β4
Example IV
Find the values of x and y. given that
π₯π
1 + ππ¦=3π₯ + 4π
π₯ + 3π¦
Solution
π₯π(1 β ππ¦)
(1 + ππ¦)(1 β ππ¦)=(3π₯ + 4π)
π₯ + 3π¦
π₯π + π₯π¦
1 + π¦2=3π₯ + 4π
π₯ + 3π¦
π₯π¦
1 + π¦2+
π₯π
1 + π¦2=
3π₯
π₯ + 3π¦+
4π
π₯ + 3π¦
βπ₯π¦
1 + π¦2=
3π₯
π₯ + 3π¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ . (1)
π₯
1 + π¦2=
4
π₯ + 3π¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(2)
From equation (1)
π₯2π¦ + 3π₯π¦2 = 3π₯ + 3π₯π¦2
β π₯2π¦ = 3π₯
β π₯2π¦ β 3π₯ = 0
π₯(π₯π¦ β 3) = 0
π₯ = 0 ππ π₯π¦ = 3
From eqn (2)
π₯2 + 3π₯π¦ = 4 + 4π¦2 β¦β¦β¦(3)
When x = 0, 0 = 4 + 4π¦2
β1 = π¦2
π¦ = Β±π
When xy = 3
π¦ =3
π₯
Substituting y = 3
x and xy = 3 in Eqn (3);
2
2
2
2
33(3) 4 4
369 4
xx
xx
π₯2 + 9 = 4 +36
π₯2
π₯2 β36
π₯2+ 5 = 0
πππ‘ π₯2 = π
π β36
π+ 5 = 0
π2 β 36 + 5π = 0
π2 + 5π β 36 = 0
(π + 9)(π β 4) = 0
(π₯2 + 9)(π₯2 β 4) = 0
π₯2 β 4 = 0 β π₯ = Β±2
π₯ = 2, π₯ = β2
When π₯ = 2, π¦ =3
2
When π₯ = β2, π¦ = β3
2
π₯2 + 9 = 0
π₯2 = β9
π₯ = Β±3π
π€βππ π₯ = 3π
π¦ =3
3π=1
π
π¦ = βπ
π€βππ π₯ = β3π
π¦ = +π
Example V
If z is a complex number such that π§ =π
2βπ+
π
1+3π.
Where p and q are real. If |z| = 7, arg P = π
2 . Find the
value of p and q.
Solution
π§ =π
2 β π+
π
1 + 3π
z =P(2 + i)
(2 β i)(2 + i)+
π(1 β 3i)
(1 + 3i)(1 β 3i)
z =2p + Pi
5+q β 3qi
10
147
2(2 ) 3
10
4 2 3
10
4 (2 3 )
10
p pi q qiz
p pi q qiz
p q p q iz
arg π§ = tanβ1
(
2π β 3π10
4π + π
10
)
=π
2
π‘ππβ1(2π β 3π
4π + π) =
π
2
2p β 3q
4P + q= β
4P + q = 0
q = β4P
|z| = 7
β(4P+q
10)2+ (
2πβ3π
10)2= 7 β¦β¦β¦β¦β¦ (1)
Substituting q = -4p in Eqn (1)
2
2 140 7
10
p
14π
10= 7
π = 5
q = β4 Γ 5
π = β20
Example VI
Given that (1 + 5i)p β 2q = 3 + 7i, find π and π
(a) When p and q are real
(b) When p and q are conjugate complex numbers
Solution
(a) (1 + 5π)π β 2π = 3 + 7π
π + 5ππ β 2π = 3 + 7π
π β 2π + 5ππ = 3 + 7π
π β 2π = 3 β¦β¦β¦β¦β¦β¦β¦β¦. (1)
5P = 7 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (2)
From Eqn (2),
P =7
5
7
5β 2π = 3
7
5β 3 = 2π
β8
5= 2π
β8
10= π
π = β4
5
β π =7
5, π = β
4
5
(b) Let π = π₯ + ππ¦
π = π₯ β ππ¦
(1 + 5π)(π₯ + ππ¦) β 2(π₯ β ππ¦) = 3 + 7π
π₯ + ππ¦ + 5π₯π β 5π¦ β 2π₯ + 2π¦π = 3 + 7π
(π₯ β 5π¦ β 2π₯) + (π¦ + 5π₯ + 2π¦)π = 3 + 7π
(βπ₯ β 5π¦) + (3π¦ + 5π₯)π = 3 + 7π
βπ₯ β 5π¦ = 3
π₯ = β3 β 5π¦ ........................ (1)
3π¦ + 5π₯ = 7 .......................... (2)
Substituting Eqn (1) in Eqn (2)
3π¦ + 5(β3 β 5π¦) = 7
3y β 15 β 25y = 7
β22y = 22
π¦ = β1
π₯ = β3 β 5(β1)
π₯ = β3 + 5
π₯ = 2
π = π₯ + ππ¦
π = 2 β π
π = 2 + π
Square root of Complex Numbers
Example I
Find the square root of 35 β 12π
Solution
Let β35 β 12π = π₯ + ππ¦
(β35 β 12π)2= (π₯ + ππ¦)2
35 β 12π = π₯2 + 2π₯π¦π + π2π¦2
35 β 12π = π₯2 β π¦2 + 2π₯π¦π
β π₯2 β π¦2 = 35
2π₯π¦ = β12
π₯π¦ = β6
π¦ = β6
π₯
π₯2 β36
π₯2= 35
π₯4 β 36 = 35π₯2
π₯4 β 35π₯2 β 36 = 0
Let π₯2 = π
m2 β 35π β 36 = 0
148
(π β 36)(π + 1) = 0
(π₯2 β 36)(π₯2 + 1) = 0
But π₯ is real
β π₯2 β 36 = 0
π₯ = Β±6
When π₯ = 6 π¦ = β6
6
π¦ = β1
π€βππ π₯ = β6, π¦ = 1
β β35 β 12i = 6 β i
or β35 β 12π = β6 + π
Example VIII
Find the square root of 5 β 12π
solution
Let β5 β 12π = π₯ + ππ¦
5 β 12π = (π₯ + ππ¦)2
5 β 12π = π₯2 + 2π₯π¦π + π¦π2
5 β 12π = π₯2 β π¦2 + 2π₯π¦π
Equating real to real and imaginary to imaginary;
β π₯2 β π¦2 = 5 β¦β¦β¦β¦β¦β¦β¦β¦.(1)
2π₯π¦ = β12
π₯π¦ = β6
π¦ = β6
π₯ β¦β¦β¦β¦β¦β¦β¦β¦.(2)
Substituting Eqn (2) in Eqn (1)
π₯2 β36
π₯2= 5
(π₯2)2 β 36 = 5π₯2
πππ‘ π = π₯2
π2 β 36 = 5π
π2 β 5π β 36 = 0
(π β 9)(π + 4) = 0
(π₯2 β 9)(π₯2 + 4) = 0
π₯2 = 9
π₯ = Β±3
π€βππ π₯ = 3, π¦ = β2
π€βππ π₯ = β3, π¦ = 2
β5 β 12π = 3 β 2π
ππ β5 β 12π = 3 + 2π
Example IX
Find the roots of π§2 β (1 β π)π§ + 7π β 4 = 0
Solution
2 4
2
b b acz
a
π§ =(1 β π) Β± β(1 β π)2 β 4(1)(7π β 4)
2 Γ 1
π§ =1 β π Β± β1 β 2π β 1 β 28π + 16
2
1 16 30
2
i i
But 16 30i a bi
16 β 30i = a2 + 2abi β b2
a2 β b2 = 16
2ab = -30
ab = -15
15a
b
2
21516b
b
2
2
22516b
b
Let m = b2
22516m
m
m2 + 16m β 225 = 0
m = 9, m = -25
b2 = 9
b = Β±3
ab = -15
a = 5
When b = -3, a = 5
When b = 3, a = -5
a + bi = 5 β 3i, -5 + 3i
16 30 (5 3 )i i
π§ =1 β π Β± (5 β 3π)
2
π§ = 3 β 2π
π§ = β2 + π
Example X
Show that 1 + 2π is a root of the equation
2π§3 β π§2 + 4π§ + 15 = 0
Solution
π§ = 1 + 2π
π§2 = (1 + 2π)2
= 1 + 4π + 4π2
= β3+ 4π
π§3 = π§ Γ π§2 = (1 + 2π)(β3 + 4π)
= β3+ 4π β 6π β 8
= β11 β 2π
149
2z3 β z2 + 4z + 15 2(β11 β 2π) β (β3 + 4π) + (4(1 + 2π) + 15
= β22 β 4π + 3 β 4π + 4 + 8π + 15
= β22 + 22 β 8π + 8π
= 0 + 0π
= 0
β 1+ 2π is a root of the equation.
Since π§ = 1 + 2π is a root of the equation
2π§3 β π§2 + 4π§ + 15 = 0
The complex conjugate π§Μ = 1 β 2i must also be a
root of the above equation
β 1β 2π = π§ is also a root of the equation
2π§3 β π§2 + 4π§ + 15 = 0
2π§3 β π§2 + 4π§ + 15 = 0
π§ = 1 + 2π
π§ = 1 β 2π
Sum of roots = 1 + 2π + 1 β 2π
= 2
Product of roots = (1)2 β (2π)2
= 1 + 4
= 5
π§2 β (π π’π ππ ππππ‘π
) π§ + πππππ’ππ‘ = 0
π§2 β 2π§ + 5 = 0
π§2 β 2π§ + 5 is a factor of 2π§3 β π§2 + 4π§ + 15
(2π§ + 3)(π§2 β 2π§ + 15) = 0
π§ = β3
2 π§ = 1 + 2π πππ π§ = 1 β 2π
Example XI
Given that 2 + 3π is a root of the equation
π§3 β 6π§2 + 21π§ β 26 = 0. Find the other roots
Solution
π§ = 2 + 3π is a root β π§ = 2 β 3π is also a root of
the equation π§3 β 6π§2 + 21π§ β 26 = 0
Sum of roots = 2 + 3π + 2 β 3π
= 4
Product of roots = (2 + 3i)(2 β 3i)
= 22 β (3π)2
= 4 + 9
= 13
β π§2 β 4π§ + 13 is a factor of
π§3 β 6π§2 + 21π§ β 26 = 0
β (π§ β 2)(π§2 β 4π§ + 13) = 0
β π§ = 2, π§ = 2 + 3π πππ π§ = 2 β 3π are roots of
equation of π§3 β 6π§2 + 21π§ β 26 = 0
Example XII
Show that 1 + π is a root of the equation
π§4 + 3π§2 β 6π§ + 10 = 0. Hence find other roots
Solution
π§ = 1 + π
z2 = 1 + 2i + i2
π§2 = 1 + 2π β 1
π§2 = 2π
π§3 = π§2. π§
= 2π(1 + π)
= 2π β 2
π§4 = (π§2)2 = (2π)2 = 4π2
= β4
β π§4 + 3π§2 β 6π§ + 10
= (β4) + 3(2π) β 6(1 + π) + 10
= β4 + 6π β 6 β 6π + 10
= β10 + 10 + 6π β 6π
= 0 + 0π
= 0
π§ = 1 + π is a root of the equation
β 1β π is also a root of the equation
Sum of the roots= 1 + π + 1 β π
= 2
Product of roots = (1)2 β π2 = 2
π§2 β (π π’π ππ ππππ‘π )π§ + πππππ’ππ‘ = 0
π§2 β 2π§ + 2 = 0
β π§2 β 2π§ + 2 is a factor of π§4 + 3π§2 β 6π§ + 10
2π§3 β π§2 + 4π§ + 15 π§2 β 2π§ + 5
2π§3 β 4π§2 + 10π§
2π§ + 3
0
3π§2 β 6π§ + 15 3π§2 β 6π§ + 15
π§3 β 6π§2 + 21π§ β 26 π§2 β 4π§ + 13
π§3 β 4π§2 + 13π§
π§ β 2
0
β2π§2 + 8π§ β 26 β2π§2 + 8π§ β 26
150
(π§2 β 2π§ + 2)(π§2 + 2π§ + 5) = 0
β π§2 + 2π§ + 5 = 0
z2 β 2z + 2 = 0
For z2 + 2z + 5 = 0, 22 (2) 4 1 5
4 1z
π§ =β2 Β± β16π2
2
π§ =β1 Β± 4π
2
π§ = β1 + 2π
π§ = β1 β 2π
-1 + 2i, -1 β 2i, 1 + i, 1 β π are roots of the
equation π§4 + 3π§2 β 6π§ + 10 = 0
Example XIII
Show that 1 β π is a root of the equation
4π§4 β 8π§3 + 9π§2 β 2π§ + 2 = 0. Find the other
roots.
Solution
π§ = 1 β π
π§2 = (1 β π)2
= 1 β 2π + π2
= β2π
π§3 = π§2. π§
= β2π(1 β π)
= β2π + 2π2
= β2 β 2π
π§4 = (π§2)2 = (β2π)2
= β4
4π§4 β 8π§3 + 9π§2 β 2π§ + 2 =
4(β4) β 8(β2 β 2π) + 9(β2π) β 2(1 β π) + 2
= β16 + 16 + 16π β 18π β 2 + 2π + 2
= 0 + 0π = 0
Since π§ = 1 β π is a of the equation and it implies
that 1 + i is also a root.
Sum of roots= 1 β π + 1 + π
= 2
Product of the roots = (1 + i)(1 β i)
12 β π2 = 2
β π§2 β (2π§) + 2 = 0
β π§2 β 2π§ + 2 is a factor of
π§4 β 8π§3 + 9π§2 β 2π§ + 2 = 0.
(π§2 β 2π§ + 1)(4π§2 + 1) = 0
4π§2 = β1 π§2 = β1
4
β π§2 =1
4π2 , 2 = Β±
1
2π
Example XIV
Given that z = 2 β i is a root of the equation
π§3 β 3π§2 + π§ + π = 0, k is real. Find other roots.
Solution
π§ = 2 β π
π§2 = (2 β π)2
= 4 β 4π + π2
= 3 β 4π
π§3 = (2 β π)(3 β 4π)
= 6 β 8π β 3π + 4π2
= 2 β 11π
β (2 β 11π) β 3(3 β 4π) + 2 β π + π = 0
2 β 11π β 9 + 12π + 2 β π + π = 0
β11π + 11π + 4 β 9 + π = 0
0 β 5 + π = 0
π = 5
β π§3 β 3π§2 + π§ + 5 = 0
π§ = 2 β π
π§ = 2 + π
π§ = 2 β π
π§ = 2 + π
Sum of roots = 4
Product of roots = 5
π§2 β 4π§ + 5 = 0 is a factor of
π§3 β 3π§2 + π§ + 5 = 0
(π§ + 1)(π§2 β 4π§ + 5) = 0
π§4 + 3π§2 β 6π§ + 10 π§2 β 2π§ + 2
π§2 + 2π§ + 5
π§4 β 2π§3 + 2π§2
2π§3 + π§2 β 6π§ +10
0
2π§3 β 4π§2 + 4π§
5π§2 β 10π§ + 10
5π§2 β 10π§ + 10 4π§4 β 8π§3 + 9π§2 β 2π§ + 2 π§2 β 2π§ + 2
4π§4 β 8π§3 + 8π§2
4π§2 + 1
z2 β 2z + 2
z2 β 2z + 2
π§3 β 3π§2 + π§ + 5 π§2 β 4π§ + 5
π§3 β 4π§2 + 5π§
π§ + 1
π§2 β 4π§ + 5
151
(π§ + 1) = 0 π§ = β1
β π§ = β1, π§ = 2 + π, π§ = 2 β π are roots of the
equation z3 β 3z2 + z + k = 0 where k = 5
Example XIV
Solve for π§1and π§2 in the simultaneous equations
below
π§1 + (1 β π)π§2 = 0
3π§2 β 3π§1 = 2 β 5π
Solution
π§1 + (1 β π)π§2 = 0β¦β¦β¦β¦β¦β¦β¦β¦(1)
3π§2 β 3π§1 = 2 β 5π β¦β¦β¦β¦β¦β¦β¦β¦β¦ . (2)
From eqn (1)
π§1 = β(1 β π)π§2
π π’ππ π‘ππ‘π’π‘π ππ πππ (2)
3π§2 β 3[(β1(1 β π)π§2)] = 2 β 5π
3π§2 + 3(1 β π)π§2 = 2 β 5π
3π§2 β 3ππ§2 + 3π§2 = 2 β 5π
6π§2 β 3ππ§2 = 2 β 5π
π§2(6 β 3π) = 2 β 5π
π§2 =2 β 5π
6 β 3π
π§2 =(2 β 5π)(6 + 3π)
(6 β 3π)(6 + 3π)
π§2 =12 + 6π β 30π + 15
36 + 9
π§2 =27 β 24π
45
π§2 =9 β 8π
15
π§2 =9
15β8π
15
π§1 = β(1 β π)π§2
π§1 = β((1 β π) (9 β 8π
15))
π§1 = β(9 β 8π β 9π β 8
15)
π§1 = β(1 β 17π
15)
π§1 =β1+ 17π
15
π§1 = β1
15+17π
15
Example XV
Solve the equation z 3 β 1
Solution
π§3 β 1 = (π§)3 β (1)3
= (π§ β 1)(π§2 + π§ + 1)
Since π3 β π3 = (π β π)(π2 + ππ + π2)
π§3 β 1 = (π§ β 1)(π§2 + π§ + 1) = 0
π§ = 1
π§2 + π§ + 1 = 0
π§ =(β1) Β± β(1)2 β 4(1)(1)
2 Γ 1
π§ =β1 Β± β3π2
2
π§ = β1
2+(β3)π
2
π§ = β1
2ββ3
2π
π§ = 1, π§ = β1
2+β3π
2, π§ = β
1
2ββ3π
2
Alternatively we can use Demovreβs theorem
π§3 β 1 = 0
π§3 = 1
π§3 = 1 + 0π
π§ = (1 + 0π)13
πππ‘ π = 1 + 0π
|π| = β1 = 1
argπ = π‘ππβ1 (0
1) = 0
π = π[cos(0) + π sin(0)]
π = 1(cos0 + π sin 0) 1
3z P
π§ = 113(cos(0 + 360π) + π sin(0 + 360π))
πΉππ π = 0, 1, 2β¦
(Depending on the number of roots you want)
For n = 0, π§ = 11
3(cos(0 + 360) + π sin(0 + 360))1
3
π§ = 113(cos120 + π sin120)
π§ = 1(β1
2+β3
2π)
π§ = β1
2+β3π
2
For n = 1, 1 1
3 31 cos(0 360 1) sin(0 36 )][ 0 1z i
π§ = 1(cos120 + π sin120)
152
π§ = β1
2+β3
2π
For n = 2, 1 1
3 31 cos(0 360 2) sin(0 36 )][ 0 2z i
π§ = 1(cos240 + π sin240)
π§ = β1
2ββ3
2π
β π§ = 1, π§ = β1
2+β3
2π, π§ = β
1
2ββ3
2π
Example XVI
Solve: π§3 + 27 = 0
Solution
π§3 + 33 = (π§ + 3)(π§2 + 3π§ + 9)
From π3 + π3 = (π + π)(π2 + ππ + π2)
β π§3 + 33 = (π§ + 3)(π§2 + 3π§ + 9)
π§ = β3
π§2 + 3π§ + 9 = 0
π§ =β3 Β± β32 β 4(1)(9)
2
π§ =β3 Β± β27π2
2
π§ = β3, π§ = β3
2+3β3π
2
z = 3 3 3
2 2i
Alternatively, we can use Demovreβs theorem
π§3 + 27 = 0
π§3 = β27
π§ = (β27 + 0π)13
πππ‘ π = β27 + 0π
|π| = β(β27)2 + ππ)
= 27
arg π = 180
π = 27(cos180 + π sin180)
π§ = π13 = 27
13(cos 180 + π sin180)
13
π§ = 2713(cos(180 + 360π) + π sin(180 + 360π))
13
When n = 0, π§ = 271
3[(cos 180 + π sin 180)]1
3
π§ = 3(cos 60 + π sin60)
= 3(1
2+πβ3
2)
=3
2+3β3
2π
For n = 1, π§ = 271
3[(cos(180 + 360 Γ 1) +
π sin(180 + 360 Γ 1)])1
3
π§ = 3(cos180 + π sin180)
z = -3
For n = 2, π§ = 271
3[(cos(180 + 360 Γ 2) +
π sin(180 + 360 Γ 2)])1
3
π§ = 3(cos300 + π sin300)
=3
2+ β
3β3
2π
=3
2β3
2β3π
π§ = β3, π§ =3
2+3β3
2π and π§ =
3
2β3β3
2π
Example XVII
Solve the equation
π§4 + 1 = 0
π§4 = β1+ 0π
π§4 = (β1 + 0π)14
πππ‘ π = β1 + 0π
|π| = 1
argπ = 180
π = 1(cos 180 + π sin180)
π§ = π14 = 1
14[(cos(180 + 360π)
+ π sin(180 + 360π))]14
πΉππ π = 0, π§ = 11
4(cos 45 + π sin 45)
π§ =β2
2+β2
2π
For n = 1
π§ = 114(cos540 + π sin540)
14
π§ = 1(cos 135 + π sin 135)
π§ =ββ2
2+πβ2
2
For n = 2, π§ = 11
4(cos 900 + π sin 900)1
4
π§ = 1(cos 225 + π sin 225)
π§ =ββ2
2+β2
2π
For n = 3
π§ = 114(cos1260 + π sin1260)
14
π§ = 1(cos 315 + π sin 315)
π§ = (β2
2ββ2
2π)
πΉππ π§4 + 1 = 0
153
π§ =β2
2+β2
2π, β2
2ββ2
2π, β
β2
2+β2
2π, β
β2
2ββ2
2π
Example XVIII
Find the fourth roots of -16
Solution 1 1
4 4( 16) ( 16 0 )z i
Let π = β16 + 0π
π§ = π14 = (β16 + 01)
14
|π| = 16
arg π = 180
π§ = π14 = 16
14[(cos(180 + 360π)
+ π sin(180 + 360π))]14
πΉππ π = 0
π§ = 2(cos 45 + π sin45)
π§ = β2 + β2π
πππ π = 1
π§ = 2(cos540 + π sin540)
π§ = 2(cos135 + π sin135)
= ββ2 + πβ2
For n = 2, π§ = 2(cos 225 + π sin225)
= ββ2 β πβ2
For n = 3, π§ = 2(cos 315 + π sin315)
π§ = β2 β β2π
β πΉππ π§ = (β16 + 0π)14
π§ = β2 β (β2)π, ββ2 + (β2)π
β2 + (β2)π, β β2 β (β2)π
Example XIX
Find the cube roots of 27i
π§ = (0 + 27π)13
πππ‘ π = 0 + 27π
|π| = β02 + 272
= 27
arg = π‘ππβ1 (27
0) = 90
π = 27(cos 90 + π sin90)
π§ = 2713(cos90 + π sin 90)
13
π§ = 2713(cos(90 + 360π) + π sin(90 + 360π))
13
πΉππ π = 0
π§ = 3(cos 30 + π sin30)
π§ =3β3
2+3
2i
π = 1
π§ = 3(cos150 + π sin150)
= 3(ββ3
2+1
2π)
= β3β3
2+3π
2
πππ π = 2
π§ = 3(cos270 + π sin270)
= β3π
Loci in the complex plane What is a locus
A locus is a path possible position of a variable point,
that obeys a given condition. It can be given as
Cartesian equation or it can be described in words.
Example I
The complex number z is represented by the point P
on the Argand diagram.
Given that |π§ β 1 β π| = |π§ β 2| find in the simplest
form the Cartesian equation of the locus
Solution
|π§ β 1 β π| = |π§ β 2|
πππ‘ π§ = π₯ + ππ¦
|π₯ + ππ¦ β 1 β π| = |π₯ + ππ¦ β 2|
|π₯ β 1 + (π¦ β 1)π| = |π₯ β 2 + ππ¦|
β(π₯ β 1)2 + (π¦ β 1)2 = β(π₯ β 2)2 + π¦2
(π₯ β 1)2 + (π¦ β 1)2 = (π₯ β 2)2 + π¦2
π₯2 β 2π₯ + 1 + π¦2 β 2π¦ + 1 = π₯2 β 4π₯ + 4 + π¦2
β2π₯ β 2π¦ + 2 = β4π₯ + 4
2π₯ β 2 = 2π¦
π¦ = π₯ β 1
The locus is a straight line with a positive gradient
π¦ = π₯ β 1) which can be represented on the complex
plane.
πΌπ ππ₯ππ
π πππ ππ₯ππ 1
β1
154
Example II
Given that |π§ β 2| = 2|π§ + π|. Show that the locus of
P is a circle.
Solution
|π§ β 2| = 2|π§ + π|
Let π§ = π₯ + ππ¦
|π₯ + ππ¦ β 2| = 2|π₯ + ππ¦ + π|
|(π₯ β 2) + ππ¦| = 2|π₯ + (π¦ + 1)π|
β(π₯ β 2)2 + π¦2 = 2βπ₯2 + (π¦ + 1)2
(π₯ β 2)2 + π¦2 = 4(π₯2 + (π¦ + 1)2)
π₯2 β 4π₯ + 4 + π¦2 = 4π₯2 + 4π¦2 + 8π¦ + 4
0 = 3π₯2 + 3π¦2 + 4π₯ + 8π¦
π₯2 + π¦2 +4
3π₯ +
8π¦
3= 0
This is sufficient to justify that locus is a circle.
Comparing π₯2 + π¦2 +4
3π₯ +
8π¦
3= 0 With
π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = 0
2π =4
3
π =2
3
82
3
4
3
yfy
f
ππππ‘ππ(βπ, βπ)
ππππ‘ππ (β2
3, β4
3)
π = βπ2 + π2 β π
π = β4
9+16
9β 0
π = β20
9
π =2
3β5
Example III
Show the region represented by |π§ β 2 + π| < 1
Solution
Let π§ = π₯ + ππ¦
|π₯ + ππ¦ β 2 + π|
|π₯ β 2 + (π¦ + 1)π| < 1
β(π₯ β 2) + (y + 1)2 < 1
(π₯ β 2)2 + (π¦ + 1)2 < 1
Itβs a circle with centre (2, -1) and radius less than 1.
It can be illustrated on the argand diagram
In order to represent (π₯ β 2)2 + (π¦ + 1)2 < 1 on the
diagram, we can either take a point inside the circle
or outside the circle as our test point.
Taking (2,-1) as the test point.
β (2 β 2)2 + (β1 + 1)2 < 1
πΌπ ππ₯ππ
π πππ ππ₯ππ
(β2
3,β4
3)
πΌπ ππ₯ππ
π πππ ππ₯ππ
(2, β1)
155
0 + 0 < 1
0 < 1
(2, β1)(the point inside the circle satisfies our locus).
It implies that (2,-1) lies in the wanted region.
Therefore, we shade the region outside the circle.
Example IV
Given that
|π§ β 1
π§ + 1| = 2
find the Cartesian equation of the locus of z and
represent the locus by the sketch on the argand
diagram. Shade the region for which the inequalities.
|π§ β 1
π§ + 1| > 2
Solution
π§ = π₯ + ππ¦
|π₯ + ππ¦ β 1
π₯ + ππ¦ + 1| = 2
|(π₯ β 1 + ππ¦)
(π₯ + 1) + ππ¦| = 2
|π₯ β 1 + ππ¦|
|(π₯ + 1) + ππ¦|= 2
|(π₯ β 1) + ππ¦| = 2|(π₯ + 1) + ππ¦|
β(π₯ β 1)2 + π¦2 = 2β(π₯ + 1)2 + π¦2
(π₯ β 1)2 + π¦2 = 4((π₯ + 1)2 + π¦2)
π₯2 β 2π₯ + 1 + π¦2 = 4(π₯2 + 2π₯ + 1 + π¦2)
3π₯2 + 3π¦2 + 10π₯ + 3 = 0
π₯2 + π¦2 +10
3π₯ + 1 = 0
The locus is a circle comparing
π₯π + π¦2 +10
3π₯ + 1 = 0 with
π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = 0
2π =10
3, π =
5
3, 2π = 0 and π = 0
Center 53
( , 0)
π = βπ2 + π2 β π
π = β25
9+ 0 β 1 =
4
3
For |π§β1
π§+1| > 2
β π₯2 + π¦2 +10
3π₯ + 1 > 0
Example V
Shade the region represented by |π§ β 1 β π| < 3
Solution
Note: Shade the region represented by |z β 1 β i| < 3.
Implies that we shade the wanted region.
Let π₯ + ππ¦
|π₯ + ππ¦ β 1 β π| < 3
|π₯ β 1 + π(π¦ β 1)| < 3
β(π₯ β 1)2 + (π¦ β 1)2 < 3
(π₯ β 1)2 + (π¦ β 1)2 < 9
It is a circle with centre (1, 1) and radius less than 9
Taking (1, 1) as our test point
(1 β 1)2 + (1 β 1)2
(0 + 0) < 9
βThe region inside the circle is the wanted region.
Example VI
πΌπ ππ₯ππ
π πππ ππ₯ππ
(β5
3, 0)
π°π ππππ
πΉπππ ππππ
(1,1)
3
156
Show that when
Re (π§ + π
(π§ + 2)) = 0,
the point P(x, y) lies on a circle with centre
β1,β1
2) and radius
1
2β5
Solution
Re (π₯ + ππ¦ + π
π₯ + ππ¦ + 2) = 0
π π (π₯ + (π¦ + 1)π
π₯ + 2 + ππ¦) = 0
Re ( ( 1) ) 2
(( 2) )( 2 )
x y i x iy
x iy x iy
= 0
π π (π₯(π₯ + 2) β π₯π¦π + (π¦ + 1)(π₯ + 2)π + π¦(π¦ + 1)
(π₯ + 2)2 + π¦2)
2 2
2 2
2 [( 1)( 2) ]Re
( 2)
x x y y y x xy i
x y
2 2
2 2 2 2
2 [( 1)( 2) ]Re 0
( 2) ( 2)
x x y y y x xy i
x y x y
βπ₯2 + 2π₯ + π¦2 + π¦
(π₯ + 2)2 + π¦2= 0
π₯2 + π¦2 + 2π₯ + π¦ = 0
Comparing with
x2 + y2 + 2x + y = 0 with
π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = 0
2π = 2, g = 1
2fy = y
π =1
2
ππππ‘ππ (β1,β1
2)
πππππ’π = βπ2 + π2 β π
= β1 +1
4β 0
=β5
2
=1
2β5
Example VII
Given that π§ = π₯ + ππ¦ . where x and y are real.
Show that Im(π§+π
π§+2) = 0
is equation of a straight line
Solution
Im(π₯+ππ¦+π
π₯+ππ¦+2) = 0
Im
( 1 ) 2
( 2 ) 2
x y i x iy
x iy x iy
= 0
πΌπ(π₯(π₯ + 2) β π₯π¦π + (π¦ + 1)(π₯ + 2)π + π¦(π¦ + 1)
(π₯ + 2)2 + π¦2)
= 0
ββπ₯π¦ + (π¦ + 1)(π₯ + 2)
(π₯ + 2)2 + π¦2= 0
βπ₯π¦ + π₯π¦ + 2π¦ + π₯ + 2
(π₯ + 2)2 + π¦2= 0
2π¦ + π₯ + 2 = 0
π¦ = βπ₯
2+ 1
Which is a straight line with a negative gradient.
Loci in and diagram for arguments of complex
numbers
If arg(π§ β π΄) = πΌ is the equation of half line with
end point A inclined at an angle πΌ to the real axis
Example I
Sketch the loci defined by the equation
arg(π§ β 1 β 2π) =1
4π
Solution
π§ β 1 β 2π = π§ β (1 + 2π)
Thus if A is a point representing 1 + 2π
arg(π§ β (1 + 2π))is the angle AP makes with the
positive real axis. Hence the equation arg(π§ β 1 β
2π) =1
4π represents the half line with end point (1,
2) inclined at angle 1
4π to the real axis.
π°π ππππ
πΉπππ ππππ
π(π₯, π¦)
0
πΌ
A
157
Example II
Sketch the locus of the equation.
arg(π§ + 2) = β2
3π
Solution
arg(π§ + 2) = β2π
3
π§ + 2 = (π§ β β2)
Thus A is a point (-2, 0). Arg(z-2) is the angle AP
makes with the real axis. Hence arg(π§ β β2) =
β2
3π represents a half line with end point (-2, 0)
inclined at angle 2
3π measured clockwise from the
positive axis.
Example III
Show by shading the region represented by
1
3π β€ arg(π§ β 2) β€ π
Solution
The equations arg(π§ β 2) =1
3π πππ arg(π§ β 2) =
π represent half lines with end point (2, 0). Hence the
inequality 1
3π β€ arg(π§ β 2) β€ π
Represent the two lines and region between them
Example IV
Sketch the separate argand diagram the loci defined
by
(π) arg(π§ + 1 β 3π) = β1
6π
(ππ) arg(π§ + 2 + π) =1
2π
Solution
(arg(π§ + 1 β 3π) = β1
6π
π§ β (β1 + 3π) = β1
6π
Thus A is a point (-1, 3)
Arg(z β (-1 + 3i) is the angle AP makes with the real
axis Hence arg(π§ + 1 β 3π) = β1
6π
is equation of the half line with end point (-1, 3)
inclined at an angle of 1
6π measured clockwise from
the real axis
(ππ) arg(π§ + 2 + π) =1
2π
arg(π§ β (β2 β π)) =1
2π
Thus, point A is (β2,β1).
arg(z β (-2 β i)) is the angle AP makes with the real
axis and arg(π§ + 2 + π) =1
2π is the equation of the
π°π ππππ
πΉπππ ππππ
(1, 2)
0
1
4π
π(π₯, π¦)
π°π ππππ
πΉπππ ππππ (β2,0)
π(π₯, π¦)
2
3π
π΄
π°π ππππ
πΉπππ ππππ
1
3π
π΄
π
π°π ππππ
πΉπππ ππππ
1
6π
A(-1, 3)
P
158
line through A inclined at and angle of 1
2π to the real
axis
Sketching of loci involving πππ (πβπ
πβπ) = πΈ
Equation involving arg (π§βπ
π§βπ)are more difficult to
interprete. If arg(π§ β π) = πΌ ,
arg(π§ β π) = π½, arg (π§ β π
π§ β π) = πΎ ,
arg(π§ β π) β arg(π§ β π) = πΎ
πΌ β π½ = πΎ. πΎ = (πΌ β π½) Β± 2π if necessary
Thus πΎ is the angle which the vector AP makes with
the vector BP.
If the turn from BP to AP is anti-clockwise the πΌ is
negative
for arg (π§ β π
π§ β π) > 0
arg (π§ β π
π§ β π) < 0
For instance, if arg (π§β3
π§β1) =
1
4π, then the locus of P is
a circular arc with end point A(3, 0) and (1, 0) such
that β π΄ππ΅ =1
4π
Similarly if arg (π§+2
π§βπ) =
1
3π then the locus of P is a
circular arc with end points A (-2, 0) and B(0, +1)
such that β π΄ππ΅ =1
3π since both cases the given
arguments are positive, the arcs must be drawn so
that the turn from BP to AP is anti-clockwise.
arg (π§ + 2
π§ β π) =
1
3π
Example II
Sketch on different argand diagram the loci defined
by the equations.
(π) arg (π§ β 1
π§ + 1) =
1
3π
(π) arg (π§ β 3
π§ β 2π) =
1
4π
(π) arg (π§
π§ β 4 + 2π) =
1
2π
π°π ππππ
πΉπππ ππππ π
2
(β2, β 1)
π°π ππππ π
π½
π¦
π΅
πΎ
πΉπππ ππππ π₯
πΌ
π§ β π
π΄
π°π ππππ
π½
π΄
πΉπππ ππππ
π
π¦
π₯
πΌ
πΎ
π΅
P
B(0, 1)
Real axis
Im axis
A(-2, 0)
159
Solution
arg (π§ β 1
π§ + 1) =
1
3π
The locus of P is a circular arc with end point A(1, 0)
and B(-1, 0) such that
β π΄ππ΅ =1
3π
(π) arg (π§ β 3
π§ β 2π) =
1
4π
The locus of P is a circular arc with end points (3, 0)
(0, 2) such that β π΄ππ΅ =1
4π
(π) arg (π§
π§ β 4 + 2π) =
1
2π
arg(π§
π§ β (4 β 2π)=1
2π
arg (π§
π§ β 4 + 2π) is a circle with end points
π΄(0, 0) and B(4,β2) such that β APB =1
2π
Example
Find the locus of arg (π§
π§β6) =
π
2
Solution
πππ‘ π§ = π₯ + ππ¦
arg (π§
π§ β 6) = arg π§ β arg(π§ β 6)
β arg(π§) β arg(π§ β 6) =π
2
arg(π₯ + ππ¦) β arg(π₯ + ππ¦ β 6) =π
2
1 1
6tan ( ) tan ( )
2
y y
x x
πππ‘ π΄ = π‘ππβ1 (π¦
π₯)
tanπ΄ =π¦
π₯
π΅ = tanβ1 (π¦
π₯ β 6)
tanπ΅ =π¦
π₯ β 6
(π΄ β π΅) =π
2
tan(π΄ β π΅) = tan (π
2)
tanπ΄ β tanπ΅
1 + tanπ΄ + tanπ΅= β
π¦π₯ β
π¦π₯ β 6
1 +π¦2
π₯(π₯ β 6)
= β
π¦(π₯ β 6) β π₯π¦π₯(π₯ β 6)
π₯2 β 6π₯ + π¦2
π₯(π₯ β 6)
= β
π₯π¦ β 6π¦ β π₯π¦
π₯2 + π¦2 β 6π₯= β
β π₯2 + π¦2 β 6π₯ = 0 which is a circle.
Revision Exercise 1 1. Prove that if |Z| = r, then ZZ* = r2.
P(x, y)
A(3, 0) Real axis
Im axis
B(-1, 0)
P
A(0, 2)
B(3, 0) Real axis
Im axis
P
B(4, -2)
A(0, 0) Real axis
Im axis
160
2. Express 3 + i in modulus-argument form.
Hence find 10( 3 )i and in the
form a + ib.
3. Express -1 + i in modulus-argument form.
Hence show that (-1 + i)16 is real and that
6
1
( 1 )i is purely imaginary, giving the value
of each.
4. Simplify the following expression:
(a) 32 2
7 7
42 27 7
(cos sin )
(cos sin )
i
i
(b)
82 25 5
33 35 5
(cos sin )
(cos sin )
i
i
5. Find the expressions for cos 3ΞΈ in terms of cos
ΞΈ,
sin 3ΞΈ in terms of sin ΞΈ and tan 3ΞΈ in terms of
tan ΞΈ.
6. Express sin 5ΞΈ and cos 5ΞΈ/cos ΞΈ in terms of sin
ΞΈ.
7. Prove that 3 5
2 4
5tan 10 tan tantan5
1 10 tan 5tan
. By considering the equation tan 5ΞΈ = 0, show
that tan2(Ο/5) = 5 β 2 5
8. Find expressions for cos 6ΞΈ/sinΞΈ in terms of cos
ΞΈ and for tan 6ΞΈ in terms of tan ΞΈ.
9. Express in terms of cosines of multiples of ΞΈ:
(a) cos5ΞΈ (b) cos7ΞΈ (c) cos4ΞΈ
10. Express in terms of sines of multiples of ΞΈ:
(a) sin3ΞΈ (b) sin7ΞΈ (c) cos4ΞΈsin3ΞΈ
11. Prove that cos6ΞΈ + sin6ΞΈ = 18
(3cos4ΞΈ + 5)
12. Evaluate (a) 4
0
sin
dπ (b)
2
4 2
0
cos sin d
13. (a) Express the following complex numbers in a
form having a real denominator.
1
3 2i,
2
1
(1 )i
(b) Find the modulus and principal arguments of
each of the complex numbers Z = 1 + 2i and
W = 2 β I, and represent Z and W clearly by
points A and B in an Argand diagram. Find
also the sum and product of Z and W and
mark the corresponding points C and D in
your diagram.
14. If the complex number x + iy is denoted by Z,
then the complex conjugate number x β iy is
denoted by Z*,
(a) Express |Z*| and (Z*) in terms of |Z| and
arg(Z).
(b) If a, b, and c are real numbers, prove that if
aZ2 + bZ + c = 0, then then a(Z*)2 + b(Z*) +
c = 0
(c) If p and q are complex numbers and q β 0,
prove *
*
*
p p
q q
15. Find the values of a and b such that (a + ib)2 = i.
Hence or otherwise solve the equation z2 + 2z +
1 β i = 0, giving your answer in the form p + iq,
where p and q are real numbers.
16. If 1
(1 )2
Z i , write down the modulus and
argument for each of the numbers Z, Z2, Z3, Z4.
Hence or otherwise, show in the Argand
diagram, the points representing the number 1 +
Z + Z2 + Z3 + Z4.
17. If Z = 3 β 4i, find
(i) Z* (ii) ZZ* (iii) (ZZ)*
18. Simplify each of the following:
(a) (3 + 4i) + (2 + 3i) (b) (2 β 4i) β 3(5 β 3i)
(b) (2i)2 (c) i4
19. Simplify each of the following:
(a) (2 + i)(3 β i) (b) (5 β 2i)(6 + i)
(c) (4 β 3i)(1 β i) (d) (3 + i)(2 β 5i)
20. Express each of the following in the form a + ib
(a) 20
3 i (b)
4
1 i
(c) 2
1
i
i (d)
1
1 2i
21. Solve the following equations:
(a) x2 + 25 = 0
(b) 2x2 + 32 = 0
(c) 4x2 + 9 = 0
(d) x2 + 2x + 5 = 0
22. If 3 β 2i and 1 + i are two of the roots of the
equation ax4 + bx3 + cx2 + dx + c = 0, find the
values of a, b, c, d and e.
23. Find the square roots of the following complex
numbers:
(a) 5 + 2i
(b) 15 + 8i
(c) 7 β 24i
24. Find the quadratic equations have the roots:
(a) 3i, -3i (b) 1 + 2i, 1 β 2i
(c) 2 + i, 2 β i (d) 2 + 3i, 2 β 3i
25. Find real and imaginary parts of the complex Z
when:
(i) 1
Z
Z = 1 + 2i
(ii) 1 3
Z i Z i
Z Z
26. Find the modulus and principal argument of the
following complex numbers
(a) 3i (b) 15 (c) -3i (d) -1
7
1
( 3 )i
161
27. Find the modulus and principle argument of:
(a) 1
1
i
i
(b)
1 7
4 3
i
i
(c) 1
2
i
i
(d)
2(3 )
1
i
i
28. If Z1 and Z2 are complex numbers, solve the
simultaneous equations
4Z1 + 3Z2 = 23
Z1 + iZ2 = 6
giving your answer in the form x + iy
29. Given that 2 + i is a root of the equation
Z3 β 11Z + 20 = 0. Find the remaining roots.
30. Show that 1 + i is a root of the equation x4 + 3x2
β 6x + 10 = 0. Hence write down the quadratic
factor of x4 + 3x2 β 6x + 10 and find all the roots
of the equation.
31. The complex number satisfies the equation
22
Zi
Z
. Find the real and imaginary parts
of Z and the modulus and argument of Z.
32. If Z1 = 13 13
24 244 cos sini and Z2=
5 5
24 242 cos sini , find and Z1Z2 in the
form a + ib.
33. If Z1=23
2cos + 23
sini and Z2 =
, find:
(i) 1
2
Z
Z (ii) arg 1
2
Z
Z
(iii) 2
1
Z
Z
(iv) arg
34. One root of the equation Z2 + aZ+b=0 where a
and b are real constants, is 2+3i. Find the values
of a and b.
35. If Z1 and Z2 are two complex numbers such that
|Z1 β Z2| = Z1 + Z2|, show that the difference of
their arguments is 3
or 2 2
36. (a) Find the modulus and argument of
(b) If 1
1 7
1
iZ
i
and . Find the
moduli of Z1, Z2, Z1 + Z2 and Z1Z2.
37. Use Demoivreβs theorem to show that:
38. Use Demoivreβs theorem to show that:
cos 4ΞΈ = cos4ΞΈ β 6cos2ΞΈ sin2ΞΈ + sin4ΞΈ
sin4ΞΈ = 4cos3ΞΈ sinΞΈ β 4cos ΞΈ sin3ΞΈ
39. Show that
40. Use Demoivreβs theorem to find the value of
-3 1
-3 1
41. Find the two square roots of I and the four
values of 1
4(-16) .
42. Find the three roots of the equation (1 β Z)3 = Z3
43. If W is a complex cube root of unity, show that
(1 + W β W2)3 β (1 β W + W2)3 = 0
44. Use Demoivreβs theorem to find the four fourth
roots of 8(-1 + i 3 ) in the form a + ib, giving a
and b correct to 2 decimal places.
45. Use Demoivreβs theorem to show that
cos5
cos
x
x = 1 β 12sin2x + 16sin4x
46. Prove that if 6
8
Z i
Z
is real, the locus of the point
representing the complex number Z in the
Argand diagram is a straight line.
47. Prove that if 2
2 1
Z i
Z
is purely imaginary, the
locus of the point representing Z in the Argand
diagram is a circle and find its radius.
48. If Z is a complex number and = 2, find the
equation of the curve in the Argand diagram on
which the point representing it lie.
49. The complex numbers Z β 2 and Z β 2i have
arguments which are
(i) equal and
(ii) differ by and each argument lies
between βΟ and Ο. In each case, find the
locus of the point which represents Z in the
Argand diagram and illustrate by a sketch.
50. Show by shading on an Argand diagram the
region in which both |Z β 3 β i| β₯ |Z β 3 β 5i|
Answers 1.
2. (a) 1, (b) -i (c) 1 3
2 2i (d)
1 3
2 2i
3. 6 6
2 cos sini ; 512 β 512 3 i , 3 1
256 256i
4. 3 3
4 42 cos sini ; 256 β 1
8i
5. (a) 1, (b) -1
6. 4cos3ΞΈ β 3cos ΞΈ β 4sin3ΞΈ, 3
3
3tan tan
1 3tan
7. 16sin5ΞΈ β 20sin3ΞΈ + 5sin ΞΈ, 1 β 12sin2ΞΈ +
16sin4ΞΈ
8. .
1
2
Z
Z
3 34 4
6 cos sini
2
1
Z
Z
2(2 ) (3 1)
3
i i
i
2
17 7
2 2
iZ
i
5 3
7 5
(cos3 sin 3 ) (cos sin )cos13 sin13
(cos5 sin 5 ) (cos 2 sin 2 )
i ii
i i
2 2
1 sin sincos ( ) sin ( )
1 sin sin
ni
n i ni
1
Z i
Z
1
2
162
9. 32cos6ΞΈ β 48cos4ΞΈ + 18cos2ΞΈ β 1, 32cos5ΞΈ β
32cos3ΞΈ + 6cos = 3 5
2 4
6 tan 20 tan 6 tan
1 15 tan 15 tan
10. (a) 1
16(cos5 5cos3 10cos ),
(b) 1
64(cos7 7cos5 21cos3 35cos )
(c) 1
16(2cos cos3 cos5 )
11. (a) 1
4(3sin sin3 ),
1
64(35sin 21sin3 7sin sin7 ),
(c) 1
64(3sin sin3 sin5 sin7 )
12. (a) 3
8
, (b) 32
13. (a) 3 2 1
,13 2
ii
(b) 5 , 63.4Β°, 5 , -26.6Β°, 3 + i,
4 + 3i.
15. 1 1
,2 2
a b or 1 1
,2 2
a b
2
1
21
iZ
or 1
22
1i
Z
16. 2
2, 45Β°;
1
2, 90Β°;
2
4, 135Β°; 1
4, 180Β°
17. (i) 3 + 4i (ii) 25 (iii) -7 + 24i
18. (a) 5 + 7i (b) -13 + 5i (c) -4 (d) 1
19 (a) 7 + i (b) 32 β 7i (c) 1 β 7i (d) 11 β 13i
20. (a) 6 β 2i (b) 2 β 2i (c) -1 + i (d) 1 2
5 5i
21 (a) x Β± 5i (b) x = Β±4i (c) Β± 3
2i (d) x = -1 Β± 2i
22. a = 1, b = -8, c = 27, d = -38, e = 26
23. (a) Β±(3 + 2i) (b) Β±(4 + i) (c) Β±(4 β 3i)
24. (a) x2 + 9 = 0 (b) x2 β 2x + 5 = 0
(c) x2 β 4x + 5 = 0 (d) x2 β 4x + 13 = 0
25 (i) -1, Β½ (ii) 1 2
5 5,
26. (a) 3, Ο/2 (b) 15, 0 (c) 3, -Ο/2 (d) 1, Ο
27. (a) 1, -Ο/2 (b) 3
42, , (c) 10
5, 1.25
28. 2 + 3i 19. 2 β i, -4
31. (i) Re(Z) = -3, Im(Z) = -1 (ii) 10 , -2.82 rads
32. 1 3i ; 4 2 4 2 i
33. (i) 1/3 (ii) 7
12
(iii) 3 (iv) 7
12
34. -4, 13
36. (a) 5, 0.6435 rad (b) 5, 6.5, 2.061, 32.5
41. (1 )
2
i , 2 2i
42. 1 1
2 2, (1 3)i . 44. (1.73 ), (1 1.73 )i i
47. centre ΒΌ + i, radius 1
47
48. 2
25 16
3 9x y
49. (i) x + y = 2 (ii) (x β 1)2 + (y β 1)2 = 2
Exercise 2 Show on the Argand diagram the region represented
by the following:
1. arg z = 14Ο,
2. arg(z β i) = 13Ο
3. arg(z + 1 β 3i) = 16Ο
4. arg(z β 3 + 2i) = Ο
5. arg(z + 2 + i) = 12Ο
6. arg(z β 1 β i) = 1
4 Ο
7. |z + 1| = |z β 3|,
8. |z| = |z β 6i|
9. 1
z i
z
= 1
10. (a) arg 13
1
1
z
z
11. (a) arg 14
3
2
z
z i
(b)
In questions 12 to 24 find the Cartesian equation of
the locus of the point P representing the complex
number z. Sketch the locus of P each case.
12. 2|z + 1| = |z β 2|
13. |z + 4i| = 3|z β 4|
14. 4
z
z = 5
15. 5 2
z i
z i
= 1
16. = 5
17.
18. z β 5 = Ξ»i(z + 5), where Ξ» is a real parameter
19. 2
2
z i
z
= Ξ»i, where Ξ» is a real number.
20. z = 3i + Ξ»(2 + 5i), where Ξ» is a real parameter.
21. Im(z2) = 2
22. Re (z2) = 1
23. 1
Re zz
= 0
24. 9
Im zz
= 0
In questions 27 to 34 shade in separate Argand
diagrams the regions represented by:
25. |z β i | β€ 3
26. |z β 4 + 3i| < 4
27. 0 β€ arg z β€ 13
28. 14Ο < arg z < 3
4Ο
124 2
z
z i
6
z
z
23
1
1
z
z i
163
29. 1 16 6
arg( 1)z
30. 1 22 3
arg( )z i
31. |z| > |z + 2|
32. |z + i| β€ |z β 3i|
33. Represent each of the following loci in an
Argand diagram.
(a) arg(z β 1) = arg(z + 1)
(b) arg z = arg(z β 1 I i)
(c) arg(z β 2) = Ο + arg z
(d) arg(z β 1) = Ο + arg(z β i)
34. Find the least value of |z + 4| for which
(a) Re(z) = 5 (b) Im(z) = 3
(c) |z| = 1 (d) arg z = 14Ο
35. Given that the complex number z varies such
that |z β 7| = 3, find the greatest and least values
of |z β i|.
36. Given that the complex number w and z vary
subject to the conditions |z β 12| = 7 and |z β i| =
4, find the greatest and least values of |w β z|.
37. In an Argand diagram, the point P represents the
complex number z, where z = x + iy. Given that
z + 2 = Ξ»i(z + 8), where Ξ» is a real parameter,
find the Cartesian equation of the locus of P as Ξ» varies. If also z = ΞΌ(4 + 3i), where Ξ» is real,
prove that there is only one possible position for
P.
38. (i) Represent on the same Argand diagram the loci
given by the equations |z β 3| = 3 and |z| = |z β 2|.
Obtain the complex numbers corresponding to the
point of intersection of these loci. (ii) Find a
complex number z whose argument is Ο/4 and
which satisfies the equation |z + 2 + i| = |z β 4 + i|.
Answers 12. x2 + y2 + 4x = 0, 13. x2 + y2 β 9x β 9x β y + 16
15. 5x + 3y = 14. 16. 2x2 + 2y2 + 25x + 75 = 0
17. 5x2 + 5y2 β 26x + 8y + 1 = 0.
18. x2 + y2 = 25, excluding (-5, 0)
19. x2 + y2 β 2x + 2y = 0, excluding (2, 0).
20. 5x β 2y + 6 = 0 21. xy = 1. 22. x2 β y2 = 1
23. x(x2 + y2 β 1) = 0, excluding (0, 0)
24. y(x2 + y2 β 9) = 0, excluding (0, 0).
34. (a) 9, (b) 3, (c) 3, (d) 4.
35. 5 2 3 , 5 2 3 . 38. 24, 2.
37. x2 + y2 + 10x + 16 = 0
38.(i) 1 5i (ii) 1 + i.
Revision Exercise 3 Show on the Argand diagram the region represented
by the following:
1.
2. = 1
3. Express the complex number 1
11 2
3 4
iz
i
in the
form x + iy where x and y are real. Given that z2
= 2 β 5i, find the distance between the points in
the Argand diagram which represent z1 and z2.
Determine the real numbers Ξ± and Ξ² such that
Ξ±z1 + Ξ²z2 = -4 + i.
4. (i) Find two complex numbers z satisfying the
equation z2 = -8 β 6i.
(ii) Solve the equation z2 β (3 β i)z + 4 = 0 and
represent the solutions on an Argand diagram
by vectors OA and OB , where O is the
origin. Show that triangle OAB is right-
angled.
5. If z and w are complex numbers, show that: 2 2 2 2| | | | 2{| | | | }z w z w z w
Interpret your results geometrically.
6. A regular octagon is inscribed in the circle |z| =
1 in the complex plane and one of its vertices
represents the number1
(1 )2
i . Find the
numbers represented by the other vertices.
7. (i) Two complex numbers z1 and z2 each have
arguments between 0 and Ο. If z1z2 = i β 3
and 1
2
z
z = 2i, find the values of z1 and z2
giving the modulus and argument of each.
(ii) Obtain in the form a + ib the solutions of
the equation z2 β 2z + 5 = 0, and represent the
solutions on an Argand diagram by the points
A and B.
The equation z2 β 2pz + q = 0 is such that p
and q are real, and its solutions in the Argand
diagram are represented by the points C and
D. Find in the simplest form the algebraic
relation satisfied by p and q in each of the
following cases:
(a) p2 < q, p β 1 and A, B, C, D are the
vertices of a triangle;
(b) p2 > q and 1
2CAD
8. (a) If βΟ < arg z1 + arg z2 β€ Ο, show that arg(z1z2)
= arg z1 + arg z2. The complex numbers
4 3 2a i and 3 7b i are represented in
the Argand diagram by points A and B
respectively. O is the origin. Show that triangle
OAB is equilateral and find the complex
number c which the point C represents where
OABC is a rhombus. Calculate |c| and arg c.
13
1
1
z
z
2 3
2
z i
z i
164
(b) z is a complex number such that
2 1 3
p qz
q i
where p and q are real. If
arg z = Ο/2 and |z| = 7 find the values of p
and q.
9. .
10. (a) Show that (1 + 3i)3 = -(26 + 18i).
(b) Find the three roots z1, z2, z3 of the equation
z3 =-1
(c) Find in the form a + ib, the three roots z'1, z'2,
z'3 of the equation z3 = 26 + 18i.
(d) Indicate in the same Argand diagram the
points represented by zr and z'r for r = 1, 2, 3,
and prove that the roots of the equations may
be paired so that |z1 β z2| = z2 β z'2 = |z3 β z'3 |
= 3.
11. Write down or obtain the non-real cube roots of
unity, w1 and w2, in the form a + ib, where a and
b are real. A regular hexagon is drawn in an
Argand diagram such that two adjacent vertices
represent w1 and w2, respectively and centre of
the circumscribing circle of the hexagon is the
point (1, 0). Determine in the form a + ib, the
complex numbers represented by the other four
vertices of the hexagon and find the product of
these four complex numbers.
12. A complex number w is such that w3 = 1 and w
β 1. Show that:
(i) w2 + w + 1 = 0
(ii) (x + a + b)(x + wa + w2b)(x + w2a + wb)
is real for real x, a and b, and simplify this
product. Hence or otherwise find the three roots
of the equation x3 β 6x + 6 = 0, giving your
answers in terms of w and cube roots of integers.
13. (i) Find, without the use of tables, the two
square roots of 5 β 12i in the form x + iy, where
x and y are real.
(ii) Represent on an Argand diagram the loci |z β
2| = 2 and |z β 4| = 7. Calculate the complex
numbers corresponding to the points of
intersection of these loci.
14. (i) Given that (1 + 5i)p β 2q = 7i, find p and q
when (a) p and q are real (b) p and q are
conjugate complex numbers.
(ii) Shade on the Argand diagram the region for
which 3Ο/4 < arg z < Ο and 0 < |z| < 1.
Choose a point in the region and label it A. If
A represents the complex number z, label
clearly the points B, C, D and E which
represent βz, iz, z + 1 and z2 respectively.
15. (i) Show that z = 1 + i is a root of the equation z4
+ 3z2 β 6z + 10 = 0. Find the other roots of the
equation.
(ii) Sketch the curve in the Argand diagram
defined by |z β 1| = 1, Im z β₯ 0. Find the
value of z at the point P in which this curve
is cut by the line |z - 1| = |z β 2|. Find also the
value of arg z and arg(z β 2) at P.
16. (i) If z = 1 + i 3 , find |z| and |z5|, and also the
values of arg z and arg(z5) lying between βΟ
and Ο. Show that Re(z5) = 16 and find the
value of Im(z5).
(ii) Draw the line |z| = |z β 4| and the half line
arg(z β i) = Ο/4 in the Argand diagram.
Hence find the complex number that satisfies
both equations.
17. (i) Without using tables, simplify 4
9 9
5
9 9
(cos cos )
(cos sin )
i
i
.
(ii) Express z1 = 7 4
3 2
i
i
in the form p + qi, where
p and q are real. Sketch in an Argand
diagram the locus of the points representing
complex numbers z such that |z β z1| = 5 .
Find the greatest value of z subject to this
condition.
18. (i) Given that z = 1 β i, find the values of r(>0)
and ΞΈ, -Ο < ΞΈ < Ο, such that z = r(cos ΞΈ + i sin
ΞΈ). Hence or otherwise find 1/z and z6,
expressing your answers in the form p + iq,
where q, r Ο΅ β.
(ii) Sketch on an Argand diagram the set of points
corresponding to the set A, where A = {z:z Ο΅ β,
arg (z β i) = Ο/4}. Show that the set of points
corresponding to the set B, where B = {z:z Ο΅ β,
|z + 7i| = 2|z β 1|}, forms a circle in the Argand
diagram. If the centre of this circle represents
the numbers z1, show that z1 Ο΅ A. 19. Use De Moivreβs theorem to show that
cos 7ΞΈ = 64cos7ΞΈ β 112cos5ΞΈ + 56cos3ΞΈ β 7cosΞΈ
20. (i) If (1 + 3i)z1 = 5(1 + i), express z1 and z12 in
the form x + iy, where x and y are real.
Sketch in an Argand diagram the circle |z β
z1| = |z1| giving the coordinates of its centre.
(ii) If z = cos ΞΈ + i sinΞΈ, show that:
12 sinz i
z
12 sinn
nz i n
z
Hence or otherwise, show that
16sin5ΞΈ = sin 5ΞΈ β 5sin 3ΞΈ + 10sin ΞΈ
21. . 22. (i) Given that x and y are real, find the values of
x and y which make satisfy the equation
2 40
2
y i y
x y x i
165
(ii) Given that z = x + iy, where x and y are real,
(a) Show that Im 02
z i
z
, the point (x, y)
lies on a straight line (b) Show that, when
Re 02
z i
z
, the point (x, y) lies on a circle
with centre (-1, -Β½) and radius 12
5
23. (i) Find |z| and arg z for which the complex
numbers z given by (a) 12 β 5i, (b) 1 2
2
i
i
,
giving the argument in degrees (to the
nearest degree) such that -180Β° < arg z β€
180Β°.
(ii) By expressing 3 β i in modulus-argument
form, or otherwise, find the least positive
integer n such that ( 3 β i)n is real and
positive.
(iii) The point P in the Argand diagram lies
outside or on the circle of radius 4 with
centre at (-1, -1). Write down in modulus
form the condition satisfied by the complex
number z represented by point P.
24. Sketch the circle C with Cartesian equation x2 + (y
β 1)2 = 1. The point P representing the non-zero
complex number z lies on C. Express |z| in terms of
π, the argument of z. Given that z' = 1/z, find the
modulus and argument of z' in terms of π. Show
that, whatever the position of P on the circle C, the
point P' representing z' lies on a certain line, the
equation of which is to be determined.
25. (a) The sum of the infinite series 1 + z + z2 + z3
+ β¦ for values of z such that |z| < 1 is 1/(1 β z).
By substituting z = Β½(cos ΞΈ + isin ΞΈ) in this
result and using De Moivreβs theorem, or
otherwise, prove that
2
1 1 1 2sinsin sin 2 sin ...
2 5 4cos2 2nn
Answers
3. 1 + 2i; 5 2 ; -2, -1
4. (i) Β±(1 β 3i), (ii) 2 β 2i, 1 + i
5. sum of squares of a parallelogram = sum of squares of
sides
6. Β±1, Β±i, 1
2 (1 β i), 1
2
(1 + i)
7. (i) -1 + i 3 , 2, 2Ο/3; 3
2 + 1
2i, 1, Ο/6
(ii) 1 Β±2i; (a) p2 = q β 4, (b) 2p = q + 5
8. (a) 3 3 5 ; 2 13, 2.38 rad, (b) 5, 20i .
9. (i) -1 β i, 3Ο/4, (ii) 2 β i, 2; -10.
10.(b) -1, 1 3
2 2i , (c) -1 β 3i,
1 1
2 2(1 3 3) (3 3)i
11. 1
2
3;
2i 1 3;i , 5
2
3
2i ; 28
12. (ii) x3 β 3abx + a3 + b3; 3 32 4 , 3 322 4 ,
3 32 2 4
13. (i) Β±(3 β 2i), (ii) 3 Β± 3i .
14. (i) (a) 7/5, -4/5; (b) 2 Β± i
15. (i) 1 β i, -1 Β± 2i, (ii) 1
2(3 3)i ; Ο/6, 2Ο/3
16. (i) 2, 32, Ο/3, -Ο/3; 16 3 , (ii) 2 + 3i
17. (i) -1, (ii) 1 + 2i, 2 5
18. (i) 2 , -Ο/4; 1 1
2 2i , 8i.
20. (i) 2 β i, 3 β 4i; (2, -1)
21. (ii) 3x2 + 3y2 + 10x + 3 = 0.
22. (i) x = 1, y = 2 or x = -1, y = -2
23. (i) a) 13, -23Β°, (b) 1, 90Β°; (ii) 12;
(iii) |z + 1 + i| β₯ 4
24. 2sin ; 1 1
2 2cosec , ; y . 25. 2 2 2 .