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CHAPTER 1: COMPLEX NUMBERS
DQT 101
CHAPTER 1 : COMPLEX NUMBERS
1.1
Introduction and Definitions
A collection of real number system was not complete. We realize that a
quadratic equation a 2  bx  c  0 , could not be solved if b2 – 4ac <0.
Consider a quadratic equation x 2  1  0, a  c  1, b  0 then we have
x    1 and x    1 , Since we know that the square of any real
number must be greater than or equal than zero, we realize that the real
number system was not complete. This leads to a study of complex
numbers which are useful in variety of applications. A complex number
system allowed the possibility of negatives squares and had the real
numbers as a subset, R. In above quadratic equation we do not have any
problems in solving this equation as complex number system allowed a
negatives squares. We introduce a symbol, j , with the property that
j 2  1 and it follows j   1 . We can use this to write down the square
root of any negative numbers. Then the solution of the equation are given
by x = j and x = -j.
Example: Write down the expression of the square roots of a) 9 b) -9
Solutions: a) 9   3
b) -9 = 9  -1,  9  9  1  9   1  3   1  3 j
Simplify j 3 , j 4 , j 8
Definition 1.1 : Complex Number
If z is a complex number then we write
z  a  bj
where a, b Є R , a is a real part and b is the imaginary part.
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.1
Express in the form z  a  bj .
a.  25
c. 5 +  36
b.
d.
 20
27 +
 18
Solutions:
a.  25 = 25(1) = 25  1 = 5j
b.
 20 =
4(5)( 1) = (2 5 )j
c. 5 +  36 = 5 + 6j
d.
27 +
 18 = 3 3 + (3 2 )j
Definition 1.2 : Equal Complex Numbers
Two complex numbers z1  a  bj and z2  c  dj are equal if and only if
a = c and b = d.
Example 1.2
Given 2 x  3 yj  4  3 j. Find x and y.
Solution:
x = 2 and y = 1
1.2
Operations of complex numbers
Definition 1.3 :
If z1  a  bj
and
z 2  c  dj then,
z1  z2  (a  c)  (b  d ) j
ii) z1  z2  (a  c)  (b  d ) j
i)
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
iii) z1z2 
 (a  bj )( c  dj )
 ac  adj  bcj  bdj 2
 (ac  bd )  (ad  bc) j
Example 1.3
Given z1  2  3 j and z2  1  j . Find
i) z1  z2
ii) z1  z2
iii) z1 * z2
iv) Determine the value of z = 7 (3 + j4)  (2 + j) (5  6j)
Solutions:
i) z1 + z2 = (2 + 3j) + (1 – j)= (2 + 1) + (3 – 1)j= 3 + 2j
ii) z1 – z2 = (2 + 3j) - (1 – j)= (2 - 1) + (3 + 1)j= 1 + 4j
iii) z1 * z2 = (2+3j)(1-j) = 2 – 2j + 3j -3j2 = 5 + j
iv)
z



7 3  4 j   2  j 5  6 j 
21  28 j   10  5J  12 j  6 j 2 
21  28 j   10  6  5  12 j 



5  35 j
21  28 j   16   7 j 
21  16  (28  7) j
Definition 1.4 : Complex Conjugate
If z = a + bj, its complex conjugate , denoted by z , is
z  a  bj
* Show that z z  a 2  b2
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.4
Find the complex conjugate of :
i) z = 4 + 5j
ii) z = 4 - 5j
iii) z = -j
iv) z = j
Solutions:
i) z = 4 - 5j
iii) z = j
ii) z = 4 + 5j
iv) z = -j
Properties of conjugate complex number:
a) z  z
b) z1  z2  z1  z2
d) z1  z 2  z1  z 2
c) z1  z 2  z1  z 2
 z1  z1
 
 z2  z2
e) 
h)
f)
zz
 Re( z )
2
g) z n  z  : n integer
1 1
 
z
z
i)
n
zz
 Im( z )
2
Definition 1.5 : Division of complex numbers
If z1  a  b
and
z 2  c  d then
z1 a  bj

z 2 c  dj
a  bj c  dj

c  dj c  dj
ac  bd  (bc  ad ) j

c2  d 2

Example 1.5:
Determine the value of :
i)
2 j
1 j
ii)
1 j
3 j
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
iii) z 
7
1  j2

3 j 3 4 j
Solutions:
2 j 2
i)

1  j  1 
j  1 

j  1 
j  2  2 j  j  j2 1 3
  j

j 
1  j2
2 2
1 j 1
ii)

3  j  3 
j  3 

j  3 
j  3  j  3 j  j2 1 2
  j

j 
9  j2
5 5
iii)
z






1.3
7
1  j2

3 j 3 4 j
1  j 23  4 j 
73  j 

3  j 3  j  3  4 j 3  4 j 
21  7 j   3  j6  4 j  8 j 2 
9  j3  3 j  j 2  9  12 j  j12  16 j 2 
21  7 j   3  8  j10
9  1
9  16
2.1  j(0.7)   0.2  j(0.4) 
2.3  1.1 j
Argand Diagram
The complex number z = a + bj is plotted as a point with coordinates
(a,b). Such a diagram is called an Argand diagram.
y (imaginary axis)
b
z = a + bj
● (a , b)
a
x(real axis)
Figure 1.1 : An Argand diagram in which the point (a,b) represents
the complex number
z = a + bj.
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.6
Plot the following complex numbers on an Argand diagram.
i) 3j
ii) 5 + 2j
iii) -4
iv) -1 – 2j
Solutions:
3 ● (0,3)
2
● (5,2)
1
(0,-4)
●
-4 -3 -2 -1 0
1
2
3
4
5
6
-1
(-1,-2) ●-2
Example 1.7
Given that z1  1  3 j dan z2  3  j . Plot z1  z2 in an Argand diagram.
y
4
3
(4,4)
(1, 3)
z1  z2
2
1
(3,1)
x
1
2
3
4
5
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
1.4 The Modulus and Argument of a Complex Number
Consider an Argand diagram in figure 1.2, the distance of the point (a,b)
imaginary axis
(a,b) , z = a + bj
b
z r
a
real axis
Definition 1.6 : Modolus of complex numbers
The modulus of a complex number z = a + bj written as z is
r = z = a 2  b2
Example 1.8
Find the modulus of the following complex numbers.
i) 3 – 4j ii) 5 + 2j iii) -1 – j iv) 7j
Solutions:
i) |3 – 4j| = 3 2  (4) 2 = 25 = 5
ii) |5 + 2j| = 52  2 2 = 29
iii) |-1 – j| = 2
iv) |7j| = 7
Properties of Modulus
i) | z || z |
ii) zz | z |2
z1 | z1 |

, z2  0.
z2 | z2 |
iii) | z1 z2 || z1 || z2 |
iv)
v) | z n || z |n
vii) | z1  z2 || z1 |  | z2 |
vi) | z1  z2 || z1 |  | z2 |
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Definition 1.7 : Arguments of complex numbers
The argument of the complex number, z = a + bj often given the symbol θ
b
is define as   tan 1   .
a
y
b
(a,b)
z
θ
x
a
Example 1.9
Find the arguments of the following complex numbers;
i) 3 – 4j
ii) 5 + 2j
iii) -1 – j
iv) 1 + 7j
Solutions
If θ is the argument of the following complex numbers, therefore,
 4
o
o
 = -53.13 = 306.87
3


i) θ = tan-1 
2
ii) θ = tan-1   = 21.80o
5
 1
o
 = 225

1
 
iii) θ = tan-1 
7
iv) θ = tan-1   = 81.87o
1
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
1.5
The polar form of complex numbers
(a,b)
|z| = r
b
θ
a
Using trigonometry we can write,
b
; b  r sin 
r
a
cos   ; a  r cos 
r
b
b
tan   ;   tan 1  
a
a
sin  
and; |z| = r = a 2  b2
Therefore, we can then write z = a + bj in polar form as:
z = r cos θ + j r sin θ
= r (cos θ + j sin θ)
or
= (r, θ)
Example 1.10
State the complex numbers of z when
i) z = 1 + j
ii) z = 3 – 4j
Solutions
i) r = 12  12 = 2 and θ = tan 1 1 = 45o
Hence, z = 2 (cos 45o + j sin 45o) or z = ( 2 ,45o)
 4
o
o
 = - 53.13 @ 306.87
 3 
ii) r = 3 2  (4) 2 = 5 and θ = tan 1 
Hence, z = 5(cos 306.87o + j sin 306.87o) or z = (5, 306.87o)
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Exercise 1.11
State the following complex numbers into the form of z = a + bj.
i) z = 4(cos 30o + j sin 30o)
ii) z = 3 (cos 135o + j sin 135o)
Solutions
i) r = 4 dan θ = 30o
 3
a = r cos θ = 4 cos 30o = 4  = 2 3
 2 
1
b = r sin θ = 4 sin 30o = 4  = 2
2
Hence, z = 4(cos 30 + j sin 30o) = 2 3  2 j
o
ii) r = 3 dan θ = 135o
a = r cos θ = 3 cos 135o = 3  

b = r sin θ = 3 sin 135o = 3 
1 
3
= 
2
2
1 
=
 2
3
2
Hence, z = 3 (cos 135o + j sin 135o) = 
3
3

j
2
2
Theorem 1
If z1 and z2 are two complex numbers in polar form where
z1  r1 (cos 1  j sin 1 ) dan z2  r2 (cos 2  j sin 2 ) therefore
i) z1z 2 = r1r2 [cos(1  2 )  j sin( 1  2 )]
ii)
r
z1
= 1 [cos(1   2 )  j sin( 1   2 )]
r2
z2
Example 1.12
Given that z1 = 4(cos 30o + j sin 30o) and z 2 = 2(cos 60o + j sin 60o). Find
i) z1z 2
ii)
z1
z2
Solutions
i) z1z 2 = 4.2 [cos (30o + 60o) + j sin (30o + 60o)]
= 8 (cos 90o + j sin 90o)
=8j
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
ii)
z1
4
= [cos (30o - 60o) + j sin (30o - 60o)]
2
z2
= 2 [cos (-30o) + j sin (-30o)]
 3 1
= 2 

j 
2 
 2
= 3 j
1.6
The exponential form of complex numbers
Definition 1.8 :
Complex number z = r(cos θ + j sin θ) can also be written in
exponential or Euler form as z = reθj where θ is measured in
radians and eθj = cos θ + j sin θ.
Example 1.13
State the following angles θ in radians.
i) 45o
ii) 120o
iii) 270o
Solutions
i) 450  450.

180

0
4
2
180
3

3
iii) 2700  2700. 0 
180
2
ii) 1200  1200.


0

Example 1.14
Express the complex number z = 1 + j in exponential form.
Solutions
From Example 1.10 (i), we know that z = 1 + j = 2 (cos 45o + j sin 45o)
which r = 2 dan θ = 45o.
from Example 1.12 (i), we have 450 

4
.
Hence, z = 1 + j = 2 (cos 45o + j sin 45o)
 2e

j
4
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Theorem 2
If z1  r1e j and z2  r2 e j , then
i) z1z 2 = r1r2 e(  ) j
1
2
1
ii)
2
z1
r e (1 2 ) j
= 1
, z2  0
r2
z2
Example 1.15:

If z1  8e
2
j

j
and z2  2e 3 , find
i) z1z 2
z1
z2
ii)
Solutions
i) z1 z 2  8.2e
  
  j
2 3
  
  j
2 3
z1 8e

ii)
z2
2
1.7
 16e

 4e 6
5
j
6
j
De Moivre’s Theorem
Theorem 3
If z = r(cos θ + j sin θ) is a complex number in polar form to any
power n, then
zn = rn(cos nθ + j sin nθ)
with any value n.
Example 1.16
If z = cos 30o + j sin 30o, find z3 dan z5.
Solutions
i) z3 = (cos 30o + j sin 30o)3
= cos 3(30o)+ j sin 3(30o)
= cos 90o + j sin 90o
=j
5
ii) z = (cos 30o + j sin 30o)5
= cos 5(30o) + j sin 5(30o)
= cos 150o + j sin 150o
=
3 1
 j
2
2
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.17
If z = 2(cos 25o + j sin 25o). Calculate
2
i) z5
ii) z-3
iii) z 3
Solutions
i) z5 = [2(cos 25o + j sin 25o)]5
= 25[cos 5(25o) + j sin 5(25o)]
= 32(cos 125o + j sin 125o)
= 32(-0.5736 + 0.8192j)
= -18.355 + 26.214j
ii) z-3 = 2-3(cos -75o + j sin -75o)
=
1
(0.2588 – 0.9659j)
8
= 0.0324 – 0.1207j
2
3
2
3
iii) z = 2 [cos
2
2
(25o) + j sin (25o)]
3
3
= 1.587( cos 16.67o + j sin 16.67o)
= 1.52 + 0.455j
1.8
Finding roots of complex numbers
Theorem 4
If z = r(cos θ + j sin θ) then, the n root of z is
1
n
1
n
z = r (kos
  360o k
  360o k
 j sin
) if θ in degrees
n
n
or
1
n
1
n
z = r (kos
  2k
  2k
 j sin
) if θ in radians
n
n
for k = 0, 1, 2, ...., n – 1.
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.18
Find the cube roots of z = 8j
Solutions
z = 8j or in polar form z = 8(cos 90o + j sin 90o)
then
1
3
1
3
z = 8 (cos
90o  360o k
90o  360o k
 j sin
) for k = 0, 1, 2.
3
3
when k = 0
1
z 3 = 2(cos 30o + j sin 30o)
= 3 j
when k = 1
1
z 3 = 2(cos 150o + j sin 150o)
=  3 j
when k = 2
1
z 3 = 2(cos 270o + j sin 270o)
= -2j
Therefore the roots of z = 8j are 3  j ,  3  j and -2j.
1.9 Expansions for cosn and sinn in terms of cosines and sines of
multiples of θ
Theorem 4
If z = cos θ + j sin θ then
i) z n 
1
= 2cos nθ
zn
ii) z n 
1
= 2j sin nθ
zn
Theorem 5 (Binomial Theorem)
If n  N , then
(a  b) n  a n  nC1a n 1b nC2 a n 2 b 2  ... nCr a n r b r  ... nCn 1ab n 1  nCn b n
n
with Cr 
n!
.
r!( n  r )!
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Example 1.19
State sin5 θ in terms of sines
Solution
Using theorem 4, z = cos θ + j sin θ , then
1
z  = 2j sin θ
z
1
(2j sin θ)5 = [z  ]5 .
z
By using Binomial expansion, then
2
3
4
1
1
1
1 1
1
[ z  ]5 = z5 – 5z4   + 10z3   - 10z2   + 5z   -  
z
z
z
z z
z
5
1
1
1
=  z 5  5  - 5  z 3  3  +10  z  

z 

z 

z
Thus,
32 j sin5 θ = 2j sin 5θ – 5(2j sin 3θ) + 10(2j sin θ)
1
5
5
sin 5θ - sin 3θ + sin θ
16
16
8
sin5 θ =
1.10 Loci in the complex number
Definition 1.9 :
A locus in a complex plane is the set of points that have a specified
property. A locus of a point in a complex plane could be a straight
line, cirle, ellipse and etc.
Example 1.20
If z = a + bj, find the equation of the locus defined by:
i)
z2j
1
z 1
ii) z  (2  3 j)  2
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CHAPTER 1: COMPLEX NUMBERS
DQT 101
Solutions
i)
z2j
a  bj  2 j

1
z 1
a  bj  1
a  bj  2 j  a  bj  1
a  (b  2) j  a  1  bj
a 2  (b  2) 2  ( a  1) 2  b 2
a 2  b 2  4b  4  a 2  2a  1  b 2
2a  4b  3  0
1
The locus is a straight line with slope .
2
ii) z  (2  3 j)  a  bj  2  3 j)  2
a  2  (b  3) j  2
( a  2) 2  (b  3) 2  2
(a  2) 2  (b  3) 2  2 2
The locus is a circle of radius 2 and centre (2,3).
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