Chapter 1: complex number

Conjugates, Modulus (Norm) and Argument of Complex Number 
Conjugates of 
(
z
)
 
The conjugate of 
, 
z
is denoted by 
z
or a 
z

, which is same 
as 
z
but with the sign of the imaginary part flipped.
The 
geometric interpretation of the complex number and it 
is conjugate is shown in 
Figure 5
, 
z
is the reflection of 
z
in the real axis. 
Given complex number 
Z
i
a
b
 
 then the 
conjugate of 
z
denoted as 
( )
( )
z
I
=
Re
z
z
m
i
a
b
 

(
z
is 
pronounced “
z
bar”)
. 
Properties: 
 
 
1
2
1
2
1
1
1
2
1
2
2
2
2
2
a)
b)
,
,
c)
d)
,
e)
( ), and
( ),
,
.
0
2
2
1
1 0
2
z
z
z
z
z
z
z
z
z z
z
z
z
z
z
z
z
z
=
z
z
z
z
z
i
z
=
=
n
n
R
 
z
Re
e
a
b





 












  
  
Example 7
 
(
Use conjugate types of Complex Numbers
)
:
 
Express 
2
3
2
i
i


in the form 
a
b
i

. 
Solution: 






2
3
2
2
2
3
2
6
4
3
2
3
2
3
2
3
2
3
2
3
2
9
6
6
4
6
4
3
2
4
7
4
7
9
4
13
13
13
.
2
2
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i






 









 










Modulus (Norm) and Argument of CN 
For any given complex number 
z
i
a
b
 
one defines the absolute value or modulus to be 
2
2
z
b
a


. 
So the 
absolute value
(or 
modulus
) 
z
of a complex number 
z
i
a
b
 
 is the distance from the origin to the point 
 
z
a
b

 
in complex plane. As 
Figure 6
shows, Pythagoras’ Theorem 
gives 

 

( )
( )
2
2
2
2
R
z
z
Im
z
e
i
b
a
a
b
 




 
 
b
Z
Im

 
a
Z
Re

Im
Re
0
2
2
r
a
b
z



 
arg
z


b
i
a
z
 
Figure 6
2

3

1
2
3
4
Im
2

1

2
3
4
1
3

0
Re
i
i

z
a
b
i
 
z
a
b
i
 
Figure 5 

---

IIUM, Faculty of Engineering, 
 
Department Engineering in Science 
Engineering Mathematics I
 
Semester 1, 2021/2022
 
Chapter I:
 
Complex Numbers
 
Lecturer
Associate Professor Dr. Abdurahim Okhunov
14
Properties:
1
1
2
2
2
,
0
0
0
1 2
1
2
,
1
2
1
2
a)
b)
c)
d)
e)
f)
z
z
z
z
if an
z
z
z
z
d
z z
z
z
z
z
z
onl
z
z
z
z
y
z
if




 






 
Note:







 
 
2
2
2
2
2
2
2
2
2
2
,
1
i
i
i
i
i
i
i
x
y
x
y
x x
y
y x
y
x
xy
xy
y
x
y
z z
i
y
z
x
z
x
y
 












 





So, 
2
z z
z
 
or 
2
2
z
z z
x
y

 

. 
Definition 3 
(Norm)
: 
The distance from 
z
i
a
b
 
 to origin 
0
 in the norm of 
z
: 
2
2
z
a
b


. Norm and modulus are synonyms. 
Some Operations with 
(
z
)
: 
To use complex numbers, we must know how to add, subtract, multiply, and divide them. We 
start by defining equality, addition and multiplication. 
Definition 4 
(
Equality and Basic Operations
)
: 
a)
Equality:
Two complex number 
1
z
i
a
b
 
and 
2
z
i
c
d
 
are 
equal 
i
d
i
b
a
c

 
if 
and 
only 
if 
1
2
( )
(
)
=
R
Re
z
z
e
(
a
c

) 
and 
1
2
( )
(
)
=
I
Im
z
z
m
(
b
d

), 
b)
Addition:
If 
1
z
i
a
b
 
and 
2
z
i
c
d
 
then 

 





1
2
=
=
z
+
z
b
i
i
d
c
c
b
i
a
a
d






; 
Properties:




 
1
2
2
1
1
2
3
1
2
3
a)
=
b)
c)
, wher
0
0
0
0 0
e
,
+
+
+
+
=
+
+
z
z
z
z
z
z
z
+
z
z
=
z
z
=
c)
Multiplication:

 
 
 

a
c
ac
b
d
bd
+
+
+
i
i
i
+
b
a
c
d


=

 



1
2
i
i
=
i
i
i
=
a
c
ac a
c
ac a
b
d
d
b
bd
d
b
bd
i
i
=
c
ac
+
+
+
+
+
+
bd
+
+
a
+
b
c
i
d




 
Figure 7
y 
x 
2
z
 
1
2
z
z

 
2
z
 
1
z
 
0 

---

IIUM, Faculty of Engineering, 
 
Department Engineering in Science 
Engineering Mathematics I
 
Semester 1, 2021/2022
 
Chapter I:
 
Complex Numbers
 
Lecturer
Associate Professor Dr. Abdurahim Okhunov
15
Properties:

 

 
1
2
2
1
1
2
3
1
2
3
1
1
a)
=
b)
c)
, where
1 0
,
z
z
z
z
z
z
z
z
z
z
=
z
z
=
=





d)
Division:
If 
1
z
i
a
b
 
and 
2
z
i
c
d
 
then 
1
2
z
z
is 













 

1
2
.
2
2
2
2
2
2
a
a
c
ac
z
c
a
b
b
d
bd
b
d
ac
c
a
c
c
bd
b
d
d
d
d
+
+
-
+
+
+
+
c
c
c
-
+
+
d
d
i
i
i
i
i
i
i
z
d
i
+
c









 



Note: 
1
1 2
1 2
2
2
2
2
2
z
z z
z z
z
z z
z


, which gives us another way of deriving the formula for dividing 
complex numbers. 
Example 8 
(
Addition of Complex Numbers
)
:
 
Carry out each operation and express the answer in standard form: 

 


 

)
,
)
a
b
3
2
4
2
6
5
0
i
0
i
i
i

 
 
 
Solution: 
a)
We could apply the definition of addition directly, but it is easier to use complex number 
properties. 

 

2
6
2
6
2
i
2
3
i
3
i
i




 

Remove parentheses

 

3
2 6
8
i
i
2
1
i
8

 


 

Combine like terms 
b)

 

5
0
5
0
i
4
0
i
0
i
i
4


 


 

Remove parentheses

 

4 0
i
0
i
5
4
5

   
 
Combine like terms