ملزمة الرياضيات للصف السادس العلمي الاحيائي - التطبيقي

Mathematics
Sixth grade
2021 - 2022
By
Dr. Anas Dheyab Khalaf
August 2021

Complex numbers Dr. Anas D. Khalaf
2
Chapter one
Complex numbers

3
Contents
Preface 4
1. Introduction ………………………………....... 5
2. Mathematical Operations on complex numbers …………………..……. 8
2.1 The addition and subtraction operations ………………………..... 8
2.2 The multiplication operation ………………………….... 9
3. The conjugate of a complex number (complex conjugate) ……….… 11
4. The square root of complex number ……………………………. 22
5. Solving the equation in ℂ …………………….. 25
6. The cube roots number of integer one ……………….……... 41
7. Geometric Representation of Complex Numbers …………….….……. 39
8. Polar form of complex number ………….……….. 43
9. De Moivre’s Theorem ………………….. 46

4
preface
These lecture notes form the material to the elementary course on chapter one of
mathematics book: Complex numbers for students in sixth grade oh high school. The
purpose of these lecture notes is to introduce and analyse the complex numbers in
an easiest way, and to make it more clear we provide lots of examples, remarks,
tables, and figures, and explain the solutions in details, moreover we give several
exercises for the readers as a homework.
Finally, I wish to express my thank to high-school students for the efforts that
they do to understand mathematics well.
Anas Dheyab Khalaf
General Directorate of Education in Saladin, Ministry of education, Saladin,34011, Iraq
August 2021

5
1. Introduction
If we want to solve the equation 𝑥2
+ 1 = 0, then its solution will be:
𝑥2
+ 1 = 0 ⇒ 𝑥2
= −1,
⇒ 𝑥 = ∓√−1,
It is clearly that we cannot find a real number that its square is -1, therefore there is
an impressing need to introduce new type of numbers which is Complex numbers,
we will introduce the symbol (𝒊 = √−𝟏), which we call the imaginary part of the
complex number.
Thus, we have
𝑖2
= 𝑖. 𝑖 = √−1. √−1 = −1,
𝑖3
= 𝑖2
. 𝑖 = −1. 𝑖 = −𝑖,
𝑖4
= 𝑖2
. 𝑖2
= −1. −1 = 1,
𝑖5
= 𝑖3
. 𝑖2
= −𝑖. −1 = 𝑖,
Example 1.1: write the following expression in a simpler way;
1) 𝑖6
= 𝑖2
. 𝑖2
. 𝑖2
= −1. −1. −1 = −1, 𝒐𝒓 𝒊𝟔
= 𝒊𝟐
. 𝒊𝟒
= −𝟏(𝟏) = −𝟏
2) 𝑖8
= 𝑖2
. 𝑖2
. 𝑖2
. 𝑖2
= −1. −1. −1. −1 = 1, 𝒐𝒓 𝒊𝟖
= 𝒊𝟒
. 𝒊𝟒
= 𝟏. 𝟏 = 𝟏
3) 𝑖16
= (𝑖4
)4
= (1)4
= 1
4) 𝑖17
= (𝑖4
)4
. 𝑖 = (1)4
. 𝑖 = 𝑖
5) 𝑖58
= (𝑖4
)14
. 𝑖2
= (1)14
. (−1) = −1
6) 𝑖12𝑛+93
= (𝑖4
)3𝑛
. 𝑖93
= (1)3𝑛
. (𝑖4)32
. 𝑖 = 𝑖
7) 𝑖−13
= 𝑖−13
. 1 = 𝑖−13
. (𝑖4)4
= 𝑖16−13
= 𝑖3
= −𝑖
8) 𝑖−26
= 𝑖−26
. 1 = 𝑖−26
. (𝑖4)7
= 𝑖28−26
= 𝑖2
= −1
Remark 1.1: we can use (i) to describe the square root for any real number. In
general, we have
6=4+2
58=56+2
= 4(14)+2

6
√−𝑏2 = √𝑏2. √−1 = 𝑏𝑖, ∀𝑏 ≥ 0 .
Example 1.2: Use (i) to write the following square roots:
1) √−16 = √16. √−1 = 4𝑖
2) √−25 = √25. √−1 = 5𝑖
3) √−12 = √12. √−1 = 2√3 𝑖
 The (standard) formula of complex number
The complex number represents a compensation of real part and imaginary part, such
that
𝑪 = 𝒂 + 𝒃𝒊,
Where a stands for the real part of C, while b is the imaginary part of C. Moreover,
C can be written as (a,b).
Example 1.3: Use the standard formula of complex number to write the
following numbers:
a) −5 = −5 + 0𝑖
b) √−100 = √100. √−1 = 10𝑖 = 0 + 10𝑖
c) – 1 − √−3 = −1 − √3𝑖
d)
1+√−25
4
=
1
4
+
5
4
𝑖
e) 𝑖999
= (𝑖4
)249
. 𝑖2
. 𝑖 = 1. −1. 𝑖 = 0 − 𝑖
f) 𝑖4𝑛+1
= (𝑖4
)𝑛
. 𝑖 = 1. 𝑖 = 0 + 𝑖
Definition 1.1: we say that two complex numbers 𝑐1 = 𝑎1 + 𝑏1𝑖,
𝑐2 = 𝑎2 + 𝑏2𝑖 are equal if their real and imaginary parts are equal, that
is
𝒄𝟏 = 𝒄𝟐 ⇔ 𝒂𝟏 = 𝒂𝟐, 𝒃𝟏 = 𝒃𝟐

7
Example 1.4: Find the real values of x and y that satisfy the following
equation
a) 2𝑥 − 1 + 2𝑖 = 1 + (𝑦 + 1)𝑖
Sol/
2𝑥 − 1 = 1 ⟹ 2𝑥 = 1 + 1 ⟹ 2𝑥 = 2
∴ 𝑥 = 1
2 = 𝑦 + 1 ⟹ 𝑦 = 2 − 1
∴ 𝑦 = 1
b) (2𝑦 + 1) − (2𝑥 − 1)𝑖 = −8 + 3𝑖
Sol/
2𝑦 + 1 = −8 ⟹ 2𝑦 = −8 − 1 ⟹ 2𝑦 = −9
∴ 𝑦 =
−9
2
−2𝑥 + 1 = 3 ⟹ −2𝑥 = 3 − 1 ⟹ 2𝑥 = −2
∴ 𝑥 = −1
2. Mathematical Operations on complex numbers
2.1 The addition and subtraction operations
Definition 2.1: let 𝑐1 = 𝑎1 + 𝑏1𝑖, 𝑐2 = 𝑎2 + 𝑏2𝑖 be two complex
numbers, then
𝑐1 + 𝑐2 = (𝑎1 + 𝑎2)+(𝑏1 + 𝑏2)𝑖.
Example 2.1:
a) If we have 3+4√2𝑖 and 5-2√2𝑖, then
R R
I I
R
I
R I R I
R
I

8
(3+4√2𝑖 )+ (5-2√2𝑖 ) = (3+5)+(4√2 −2√2)𝑖
=8+2√2𝑖
b) If we have 3 and 2-5𝑖, then
(3+2) + (0-5)𝑖 = 5-5𝑖
c) 1 − 𝑖, 3𝑖
1-𝑖 + 3𝑖 = (1 + 0) + (−1 + 3)𝑖 = 1 + 2𝑖.
Remark 2.1: for the subtraction of complex number, we can use similar
method of addition.
Example 2.2: calculate the following
(7 − 13𝑖) − (9 + 4𝑖) =7 − 13𝑖 − 9 − 4𝑖 = (7 − 9) − (13 − 4)𝑖
=−2 − 17𝑖
Example 2.3: solve the following equation (2 − 4𝑖) + 𝑥 = −5 + 𝑖
𝑥 = −5 + 𝑖 − 2 + 4𝑖 ⟹ 𝑥 = (−5 − 2) + (1 + 4)𝑖
∴ 𝑥 = −7 + 5𝑖
2.2 The multiplication operation
Definition 2.2: let 𝑐1 = 𝑎1 + 𝑏1𝑖, 𝑐2 = 𝑎2 + 𝑏2𝑖 be two complex
numbers, then
𝑐1. 𝑐2 = (𝑎1 + 𝑏1𝑖)(𝑎2 + 𝑏2𝑖)
= 𝑎1𝑎2 + 𝑎1𝑏2𝑖 + 𝑏1𝑎2𝑖 + 𝑏1𝑏2𝑖2
= (𝑎1𝑎2 − 𝑏1𝑏2) + (𝑎1𝑏2 + 𝑏1𝑎2)𝑖,
where 𝑖2
= −1.

9
Example 2.4: find the multiplications of the following numbers
a) (2 − 3𝑖)(3 − 5𝑖)
(2 − 3𝑖)(3 − 5𝑖) = 6 − 10𝑖 − 9𝑖 + 15𝑖2
= 6 − 15 − 19𝑖
= −9 − 19𝑖
b) (2 + 𝑖)(3 + 6𝑖)
(2 + 𝑖)(3 + 6𝑖) = 6 + 12𝑖 + 3𝑖 + 6𝑖2
= 6 − 6 + 15𝑖
= 0 + 15𝑖 = 15𝑖
c) (3 + 4𝑖)2
(3 + 4𝑖)2
= 9 + 24𝑖 + 16𝑖2
= 9 − 16 + 24𝑖
= −7 + 24𝑖
d) −
5
2
(4 + 3𝑖)
−
5
2
(4 + 3𝑖) = −
5
2
4 −
5
2
3𝑖 = −10 −
15
2
𝑖
e) (1 + 𝑖)4
− (1 − 𝑖)4
(1 + 𝑖)4
− (1 − 𝑖)4
= ((1 + 𝑖)2
)2
− ((1 − 𝑖)2
)2
= (1 + 2𝑖 + 𝑖2
)2
− (1 − 2𝑖 + 𝑖2
)2
= (1 − 1 + 2𝑖)2
− (1 − 1 − 2𝑖)2

10
= (2𝑖)2
− (2𝑖)2
= 4𝑖2
− 4𝑖2
= −4 + 4 = 0 + 0𝑖
3. The conjugate of a complex number (complex conjugate)
Definition 3.1: let 𝑐 = 𝑎 + 𝑏𝑖 be a complex number, then the complex
conjugate of 𝑐 is defined by
𝑐̅ = 𝑎 − 𝑏𝑖, ∀𝑎, 𝑏 ∈ ℝ.
Remark 3.1: The properties of complex conjugate
 
 
1 2 1 2
1 2 1 2
2 2
1 1
2
2 2
1)
2)
3)
4)
5)( ) , 0.
c c c c
c c c c
c c
if c a bi c c a b
c c
c
c c
  
  

      
 

11
Example 3.1: if 1 2
1 , 3 2
c i c i
    , prove that
 
 
1 2 1 2
1 2 1 2
1 1
2 2
1)
2)
3)
c c c c
c c c c
c c
c c
  
  
 

 
 
Sol/
1-
 
     
1 2
1 2
1 2
. . .:
1 3 2 4 4
. . .: (1 ) (3 2 ) 1 3 2
4
. . . .
L H S c c
c c i i i i
R H S c c i i i i
i
L H S R H S

        
        
 
 

12
2-
 
   
   
1 2
2
1 2
2
. . .:
(1 )(3 2 ) 3 2 3 2
3 2 5 5
. . .:
(1 ) (3 2 ) (1 )(3 2 )
3 2 3 2 3 2 5
. . . .
L H S c c
i i i i i
i i i
R H S c c
i i i i
i i i i i
L H S R H S

      
      

      
        
 
3-
1
2
2 2
1
2
2 2
. . .:
3 2 3 2 1 3 3 2 2
1 1 1 1 1
1 5 1 5 1 5
2 2 2 2 2
. . .:
3 2 3 2 3 2 1 3 3 2 2
1 1 1 1 1
1
1 5 1 5
2 2 2
. . . .
c
L H S
c
i i i i i
i i i
i
i i
c
R H S
c
i i i i i i
i i i
i
i
i
L H S R H S
 
 
 
     
     
   
     
   
     

   
    
   
   
      
    
   


  
 

13
Remark 3.2: the multiplicative inverse of c is
1
c .
Example 3.2: find the multiplicative inverse of 2 2
c i
  and put it in
the standard formula of complex number.
Sol/
2 2
1 1 1 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 1 1
8 8 8 4 4
i i
c i i i
i i
i
 
   
   

    
Example 3.3: put the following numbers in the standard formula of
complex number:
1)
2−𝑖
3+4𝑖
=
2 − 𝑖
3 + 4𝑖
.
3 − 4𝑖
3 − 4𝑖
=
6 − 8𝑖 − 3𝑖 − 4
9 + 16
=
2 − 11𝑖
25
=
2
25
−
11
25
𝑖
2)
12+𝑖
𝑖
=
12 + 𝑖
𝑖
.
−𝑖
−𝑖
=
−12𝑖 − 𝑖2
1
=
1 − 12𝑖
1
= 1 − 12𝑖
3)
𝑖
2+3𝑖
=
𝑖
2 + 3𝑖
×
2 − 3𝑖
2 − 3𝑖
=
2𝑖 + 3
4 + 9
=
3 + 2𝑖
13
=
3
13
+
2
13
𝑖
4) (
3+𝑖
1+𝑖
)
3

14
= (
3 + 𝑖
1 + 𝑖
×
1 − 𝑖
1 − 𝑖
)
3
= (
3 − 3𝑖 + 𝑖 − 𝑖2
1 + 1
)
3
= (
4 − 2𝑖
2
)
3
= (
4
2
−
2𝑖
2
)
3
= (2 − 𝑖)3
= (2 − 𝑖)2
. (2 − 𝑖) = (4 − 4𝑖 − 1)(2 − 𝑖)
= (3 − 4𝑖)(2 − 𝑖) = 6 − 3𝑖 − 8𝑖 − 4 = 2 − 11𝑖
5)
2+3𝑖
1−𝑖
×
1+4𝑖
4+𝑖
=
2 + 8𝑖 + 3𝑖 − 12
4 + 𝑖 − 4𝑖 + 1
=
−10 + 11𝑖
5 − 3𝑖
=
−10 + 11𝑖
5 − 3𝑖
×
5 + 3𝑖
5 + 3𝑖
=
−50 − 30𝑖 + 55𝑖 − 33
52 + 32
==
−83 + 25𝑖
34
=
−83
34
+
25
34
𝑖
6) (1 + 𝑖)3
+ (1 − 𝑖)3
= (1 + 𝑖)2(1 + 𝑖) + (1 − 𝑖)2(1 − 𝑖)
= (1 + 2𝑖 − 1)(1 + 𝑖) + (1 − 2𝑖 + 1)(1 − 𝑖)
= (1 + 2𝑖 − 1 + 𝑖 − 2 − 𝑖) + (1 − 2𝑖 + 1 − 𝑖 − 2 − 𝑖)
= 2𝑖 − 2 − 2𝑖 − 2 = −4 + 0𝑖
Example 3.4: prove that:
1)
1
(2−𝑖)2 −
1
(2+𝑖)2 =
8
25
𝑖.

15
L.H.S:
1
(2−𝑖)2 −
1
(2+𝑖)2 =
1
4−4𝑖−1
−
1
4+4𝑖−1
=
1
3−4𝑖
×
3+4𝑖
3+4𝑖
−
1
3+4𝑖
×
3−4𝑖
3−4𝑖
=
3 + 4𝑖
9 + 16
−
3 − 4𝑖
9 + 16
=
3 + 4𝑖 − 3 + 4𝑖
25
=
8
25
𝑖
Hence, proved.
2) (1 − 𝑖)(1 − 𝑖2)(1 − 𝑖3) = 4
L.H.S: (1 − 𝑖)(1 − 𝑖2)(1 − 𝑖3) = (1 − 𝑖)(1 − (−1))(1 − 𝑖2
. 𝑖)
= 2(1 − 𝑖)(1 + 𝑖) = 2(1 + 𝑖 − 𝑖 − 𝑖2)
= 2(1 − (−1)) = 2.2 = 4 = R.H.S
Remark 3.3: the expression 𝑥2
+ 𝑦2
can be analyzed into product of two
factors as follows:
𝑥2
+ 𝑦2
= 𝑥2
− 𝑦2
𝑖2
= (𝑥 − 𝑦𝑖)(𝑥 + 𝑦𝑖)
Example 3.5: rewrite each of the following as product of two factors of
the form 𝑎 + 𝑏𝑖
a) 𝑥2
+ 𝑦2
= 𝑥2
− 𝑦2
𝑖2
= (𝑥 − 𝑦𝑖)(𝑥 + 𝑦𝑖)
b) 9𝑥2
+ 49𝑦2
= 9𝑥2
− 49𝑦2
𝑖2
= (3𝑥 − 7𝑦𝑖)(3𝑥 + 7𝑦𝑖)
c) 85 = 81 + 4 = 81 − 4𝑖2
= (9 − 2𝑖)(9 + 2𝑖)
d) 125 = 100 + 25 = 100 − 25𝑖2
= (10 − 5𝑖)(10 + 5𝑖)

16
Example 3.6: find the real values of 𝑥, 𝑦 that satisfy
1) 𝑦 + 5𝑖 = (2𝑥 + 𝑖)(𝑥 + 2𝑖).
Sol/
𝑦 + 5𝑖 = 2𝑥2
+ 4𝑥𝑖 + 𝑥𝑖 − 2 ⇒ 𝑦 + 5𝑖 = 2𝑥2
− 2 + 5𝑥𝑖
𝑦 = 2𝑥2
− 2 … (1) and 5 = 5𝑥 …. (2),
From equation (2), we have
𝑥 =
5
5
= 1, then using (1), we get
𝑦 = 2(1)2
− 2 = 0.
2) 𝟖𝒊 = (𝒙 + 𝟐𝒊)(𝒚 + 𝟐𝒊) + 𝟏
8𝑖 = 𝑥𝑦 + 2𝑥𝑖 + 2𝑦𝑖 − 4 + 1 ⇒ 8𝑖 = 𝑥𝑦 − 3 + 2(𝑥 + 𝑦)𝑖
⇒ 𝑥𝑦 − 3 = 0 … (1), (𝑥 + 𝑦) = 4 … (2)
From Eq. (1), we have
𝑥 =
3
𝑦
, and by Eq. (2), yields
[(
3
𝑦
+ 𝑦) = 4] × 𝑦 ⇒ 3 + 𝑦2
= 4𝑦 ⇒ 𝑦2
− 4𝑦 + 3 = 0
(𝑦 − 3)(𝑦 − 1) = 0
Then, either 𝑦 = 3 ⇒ 𝑥 =
3
3
= 1, or 𝑦 = 1 ⇒ 𝑥 =
3
1
= 3.

17
3) (
𝟏−𝒊
𝟏+𝒊
) + (𝒙 + 𝒚𝒊) = (𝟏 + 𝟐𝒊)𝟐
.
⇒ (
1 − 𝑖
1 + 𝑖
×
1 − 𝑖
1 − 𝑖
) + (𝑥 + 𝑦𝑖) = 1 + 4𝑖 − 4
⇒ (
1 − 𝑖 − 𝑖 + 𝑖2
12 + 𝑖2
) + (𝑥 + 𝑦𝑖) = −3 + 4𝑖
⇒ (
−2𝑖
2
) + (𝑥 + 𝑦𝑖) = −3 + 4𝑖 ⇒ 𝑥 + 𝑦𝑖 − 𝑖 = −3 + 4𝑖
⇒ 𝑥 = −3, and 𝑦 − 1 = 4 ⇒ 𝑦 = 5.
4)
𝟐−𝒊
𝟏+𝒊
𝒙 +
𝟑−𝒊
𝟐+𝒊
𝒚 =
𝟏
𝒊
.
(
2 − 𝑖
1 + 𝑖
×
1 − 𝑖
1 − 𝑖
) 𝑥 + (
3 − 𝑖
2 + 𝑖
×
2 − 𝑖
2 − 𝑖
) 𝑦 =
1
𝑖
×
−𝑖
−𝑖
⇒ (
2 − 2𝑖 − 𝑖 − 1
1 + 1
) 𝑥 + (
6 − 3𝑖 − 2𝑖 − 1
4 + 1
) 𝑦 =
−𝑖
1
⇒ (
1 − 3𝑖
2
) 𝑥 + (
5 − 5𝑖
5
) 𝑦 = −𝑖 ⇒ (
1 − 3𝑖
2
) 𝑥 + (1 − 𝑖)𝑦 = −𝑖
⇒
1
2
𝑥 −
3
2
𝑥𝑖 + 𝑦 − 𝑦𝑖 = −𝑖
Thus,
1
2
𝑥 + 𝑦 = 0 … (1) and −
3
2
𝑥 − 𝑦 = −1 … (2), it follows
1
2
𝑥 + 𝑦 = 0 … (1)
−
3
2
𝑥 − 𝑦 = −1 … (2)
−
2
2
𝑥 = −1 ⇒ −𝑥 = −1 ⇒ 𝑥 = 1, and using Eq. (1), we get

18
1
2
(1) + 𝑦 = 0 ⇒ 𝑦 = −
1
2
.
Example 3.7: let
𝑥−𝑦𝑖
1+5𝑖
,
3−2𝑖
𝑖
be conjugate numbers, then find the values of
𝑥, 𝑦 ∈ ℝ.
Sol/ since the numbers are conjugate to each other, then
(
𝑥 − 𝑦𝑖
1 + 5𝑖
)
̅̅̅̅̅̅̅̅̅̅̅
=
3 − 2𝑖
𝑖
⇒
𝑥 + 𝑦𝑖
1 − 5𝑖
=
3 − 2𝑖
𝑖
because we have (
𝑐1
𝑐2
̅) =
𝑐1
𝑐2
̅̅̅
̅.
⇒ (𝑥 + 𝑦𝑖)𝑖 = (1 − 5𝑖)(3 − 2𝑖) ⇒ 𝑥𝑖 − 𝑦 = 3 − 2𝑖 − 15𝑖 − 10
⇒ 𝑥𝑖 − 𝑦 = −7 − 17𝑖
⇒ 𝑥 = −17, 𝑦 = 7
Example 3.8: find the values of 𝑥, 𝑦 ∈ ℝ that satisfy
𝑦
1+𝑖
=
𝑥2+4
𝑥+2𝑖
.
Sol/
𝑦
1 + 𝑖
=
𝑥2
+ 4
𝑥 + 2𝑖
⇒
𝑦
1 + 𝑖
=
𝑥2
− 4𝑖2
𝑥 + 2𝑖
⇒
𝑦
1 + 𝑖
=
(𝑥 − 2𝑖)(𝑥 + 2𝑖)
𝑥 + 2𝑖
⇒
𝑦
1 + 𝑖
= 𝑥 − 2𝑖 ⇒ 𝑦 = (𝑥 − 2𝑖)(1 + 𝑖)
R R
I I

19
⇒ 𝑦 = 𝑥 + 𝑥𝑖 − 2𝑖 + 2 ⇒ 𝑦 = 𝑥 + 2, and
𝑥 − 2 = 0 ⇒ 𝑥 = 2, thus
𝑦 = 2 + 2 = 4.
R
R I
I
R

20
Exercises (1-1)
1) Rewrite the following numbers and expressions in a simplest way, use
the standard formula for complex number:
a) 𝑖5
b) 𝑖124
c) 𝑖−7
d) 𝑖−15
e) √−25 f) 𝑖(1 + 𝑖)
g) (2 + 3𝑖)2
+ (12 + 2𝑖) h) (1 + 𝑖)2
+ (1 − 𝑖)2
i)
1+𝑖
1−𝑖
j)
1+2𝑖
−2+𝑖
k)
3+4𝑖
3−4𝑖
2) Rewrite each of the following as product of two factors of the
form 𝑎 + 𝑏𝑖
a) 41 b) 29
3) Find the values of 𝑥, 𝑦 ∈ ℝ that satisfy (
1−𝑖
1+𝑖
) + (𝑥 + 𝑦𝑖) = (1 + 2𝑖)2
.
4) If
3+𝑖
2−𝑖
,
6
𝑥+𝑦𝑖
are complex conjugate, then find the values of 𝑥, 𝑦 ∈ ℝ .

21
4. The square root of complex number
To find the square-roots of complex number c we should first rewrite it in the
standard formula i.e. 𝑐 = 𝑎 + 𝑏𝑖, then suppose that (𝑥 + 𝑦𝑖)2
= 𝑎 + 𝑏𝑖, and then
solve the equation by taking the real (complex) part of left hand with the real
(complex) part of right hand.
Example 4.1: find the square roots of
1) 𝑐 = 8 + 6𝑖
Sol/
(𝑥 + 𝑦𝑖)2
= 8 + 6𝑖 ⇒ 𝑥2
+ 2𝑥𝑦𝑖 − 𝑦2
= 8 + 6𝑖
⇒ 𝑥2
− 𝑦2
= 8 … (1), and
2𝑥𝑦 = 6 ⇒ 𝑦 =
6
2𝑥
=
3
𝑥
… (2)
By (1) and (2), we have
𝑥2
− (
3
𝑥
)
2
= 8 ⇒ [𝑥2
−
9
𝑥2
= 8] × 𝑥2
𝑥4
− 9 = 8𝑥2
⇒ 𝑥4
− 8𝑥2
− 9 = 0 ⇒ (𝑥2
− 9)(𝑥2
+ 1) = 0, either
(𝑥2
− 9) = 0 ⇒ 𝑥2
= 9 ⇒ 𝑥 = ∓3 or (𝑥2
+ 1) = 0 ⇒ 𝑥2
= −1 we ignore this
value because 𝑥 is real number.

22
Now , by substituting the values of 𝑥 into Eq. (2), we get
𝑦 =
3
3
= 1 or 𝑦 =
3
−3
= −1
∴ 𝑐1 = 3 + 𝑖, 𝑐2 = −3 − 𝑖
Hence, the square roots of c are 3 + 𝑖 and −3 − 𝑖 .
2) 𝒄 = −𝟔𝒊
Sol/
(𝑥 + 𝑦𝑖)2
= 0 − 6𝑖 ⇒ 𝑥2
+ 2𝑥𝑦𝑖 − 𝑦2
= 0 − 6𝑖
⇒ 𝑥2
− 𝑦2
= 0 … (1), and
2𝑥𝑦 = −6 ⇒ 𝑦 =
−6
2𝑥
=
−3
𝑥
… (2)
𝑥2
− (
−3
𝑥
)
2
= 0 ⇒ [𝑥2
−
9
𝑥2
= 0] × 𝑥2
𝑥4
− 9 = 0 ⇒ (𝑥2
− 3)(𝑥2
+ 3) = 0, either (𝑥2
− 3) = 0 ⇒ 𝑥2
= 3 ⇒
𝑥 = ∓√3 or (𝑥2
+ 3) = 0 ⇒ 𝑥2
= −3 we ignore this value because 𝑥 is real
number.

23
𝑦 =
3
√3
= √3 or 𝑦 =
3
−√3
= −√3
∴ 𝑐1 = √3 + √3𝑖, 𝑐2 = −√3 − √3𝑖
Hence, the square roots of c are √3 + √3𝑖 and −√3 − √3𝑖.
3)
𝟒
𝟏−√𝟑𝒊
Sol/
4
1 − √3𝑖
×
1 + √3𝑖
1 + √3𝑖
=
4 + 4√3𝑖
12 + (√3)
2 =
4 + 4√3𝑖
4
= 1 + √3𝑖
(𝑥 + 𝑦𝑖)2
= 1 + √3𝑖 ⇒ 𝑥2
+ 2𝑥𝑦𝑖 − 𝑦2
= 1 + √3𝑖
⇒ 𝑥2
− 𝑦2
= 1 … (1), and
2𝑥𝑦 = √3 ⇒ 𝑦 =
√3
2𝑥
… (2)
𝑥2
− (
√3
2𝑥
)
2
= 1 ⇒ [𝑥2
−
3
4𝑥2
= 1] × 4𝑥2

24
4𝑥4
− 3 = 4𝑥2
⇒ 4𝑥4
− 4𝑥2
− 3 = 0 ⇒ (2𝑥2
− 3)(2𝑥2
+ 1) = 0, either
(2𝑥2
− 3) = 0 ⇒ 2𝑥2
= 3 ⇒ 𝑥 = ∓√
3
2
or (𝑥2
+ 1) = 0 ⇒ 𝑥2
= −1 we ignore
this value because 𝑥 is real number.
𝑦 =
√3
√2 . √2(−
√3
√2
)
=
1
√2
or 𝑦 =
√3
√2 . √2(
−√3
√2
)
=
−1
√2
Hence, the square roots of c are ±(√
3
2
+
1
√2
𝑖).
5. Solving the equation in ℂ
We know that the equation 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, where 𝑎 ≠ 0; 𝑎, 𝑏, 𝑐 ∈ ℝ, has two
solutions that can be found by using the Law: 𝑥 =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
, and we also know
that if the charecrstic expression 𝑏2
− 4𝑎𝑐 is negative then the equation 𝑎𝑥2
+
𝑏𝑥 + 𝑐 = 0 dose not have real solutions, but it has two solutions in ℂ.
Example 5.1: solve the following equations in ℂ:
1) 𝑥2
+ 4𝑥 + 5 = 0
Sol/

25
𝑥 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−4 ± √42 − 4(1)(5)
2
⇒ 𝑥 =
−4±√16−20
2
=
−4±√−4
2
=
−4±2𝑖
2
= −2 ± 𝑖
∴ 𝑆 = {−2 − 𝑖, −2 + 𝑖}
2) 𝑧2
= −12
Sol/
𝑧 = ±√−12 = ±√3 × 4𝑖 = ±2√3𝑖
∴ 𝑆 = {−2√3𝑖, 2√3𝑖}, which are conjugate roots.
3) 𝑧2
− 3𝑧 + 3 + 𝑖 = 0
𝑧 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
3 ± √32 − 4(1)(3 + 𝑖)
2
=
3 ± √9 − 12 − 4𝑖
2
=
3 ± √−3 − 4𝑖
2
to solve this equation, we need to find the roots of −3 − 4𝑖 as follows
(𝑥 + 𝑦𝑖)2
= −3 − 4𝑖 ⇒ 𝑥2
+ 2𝑥𝑦𝑖 − 𝑦2
= −3 − 4𝑖
⇒ 𝑥2
− 𝑦2
= −3 … (1)
⇒ 𝑦 =
−4
2𝑥
=
−2
𝑥
… . . (2)

26
By (1) and (2), we obtain
𝑥2
− (
−2
𝑥
)
2
= −3 ⇒ [𝑥2
−
4
𝑥2
= −3] × 𝑥2
⇒ 𝑥4
− 4 = −3𝑥2
⇒ 𝑥4
+ 3𝑥2
− 4 = 0
⇒ (𝑥2
+ 4)(𝑥2
− 1) = 0
𝑥2
= 1 ⇒ 𝑥 = ±1, then from Eq. (2), we have
𝑦 =
−2
±1
= ±2
Now, going back to the roots of −3 − 4𝑖, we have found that
√−3 − 4𝑖 = ±(1 − 2𝑖)
Thus,
𝑧 =
3 ± (1 − 2𝑖)
2
Neither 𝑧 =
3
2
−
1+2𝑖
2
= 1 + 𝑖 or 𝑧 =
3
2
+
1−2𝑖
2
= 2 − 𝑖.
The roots are not conjugate.
4 ) 𝑧2
+ 2𝑧 + 𝑖(2 − 𝑖) = 0
𝑧2
+ 2𝑧 + 𝑖(2 − 𝑖) = 0 ⇒ 𝑧2
+ 2𝑧 + 2𝑖 + 1 = 0

27
𝑧 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−2 ± √22 − 4(1)(2𝑖 + 1)
2
=
−2 ± √4 − 8𝑖 − 4
2
=
−2 ± √−8𝑖
2
To solve this equation, we need to find the roots of −8𝑖 as follows
(𝑥 + 𝑦𝑖)2
= −8𝑖 ⇒ 𝑥2
+ 2𝑥𝑦𝑖 − 𝑦2
= −8𝑖
⇒ 𝑥2
− 𝑦2
= 0 … (1)
⇒ 𝑦 =
−8
2𝑥
=
−4
𝑥
… . . (2)
Using (1) and (2), we obtain
𝑥2
− (
−4
𝑥
)
2
= 0 ⇒ [𝑥2
−
16
𝑥2
= 0] × 𝑥2
⇒ 𝑥4
− 16 = 0
⇒ (𝑥2
− 4)(𝑥2
+ 4) = 0
Hence, 𝑥2
− 4 = 0 ⇒ 𝑥 = ±2, then from Eq. (2), we have
𝑦 =
−4
±2
= ±2
Now, going back to the roots of −8𝑖, we have found that
√−8𝑖 = ±(2 − 2𝑖)

28
Thus,
𝑧 =
−2 ± (2 − 2𝑖)
2
Neither 𝑧 =
−2
2
−
2+2𝑖
2
= −2 + 𝑖 or 𝑧 =
−2
2
+
2−2𝑖
2
= 0 − 𝑖.
Consequently, the roots are not conjugate.
Remark 5.1: we can reformulate the quadratic equation from its roots as follows:
 Write the roots in standard form of complex number.
 Find the sum of roots (𝑐1 + 𝑐2).
 Find the multiplication of the roots (𝑐1 × 𝑐2).
 Apply the following formula 𝑥2
− (𝑐1 + 𝑐2)𝑥 + (𝑐1 × 𝑐2) = 0.
Note that if the factors of an equation are real and one of them is known, then the
other one is its complex conjugate.
Example 5.2: find the quadratic equations that their roots are:
 ±(2 + 2𝑖)
𝑐1 + 𝑐2 = (2 + 2𝑖) + (−2 − 2𝑖) = 2 − 2 + (2 − 2)𝑖 = 0
𝑐1 × 𝑐2 = (2 + 2𝑖) × (−2 − 2𝑖) = −4 − 4𝑖 − 4𝑖 + 4 = −8𝑖
Then the quadratic equation can be given by
𝑥2
− −8𝑖 = 0

29
 𝑀 =
3−𝑖
1+𝑖
, 𝐿 = (3 − 2𝑖)2
𝑀 =
3 − 𝑖
1 + 𝑖
=
3 − 𝑖
1 + 𝑖
×
1 − 𝑖
1 − 𝑖
=
3 − 3𝑖 − 𝑖 − 1
1 + 1
=
2 − 4𝑖
2
= 1 − 2𝑖
𝐿 = (3 − 2𝑖)2
= 9 − 12𝑖 − 4 = 5 − 12𝑖
𝑀 + 𝐿 = (1 − 2𝑖) + (5 − 12𝑖) = (1 + 5) + (−2 − 12)𝑖 = 6 − 14𝑖
𝑀 × 𝐿 = (1 − 2𝑖)(5 − 12𝑖) = 5 − 12𝑖 − 10𝑖 − 24 = −19 − 22𝑖
Then the quadratic equation is
𝑥2
− (6 − 14𝑖)𝑥 + (−19 − 22𝑖) = 0
Example 5.3: write the quadratic equation that has real factors and one of its roots
is 1) 𝐴 = 3 − 4𝑖.
Sol/ since the quadratic equation has real factors and one of its roots is 3 − 4𝑖, then
by remark (5.1 ), the other root is the complex conjugate of 3 − 4𝑖 namely 3 +
4𝑖 ≔ 𝐵.
𝐴 + 𝐵 = (3 − 4𝑖) + (3 + 4𝑖) = 6 + 0𝑖 = 6
𝐴 × 𝐵 = (3 − 4𝑖)(3 + 4𝑖) = 9 + 12𝑖 − 12𝑖 + 16 = 25
𝑥2
− 6𝑥 + 25 = 0

30
2 ) 𝑨 =
√𝟐+𝟑𝒊
𝟒
Sol/ since the quadratic equation has real factors and one of its roots is 𝐴 =
√2+3𝑖
4
,
then by remark (5.1), the other root is the complex conjugate of 𝐴 namely
√𝟐−𝟑𝒊
𝟒
≔
𝐵.
𝐴 + 𝐵 = (
√2
4
+
3
4
𝑖) + (
√2
4
−
3
4
𝑖) = (
√2
4
+
√2
4
) + (
3
4
𝑖 −
3
4
𝑖)
=
√2
2
+ 0𝑖 =
1
√2
𝐴 × 𝐵 = (
√2
4
+
3
4
𝑖) (
√2
4
−
3
4
𝑖) =
2
16
−
3√2
16
𝑖 +
3√2
16
𝑖 +
9
16
=
11
16
𝑥2
−
1
√2
𝑥 +
11
16
= 0
Example 5.4: Let 3 + 𝑖 be one of the roots of the equation 𝑥2
− 𝑎𝑥 + (5 +
5𝑖) = 0, then what is the value of 𝑎 ?, and what is the other root ?
Sol/ suppose that the other root is denoted by 𝐿, then
𝐿 × (3 + 𝑖) = 5 + 5𝑖 ⇒ 𝐿 =
5 + 5𝑖
3 + 𝑖
⇒ 𝐿 =
5 + 5𝑖
3 + 𝑖
×
3 − 𝑖
3 − 𝑖

31
⇒ 𝐿 =
15 − 5𝑖 + 15𝑖 + 5
9 + 1
=
20 + 10𝑖
10
= 2 + 𝑖
And the summation of the roots is
⇒ 𝑎 = 𝐿 + 𝑀
⇒ 𝒂 = (2 + 𝑖) + (3 + 𝑖) = 5 + 2𝑖
⇒ 𝑎 = 5 + 2𝑖
6. The cube roots number of integer one
Let 𝑧 be one of the cube roots of the integer one, such that
𝑧3
= 1 ⇒ 𝑧3
− 1 = 0 ⇒ (𝑧 − 1)(𝑧2
+ 𝑧 + 1) = 0
Neither 𝑧 = 1 or 𝑧2
+ 𝑧 + 1 = 0, this equation can be solved by the quadratic
formula 𝑧 =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
as follows:
𝑧 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−1 ± √12 − 4(1)(1)
2(1)
=
−1 ± √3𝑖
2
=
−1
2
±
√3
2
𝑖
Then the cube roots of the integer one are: 1, 𝜔 ≔
−1
2
+
√3
2
𝑖, 𝜔2
≔
−1
2
−
√3
2
𝑖. The
sum of these roots is

32
1 + (
−1
2
+
√3
2
𝑖) + (
−1
2
−
√3
2
𝑖) = 0
That is 1 + 𝜔 + 𝜔2
= 0, which gives us the following relations:
 1 + 𝜔 = − 𝜔2
 1 + 𝜔2
= −𝜔
 𝜔 + 𝜔2
= −1
 1 = − 𝜔 − 𝜔2
 𝜔2
= −1 − 𝜔
 𝜔 = −1 − 𝜔2
Moreover, the multiplication of the roots is 1 . 𝜔 . 𝜔2
= 1 ⇒ 𝜔3
= 1, from this
we obtain
𝜔 =
1
𝜔2
, 𝜔2
=
1
𝜔
,
𝜔4
= 𝜔3
. 𝜔 = 1. 𝜔 = 𝜔,
𝜔6
= 𝜔3
. 𝜔3
= 1. 1 = 1,
𝜔−5
=
1
𝜔5
=
1
𝜔3. 𝜔2
=
1
𝜔2
= 𝜔,
𝜔−5
=
1
𝜔5
=
1
𝜔3. 𝜔2
=
1
𝜔2
= 𝜔,
𝜔−8
=
1
𝜔8
=
1
𝜔6. 𝜔2
=
1
𝜔2
= 𝜔.

33
Example 6.1: find the value of the following 𝜔33
, 𝜔25
, 𝜔−58
, 𝜔−325
.
Sol/
𝜔33
= (𝜔3
)10
. 𝜔3
= (1)10
. 13
= 1
𝜔25
= (𝜔3
)8
. 𝜔 = 𝜔
𝜔−58
=
1
𝜔58
=
1
(𝜔3)19. 𝜔
=
1
𝜔
= 𝜔2
𝜔−325
=
1
𝜔325
=
1
(𝜔3)108.𝜔
=
1
𝜔
= 𝜔2
.
Example 6.2: write the following expressions in a simplest way:
1)
1
(1+ 𝜔−32)12
1
(1 + 𝜔−32)12
=
1
(1 + ( 𝜔3)−11. 𝜔)12
=
1
(−𝜔2)12
=
1
(−𝜔2)12
= −
1
(𝜔3)8
= −1
2) (1 + 𝜔2
)−4
(1 + 𝜔2
)−4
= (−𝜔)−4
=
1
(−𝜔)4
=
1
𝜔3.𝜔
= 𝜔2
.
3) 𝜔9𝑛+5
= (𝜔3)3𝑛
. 𝜔5
= (1)3𝑛
. 𝜔3
. 𝜔2
= 𝜔2

34
4) (3𝜔9𝑛
+
5
𝜔5
+
4
𝜔4)
6
(3𝜔9𝑛
+
5
𝜔3
+
4
𝜔4
)
6
= (3𝜔9𝑛
+
5𝜔4
+ 4𝜔5
𝜔5𝜔4
)
6
= (3(𝜔3
)3𝑛
+
5𝜔4
+ 4𝜔5
(𝜔3)3
)
6
= (3 + 5𝜔3
. 𝜔 + 4𝜔3
. 𝜔2)6
= (3(−𝜔 − 𝜔2) + 5𝜔 + 4 𝜔2)6
= (−3𝜔 − 3𝜔2
+ 5𝜔 + 4 𝜔2)6
= (2𝜔 − 3𝜔2
+ 5𝜔 + 4 𝜔2)6
Example 6.3: Prove that:
a) (5 + 3𝜔 + 3𝜔2)2
= −4(2 + 𝜔 + 2𝜔2)3
= 4
Sol/
L.H.S: (5 + 3𝜔 + 3𝜔2)2
= (5 + 3(𝜔 + 𝜔2
))2
= (5 + 3(−1))2
= 22
= 4
R.H.S: −4(2 + 𝜔 + 2𝜔2)3
= −4(2(1 + 𝜔2) + 𝜔)3
= −4(2(−𝜔) + 𝜔)3
= −4(−2𝜔 + 𝜔)3
= −4(−𝜔)3
= −4 . −1(𝜔)3
= 4
b) (
1
2+𝜔
−
1
2+𝜔2)
2
= −
1
3
L.H.S: (
1
2+𝜔
−
1
2+𝜔2)
2
= (
(2+𝜔2)−(2+𝜔)
(2+𝜔2)(2+𝜔)
)
2
= (
2−2+𝜔2−𝜔
4+2𝜔+2𝜔2+𝜔3)
2
= (
𝜔2−𝜔
4+2(𝜔+𝜔2)+1
)
2
= (
−
1
2
−
√3
2
𝑖+
1
2
−
√3
2
𝑖
4−2+1
)
2
= (
−2√3
2
𝑖
3
)
2
= (
−√3𝑖
3
)
2
=
−3
9
= −
1
3

35
c)
𝜔14+𝜔7−1
𝜔10+𝜔5−2
=
2
3
L.H.S:
𝜔14+𝜔7−1
𝜔10+𝜔5−2
=
(𝜔3)4. 𝜔2+(𝜔3)2. 𝜔−1
(𝜔3)3. 𝜔+ 𝜔3. 𝜔2−2
=
𝜔2+ 𝜔−1
𝜔+ 𝜔2−2
=
−1−1
−1−2
=
−2
−3
=
2
3
= L.H.S
d) (1 + 𝜔2
)3
+ (1 + 𝜔)3
= −2
L.H.S: (−𝜔)3
+ (−𝜔2
)3
= −𝜔3
−𝜔6
= −1 − (𝜔3
)2
= −1 − 1 = −2 = R.H.S
Example 6.4: write the quadratic equation that its roots are:
1) 𝑨 = 𝟏 − 𝒊𝝎𝟐
, 𝑩 = 𝟏 − 𝒊𝝎
𝐴 + 𝐵 = (1 − 𝑖𝜔2) + (1 − 𝑖𝜔) = 2 − (𝜔 + 𝜔2)𝑖 = 2 + 𝑖
𝐴 . 𝐵 = (1 − 𝑖𝜔2)(1 − 𝑖𝜔) = 1 − 𝑖𝜔 −𝑖𝜔2
+ 𝜔3
𝑖2
= 1 − 𝑖𝜔 −𝑖𝜔2
− 1
= −(𝜔 + 𝜔2)𝑖 = 𝑖
𝑥2
− (2 + 𝑖)𝑥 + 𝑖 = 0
2) 𝑨 =
𝟐
𝟏−𝝎
, 𝑩 =
𝟐
𝟏−𝝎𝟐
𝐴 + 𝐵 =
2
1 − 𝜔
+
2
1 − 𝜔2
=
2(1 − 𝜔2) + 2(1 − 𝜔)
1 − 𝜔2 − 𝜔 + 𝜔3
=
2−2𝜔2
+ 2 − 2𝜔
1 − 𝜔2 − 𝜔 + 1
=
4−2(𝜔2
+ 𝜔)
2 − (𝜔2 + 𝜔)
=
6
2 − (−1)
=
6
3
= 2

36
𝐴 . 𝐵 = (
2
1 − 𝜔
) (
2
1 − 𝜔2
) =
4
1 − 𝜔2 − 𝜔 + 𝜔3
=
4
3
Then, the quadratic equation is
𝑥2
− 2𝑥 +
4
3
= 0
3) 𝑨 =
𝝎
𝟐−𝝎𝟐
, 𝑩 =
𝝎𝟐
𝟐−𝝎
𝐴 + 𝐵 =
𝜔
2 − 𝜔2
+
𝜔2
2 − 𝜔
=
𝜔(2 − 𝜔) + 𝜔2
(2 − 𝜔)
(2 − 𝜔2)(2 − 𝜔)
=
2𝜔 − 𝜔2
+ 2𝜔2
− 𝜔4
4 − 2𝜔 − 2𝜔2 + 𝜔3
=
𝜔2
+ 2𝜔 − 𝜔3
. 𝜔
5 − 2(𝜔2 + 𝜔)
=
𝜔2
+ 2𝜔 − 𝜔
5 − 2(−1)
=
𝜔2
+ 𝜔
7
= −
1
7
𝐴 . 𝐵 = (
𝜔
2 − 𝜔2
) (
𝜔2
2 − 𝜔
) =
𝜔3
5 − 2(𝜔2 + 𝜔)
=
1
7
Then, the quadratic equation is
𝑥2
+
1
7
𝑥 +
1
7
= 0
Example 6.5: let 𝑧2
+ 𝑧 + 1 = 0, then find the value of
1+3𝑧10+3𝑧11
1−3𝑧7−3𝑧8
.
Sol/ first, we need to find the value of 𝑧 by the quadratic formula
𝑎 = 1, b = 1 , c = 1,

37
𝑧 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−1 ± √12 − 4(1)(1)
2(1)
=
−1 ± √−3
2
=
−1 ± √3𝑖
2
Then 𝑧 =
−1+√3𝑖
2
= 𝜔 or 𝑧 =
−1−√3𝑖
2
= 𝜔2
.
Now, let 𝑧 = 𝜔, we obtain
1+3𝑧10+3𝑧11
1−3𝑧7−3𝑧8 =
1+3𝜔10+3𝜔11
1−3𝜔7−3𝜔8
=
1+3(𝜔3)3 . 𝜔 +3(𝜔3)3 . 𝜔2
1−3(𝜔3)2 . 𝜔−3(𝜔3)2 . 𝜔2 =
1+3 𝜔 +3 𝜔2
1−3𝜔−3𝜔2
=
1+3 (𝜔 +𝜔2)
1−3(𝜔+𝜔2)
=
−2
4
= −
1
2
Now, let 𝑧 = 𝜔2
, we obtain
1 + 3z10
+ 3z11
1 − 3z7 − 3z8
=
1 + 3(ω2
)10
+ 3(ω2
)11
1 − 3(ω2)7 − 3(ω2)8
=
1+3(ω3
)
6
. ω2
+3(ω3
)
7
. ω
1−3(ω3
)
4
. ω2−3(ω3
)
5
. ω
=
1+3 (ω2+ω )
1−3(ω2+ω)
=
−2
4
= −
1
2

38
7. Geometric Representation of Complex Numbers
In the beginning of this chapter, we have said that a complex number 𝑧 can
be represented as an ordered bear of real numbers (𝑥, 𝑦) and denoted by 𝑧(𝑥, 𝑦).
Example 7.1: represent the following
Numbers geometrically
1)(3 + 4 𝑖) + (5 + 2𝑖)
(3 + 4 𝑖) + (5 + 2𝑖)
= (3 + 5) + (4 + 2)𝑖 = 8 + 6𝑖
2)(6 − 2 𝑖) − (2 − 5𝑖)
(6 − 2 𝑖) − (2 − 5𝑖)
= (6 − 2 𝑖) + (−2 + 5𝑖)
= 4 + 3𝑖
Figure 1: Geometric Representation (𝟑 + 𝟒 𝒊) + (𝟓 + 𝟐𝒊)
Figure 2: Geometric Representation (𝟔 − 𝟐𝒊) − (𝟐 − 𝟓𝒊)

39
Example 7.2: write the additive counterpart of the following numbers, then
represent the numbers and their additive
Counterparts on Argand plane.
a) 𝑧1 = 2 + 3𝑖
𝑧1 = 2 + 3𝑖 ⇒ 𝑝1(2, 3)
−𝑧1 = −2 − 3𝑖 ⇒ 𝑝2(−2, − 3)
b) 𝑧1 = −1 + 3𝑖
𝑧1 = −1 + 3𝑖 ⇒ 𝑝1(−1, 3)
−𝑧1 = 1 − 3𝑖 ⇒ 𝑝2(1, − 3)
Example 7.3: give the complex conjugate for the following numbers then provide
the geometric representation for each of them:
a) 𝑧1 = 5 + 3𝑖
𝑧1 = 5 + 3𝑖 ⇒ 𝑝1(5, 3)
𝑧̅1 = 5 − 3𝑖 ⇒ 𝑝2(5, −3)
Figure 3: the geometric Representation for example a
Figure 4: the geometric Representation for example b
Figure 5: the geometric Representation for example a

40
b ) 𝑧 = −2𝑖
𝑧 = 0 − 2𝑖 ⇒ 𝑝1(0, −2)
𝑧̅ = 0 + 2𝑖 ⇒ 𝑝2(0, 2)
a) Example 7.4: let 𝑧1 = 4 − 2𝑖 and 𝑧2 = 1 + 2𝑖 then represent the following on
Argand:
a) −3𝑧2 = −3(1 + 2𝑖) = −3 − 6𝑖 ⇒ 𝑝1(−3, −6)
b) 2𝑧1 = 2(4 − 2𝑖) = 8 − 4𝑖 ⇒ 𝑝2(8, −4)
c) 𝑧1 − 𝑧2 = (4 − 2𝑖) − (1 + 2𝑖) = 3 − 4𝑖 ⇒ 𝑝3(3, −4)
d) 𝑧1 + 𝑧2 = (4 − 2𝑖) + (1 + 2𝑖) = 5 + 0𝑖 ⇒ 𝑝4(5, 0)
Figure 6: the geometric Representation for example b
(a) (b) (c) (d)
Figure 6: the geometric Representation for example

41
Exercises (2-2)
1) Find the value of the following:
𝜔14
, 𝜔64
, 𝜔−6
, 𝜔−8
2) Write the following expressions in a simplest way: 𝜔12𝑛+5
3) Prove that:
a) 𝜔7
+ 𝜔5
+ 1 = 0
b) (1 −
2
𝜔2
+ 𝜔2
) (1 + 𝜔 −
5
𝜔
) = 18
4) Write the quadratic equation that its roots are:
a) 𝐴 = 1 + 𝜔2
, 𝐵 = 1 + 𝜔 𝒃)
3𝑖
𝜔2
,
−3𝜔2
𝑖
5) Write the additive counterpart of the following numbers, then represent the
numbers and their additive Counterparts on Argand plane:
a) 𝑧 = 𝑖 b) 𝑧 = 3 − 2𝑖
6) Give the complex conjugate for the following numbers then provide the
geometric representation for each of them:
a) 𝑧1 = −3 + 2𝑖 b ) 𝑧2 = 1 − 𝑖
7) Let 𝑧 = 4 + 2𝑖 then represent the following on Argand:
a) 𝑧 b) 𝑧̅ c) – 𝑧

42
8. Polar form of complex number
If we have a complex number 𝑧 and we geometrically represent it by the point
𝑝(𝑥, 𝑦), then (𝑟, 𝜃) are the polar coordinates for 𝑝 such that 𝑂 represent the polar
(Origin pojnt) and 𝑂𝑋
⃑⃑⃑⃑⃑ stands for the primary side. See figure 8.
Definition 8.1: let 𝑟 be a non negative real number, then we say that 𝑟 is a modulus
for the complex number 𝑧 (Mod z ) and denoted by ‖𝑧‖. Mod z is defined by
𝑟 = ‖𝑧‖ = √𝑥2 + 𝑦2
The measure of an angle 𝜃 ( known as argument of a complex number arg(𝑧)) that
made between the victor 𝑂𝑝
⃑⃑⃑⃑⃑ and the right part of 𝑋-axis can be given by
cos 𝜃 =
𝑥
𝑟
=
𝑥
‖𝑧‖
⇒ ℝ(𝑧) = 𝑥 = 𝑟 cos 𝜃
sin 𝜃 =
𝑦
𝑟
=
𝑦
‖𝑧‖
⇒ 𝐼(𝑧) = 𝑦 = 𝑟 sin 𝜃
Remark 8.1: if 𝜃 is argument of a complex number 𝑧, then 𝜃 + 2𝑛𝜋, 𝑛 ∈ ℤ, are also
is argument of the same complex number 𝑧. Whereas, if 𝜃 ∈ [0,2𝜋) then we call 𝜃
the principle value of complex number.
Figure 8: Polar form of complex number

43
Table 1: the special angels
Example 8.1: let 𝑧 = 1 + √3𝑖, then find the modulus and principle value of 𝑧.
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √12 + (√3)2 = √12 + 3 = √4 = 2
cos 𝜃 =
𝑥
‖𝑧‖
=
1
2
⇒ cos 𝜃 has positive value
sin 𝜃 =
𝑦
‖𝑧‖
=
√3
2
⇒ sin 𝜃 has positive value too, then 𝜃 is located in first quarter.
∴ arg(𝑧) =
𝜋
3
.
Example 8.2: if 𝑧 = −1 − 𝑖, then find the modulus and principle value of 𝑧.
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √−12 + (−1)2 = √1 + 1 = √2
cos 𝜃 =
𝑥
‖𝑧‖
=
−1
√2
⇒ cos 𝜃 has negative value
sin 𝜃 =
𝑦
‖𝑧‖
=
−1
√2
⇒ sin 𝜃 has negative value too, then 𝜃 is located in third
quarter.

44
∴ arg(𝑧) = 𝜋 +
𝜋
4
=
5𝜋
4
.
Example 8.3: if 𝑧 = −1 − √3𝑖, then find the modulus and principle value of 𝑧.
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √−12 + (−√3)2 = √1 + 3 = 2
cos 𝜃 =
𝑥
‖𝑧‖
=
−1
2
sin 𝜃 =
𝑦
‖𝑧‖
=
−√3
2
⇒ sin 𝜃 has negative value too, then 𝜃 is located in third
quarter.
∴ arg(𝑧) = 𝜋 +
𝜋
3
=
4𝜋
3
.
Remark 8.2:
1) The modulus of the complex number 𝑧 = 0 is unknown because the zero victor
has no value.
2) by the polar form of the complex number, we have 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃,
then we obtain 𝑧 = 𝑟 cos 𝜃 + 𝑖𝑟 sin 𝜃 = 𝑟(cos 𝜃 + 𝑖sin 𝜃) .
Example 8.4: find the polar form for the following:
a) 𝑧 = −2 + 2𝑖
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √−22 + (2)2 = √8 = 2√2
cos 𝜃 =
𝑥
‖𝑧‖
=
−1
√2
sin 𝜃 =
𝑦
‖𝑧‖
=
1
√2
⇒ sin 𝜃 has positive value, then 𝜃 is located in second quarter.

45
∴ arg(𝑧) = 𝜋 −
𝜋
4
=
3𝜋
4
.
𝑧 = 𝑟(cos 𝜃 + 𝑖sin 𝜃) = 2√2(cos
3𝜋
4
+ 𝑖 sin
3𝜋
4
)
b) 7i
7𝑖 = 7(𝑖) = 7 (cos
𝜋
2
+ 𝑖sin
𝜋
2
)
9. De Moivre’s Theorem
We know that the complex numbers 𝑧1 and 𝑧2 can be written in a polar form
as 𝑧1 = cos 𝜃 + 𝑖sin 𝜃 , 𝑧2 = cos 𝜗 + 𝑖sin 𝜗, then 𝑧1. 𝑧2 is
𝑧1. 𝑧2 = (cos 𝜃 + 𝑖sin 𝜃) (cos 𝜗 + 𝑖sin 𝜗)
= cos 𝜃cos 𝜗 + 𝑖 cos 𝜃 sin 𝜗 + 𝑖 cos 𝜗 sin 𝜃 − sin 𝜃 sin 𝜗
= (cos 𝜃 cos 𝜗 − sin 𝜃sin 𝜗) + 𝑖(cos 𝜃 sin 𝜗 + 𝑖 cos 𝜗 sin 𝜃)
= cos(𝜃 + 𝜗) + 𝑖(sin(𝜃 + 𝜗))
And if 𝜃 = 𝜗, then we obtain
(cos 𝜃 + 𝑖sin 𝜃)2
= cos 2𝜃 + 𝑖(sin 2𝜃)
And in general we have the following theorem.
Theorem 9.1: for all 𝒏 ∈ ℕ, 𝜽 ∈ ℝ then
(𝐜𝐨𝐬 𝜽 + 𝒊𝐬𝐢𝐧 𝜽)𝒏
= 𝐜𝐨𝐬 𝒏𝜽 + 𝒊 𝐬𝐢𝐧 𝒏𝜽
Example 9.1: calculate the following
a) (cos
3𝜋
8
+ 𝑖sin
3𝜋
8
)4
Using De Moivre’s Theorem, we can write
(cos
3𝜋
8
+ 𝑖sin
3𝜋
8
)4
= cos 4
3𝜋
8
+ 𝑖sin 4
3𝜋
8
= cos
3𝜋
2
+ 𝑖sin
3𝜋
2
= 0 + 𝑖(−1) = −𝑖

46
b) (𝐜𝐨𝐬
𝟓𝝅
𝟐𝟒
+ 𝒊𝐬𝐢𝐧
𝟓𝝅
𝟐𝟒
)𝟒
Using De Moivre’s Theorem, we can write
(cos
5𝜋
24
+ 𝑖sin
5𝜋
24
)4
= cos 4
5𝜋
24
+ 𝑖sin 4
5𝜋
24
= cos
5𝜋
6
+ 𝑖sin
5𝜋
6
Here 𝜃 = 5(30) = 150, then 𝜃 is in the second quarter, therefore cos 𝜃 is
negative and sin 𝜃 is positive, hence
cos
5𝜋
6
+ 𝑖sin
5𝜋
6
= −
√3
2
+ 𝑖 (
1
2
)
c) (𝐜𝐨𝐬 𝜃 + 𝒊𝐬𝐢𝐧 𝜃)𝟖
(𝐜𝐨𝐬 𝜃 − 𝒊𝐬𝐢𝐧 𝜃)𝟒
Using De Moivre’s Theorem, and the fact that
cos 𝑛𝜃 − 𝑖sin 𝑛𝜃 = (cos 𝜃 + 𝑖sin 𝜃)−𝑛
we can write
(cos 𝜃 + 𝑖sin 𝜃)8 (cos 𝜃 − 𝑖sin 𝜃)4
= (cos 8𝜃 + 𝑖sin 8𝜃) (cos 4𝜃 − 𝑖sin 4𝜃)
= (cos 𝜃 + 𝑖sin 𝜃)8 (cos 𝜃 + 𝑖sin 𝜃)−4
= (cos 𝜃 + 𝑖sin 𝜃)4
= cos 4𝜃 + 𝑖sin 4𝜃
Here 𝜃 = 5(30) = 150, then 𝜃 is in the second quarter, therefore cos 𝜃 is
negative and sin 𝜃 is positive, hence
cos
5𝜋
6
+ 𝑖sin
5𝜋
6
= −
√3
2
+ 𝑖 (
1
2
)

47
c) (1 + 𝑖)11
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √12 + (1)2 = √2
cos 𝜃 =
𝑥
‖𝑧‖
=
1
√2
sin 𝜃 =
𝑦
‖𝑧‖
=
1
√2
∴ 𝜃 =
𝜋
4
.
𝑧11
= 𝑟11(cos 𝜃 + 𝑖sin 𝜃)11
= (√2)
11
(cos
𝜋
4
+ 𝑖 sin
𝜋
4
)11
= (√2
2
)
5
. √2 (cos
𝜋
4
+ 𝑖 sin
𝜋
4
)11
= 32√2 (cos 11
𝜋
4
+ 𝑖 sin 11
𝜋
4
)
= 32√2 (−
1
√2
+
1
√2
𝑖) = −32 − 32𝑖.
d) (1 − 𝑖)7
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √12 + (−1)2 = √2
cos 𝜃 =
𝑥
‖𝑧‖
=
1
√2
sin 𝜃 =
𝑦
‖𝑧‖
=
−1
√2
⇒ sin 𝜃 has negative value, then 𝜃 is located in fourth quarter.
∴ arg(𝑧) = 2𝜋 −
𝜋
4
=
7𝜋
4
.
𝑧7
= (√2)
7
(cos
7𝜋
4
+ 𝑖 sin
7𝜋
4
)7
= (√2
2
)
3
. √2 (cos
7𝜋
4
+ 𝑖 sin
7𝜋
4
)7
= 8√2 (cos 7
7𝜋
4
+ 𝑖 sin 7
7𝜋
4
)

48
= 8√2 (
1
√2
+
1
√2
𝑖) = 8 + 8𝑖.
Proposition 9.1: let 𝑧 be a complex number. For all 𝑛 ∈ ℤ+
, 𝜃 ∈ ℝ, then
√𝑧
𝑛
= 𝑟
1
𝑛 [cos
𝜃 + 2𝜋𝑘
𝑛
+ 𝑖 sin
𝜃 + 2𝜋𝑘
𝑛
]
where 𝑘 = 0,1,2, … , 𝑛 − 1.
Example 9.2: find the square roots of (−1 + √3𝑖) by using De Moivre’s Theorem
and last proposition.
Sol/
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √(−1)2 + (√3)2 = √4 = 2
cos 𝜃 =
𝑥
‖𝑧‖
=
−1
2
sin 𝜃 =
𝑦
‖𝑧‖
=
√3
2
⇒ sin 𝜃 has positive value, then 𝜃 is located in second quarter.
∴ arg(𝑧) = 𝜋 −
𝜋
3
=
2𝜋
3
.
𝑧 = 𝑟(cos 𝜃 + 𝑖sin 𝜃) = 2(cos
2𝜋
3
+ 𝑖 sin
2𝜋
3
)
𝑧
1
2 = 𝑟
1
2(cos 𝜃 + 𝑖sin 𝜃)
1
2, for 𝑧
1
2 has two values of 𝑘 = 0,1. Then
𝑧
1
2 = (2)
1
2 (cos
2𝜋
3
+ 2𝜋𝑘
2
+ 𝑖sin
2𝜋
3
+ 2𝜋𝑘
2
)
= √2 (cos
2𝜋 + 6𝜋𝑘
6
+ 𝑖sin
2𝜋 + 6𝜋𝑘
6
)

49
For 𝑘 = 0 ⇒ 𝑧1 = √2 (cos
2𝜋
6
+ 𝑖sin
2𝜋
6
) = √2 (cos
𝜋
3
+ 𝑖sin
𝜋
3
)
= √2 (
1
2
+
√3
2
𝑖) =
1
√2
+
√3
√2
𝑖
If 𝑘 = 1 ⇒ 𝑧2 = √2 (cos
2𝜋+6𝜋
6
+ 𝑖sin
2𝜋+6𝜋
6
) = √2 (cos
4𝜋
3
+ 𝑖sin
4𝜋
3
)
= √2 (
−1
2
−
√3
2
𝑖) =
−1
√2
−
√3
√2
𝑖
Example 9.3: find the four roots for the number (-16).
Sol/
𝑧 = −16 = 16(cos 𝜋 + 𝑖sin 𝜋)
𝑧
1
4 = (16)
1
4(cos 𝜋 + 𝑖sin 𝜋)
1
4 = 2 (cos
𝜋+2𝜋𝑘
4
+ 𝑖sin
𝜋+6𝜋𝑘
4
)
𝑘 = 0 ⇒ 𝑧1 = 2 (cos
𝜋
4
+ 𝑖sin
𝜋
4
) = 2 (
1
√2
+
1
√2
𝑖) = √2 + √2𝑖
𝑘 = 1 ⇒ 𝑧2 = 2 (cos
3𝜋
4
+ 𝑖sin
3𝜋
4
) = 2 (−
1
√2
+
1
√2
𝑖) = −√2 + √2𝑖
𝑘 = 2 ⇒ 𝑧3 = 2 (cos
5𝜋
4
+ 𝑖sin
5𝜋
4
) = 2 (−
1
√2
−
1
√2
𝑖) = −√2 − √2𝑖
𝑘 = 3 ⇒ 𝑧2 = 2 (cos
7𝜋
4
+ 𝑖sin
7𝜋
4
) = 2 (
1
√2
−
1
√2
𝑖) = √2 − √2𝑖

50
Example 9.4: find the six roots for the number (-64i).
Sol/
𝑧 = −64𝑖 = 64 (cos
3𝜋
2
+ 𝑖sin
3𝜋
2
)
𝑧
1
6 = (64)
1
6 (cos
3𝜋
2
+ 𝑖sin
3𝜋
2
)
1
6
= 2 (cos
3𝜋
2
+2𝜋𝑘
6
+ 𝑖sin
3𝜋
2
+6𝜋𝑘
6
)
= 2 (cos
3𝜋 + 4𝜋𝑘
12
+ 𝑖sin
3𝜋 + 4𝜋𝑘
12
)
𝑘 = 0 ⇒ 𝑧1 = 2 (cos
3𝜋
12
+ 𝑖sin
3𝜋
12
) = 2 (
1
√2
+
1
√2
𝑖) = √2 + √2𝑖
𝑘 = 1 ⇒ 𝑧2 = 2 (cos
7𝜋
12
+ 𝑖sin
7𝜋
12
) = 2 (−
1
√2
+
1
√2
𝑖) = −√2 + √2𝑖
𝑘 = 2 ⇒ 𝑧3 = 2 (cos
11𝜋
12
+ 𝑖sin
11𝜋
12
)
𝑘 = 3 ⇒ 𝑧4 = 2 (cos
15𝜋
12
+ 𝑖sin
15𝜋
12
) = 2 (cos
5𝜋
4
+ 𝑖sin
5𝜋
4
)
= 2 (−
1
√2
−
1
√2
𝑖) = −√2 − √2𝑖
𝑘 = 4 ⇒ 𝑧5 = 2 (cos
19𝜋
12
+ 𝑖sin
19𝜋
12
)
𝑘 = 5 ⇒ 𝑧6 = 2 (cos
25𝜋
12
+ 𝑖sin
25𝜋
12
)

51
Example 9.5: find the polar form and the five roots of (√3 + 𝑖)2
.
Sol/
𝑚𝑜𝑑 𝑧 = ‖𝑧‖ = √𝑥2 + 𝑦2 = √(√3)2 + (1)2 = √4 = 2
cos 𝜃 =
𝑥
‖𝑧‖
=
√3
2
⇒ cos 𝜃 has positive value.
sin 𝜃 =
𝑦
‖𝑧‖
=
1
2
∴ 𝜃 =
𝜋
6
.
𝑧 = 𝑟(cos 𝜃 + 𝑖sin 𝜃) = 2(cos
𝜋
6
+ 𝑖 sin
𝜋
6
)
𝑧2
= 22
(cos
2𝜋
6
+ 𝑖 sin
2𝜋
6
) = 4 (cos
𝜋
3
+ 𝑖 sin
𝜋
3
)
𝑧
2
5 = 𝑟
2
5(cos 𝜃 + 𝑖sin 𝜃)
2
5 = (2)
2
5 (cos 5.
𝜋
3
+2𝜋𝑘
2
+ 𝑖sin 5.
𝜋
3
+2𝜋𝑘
2
)
= √4
5
(cos
5𝜋 + 30𝜋𝑘
6
+ 𝑖sin
5𝜋 + 30𝜋𝑘
6
)
𝑘 = 0 ⇒ 𝑧1 = √4
5
(cos
5𝜋
6
+ 𝑖sin
5𝜋
6
) = √4
5
(−
√3
2
+
1
2
)
𝑘 = 1 ⇒ 𝑧2 = √4
5
(cos
35𝜋
6
+ 𝑖sin
35𝜋
6
)
𝑘 = 2 ⇒ 𝑧3 = √4
5
(cos
65𝜋
6
+ 𝑖sin
65𝜋
6
)
𝑘 = 3 ⇒ 𝑧4 = √4
5
(cos
95𝜋
6
+ 𝑖sin
95𝜋
6
)

52
𝑘 = 4 ⇒ 𝑧5 = √4
5
(cos
155𝜋
6
+ 𝑖sin
155𝜋
6
)
Example 9.6: solve the equation 𝑥3
+ 1 = 0, where 𝑥 ∈ ℂ.
Sol/
𝑥3
+ 1 = 0 ⇒ 𝑥3
= −1 ⇒ 𝑥3
= −1(1) = cos 𝜋 + 𝑖sin 𝜋
⇒ 𝑥 = (cos 𝜋 + 𝑖sin 𝜋)
1
3 = cos
𝜋+2𝜋𝑘
3
+ 𝑖sin
𝜋+2𝜋𝑘
3
𝑘 = 0 ⇒ 𝑥1 = cos
𝜋
3
+ 𝑖sin
𝜋
3
=
1
2
+
√3
2
𝑖
𝑘 = 1 ⇒ 𝑧2 = cos
3𝜋
3
+ 𝑖sin
3𝜋
3
= −1 + 0 𝑖 = −1
𝑘 = 2 ⇒ 𝑧3 = cos
5𝜋
3
+ 𝑖sin
5𝜋
3
=
1
2
−
√3
2
𝑖
Hence, 𝑆 = {
1
2
+
√3
2
𝑖 , −1 ,
1
2
−
√3
2
𝑖}.

53
Exercises (2-3)
1) Find the modulus and principle value of the following numbers:
a) 𝑧 = −1 + √3𝑖 b) 𝑧 = −1 + 𝑖 c) 𝑧 = 1 − 𝑖
2) find the polar form for the following:
a) –i b) -7i c) 3 d) 5i e) 2 f) -1
3) Calculate:
a) (cos
7𝜋
12
+ 𝑖 sin
7𝜋
12
)
−3
b)
(cos 2𝜃+𝑖 sin 2𝜃)5
(cos 3𝜃+𝑖 sin 2𝜃)3
c) Find the square roots of (27𝑖) by using De Moivre’s Theorem.

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