COMPLEX NUMBER

complex number

WAN AZLIZA BINTI WAN ZAKARIA
RAHIMAH BT MOHD ZAIN @ AB. RAZAK

NIK NOOR SALISAH BINTI NIK ISMAIL

[CHAPTER 4: COMPLEX NUMBER]

COMPLEX NUMBER

OBJECTIVES:
At the end of this topic, students should be able to understand about:

i. Identify real part and imaginary part
ii. Recognize that i   1
iii. Perform the operations of complex number
iv. Represent complex number using Argand Diagram
v. Complex number in other form

4.1 INTRODUCTION OF COMPLEX NUMBER

Introduction including:
 Complex Number in general form -
 Definition of 1
 Real part
 Imaginary part

REAL NUMBER the numbers like 1, 10.158, -0.34, 2/5, 3 , or any number
when squared give a negative result
IMAGINARY
NUMBER

 The "unit" imaginary number (like 1 for Real Numbers) is i, which is the square root
of −1

 Because when we square i we get −1; i2 = −1
 And we keep that little "i" there to remind us we need to multiply by 1

1

[CHAPTER 4: COMPLEX NUMBER]
4.1.1 THE CONCEPT OF A COMPLEX NUMBER

 A Complex Number is a combination of a Real Number and an Imaginary Number.
 General form of complex number written as:

a + bi

Examples of a (real number):-

1 10.15 -0.3462 2 3
5

Examples of b (Imaginary Numbers):-

3i 1.04i −2.8i 3i ( 2)i 1998i
4

Examples of complex number: 0.8 − 2.2i −2 +  i 2 + 1 i
1 + i 39 + 3i 2

Complex Number Real Part Imaginary Part Purely Real
2 Purely Imaginary
3 + 2i 3
5 5 0
−6i 0 −6

A Complex Number consists of real part and imaginary
part. But either part can be 0.

2

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 1

Simplify: 3. i9
1.  4 4.  2   3
2. 7i  i 2

SOLUTION:
1.  4  4(1)  4i 2  2i
2. 7i  i2  7i3  7i(1)  7i

3. i9  i 2  i 2  i 2  i 2  i  1 1  1  1  i

4.  2   3  2  3 1  2  3i

LET’S PRACTICE 1 3. −√−64

Simplify:
1. 2i  3i

2. 4   2 4. √−98

Ans:  6 , 4  2i , −8 , 9.9

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[CHAPTER 4: COMPLEX NUMBER]
4.2 THE OPERATIONS OF COMPLEX NUMBERS
4.2.1 ADDITION
 To add two or more complex numbers, we add each part separately.

Complex Number 2
Imaginary part

a  bi  c  di  a  c  b  d i

Complex Number 1 Real part

EXAMPLE 2

a. Perform the addition of 3  2i and 1  7i

SOLUTION:
( 3  2i ) + (1  7i )

REAL PART 3 1  4
IMAGINARY PART 2  7i  9i
ANSWER
4  9i

b. Perform the addition of 3  5i and 4  3i

SOLUTION:
(3  5i) + (4  3i)

REAL PART 3  4  7
IMAGINARY PART 5  3i  2i
ANSWER
7  2i

4

[CHAPTER 4: COMPLEX NUMBER] Ans: 6  10i
LET’S PRACTICE 2 Ans: 1  15i
Solve the following operation of complex number. Ans:  4  2i
Ans:  2  9i
a. 5  2i  1 8i Ans: 9  3i
Ans:  12  24i
b. 3  6i   2  9i Ans:  47  68i
Ans: 93  29i
c. 8i 1   5  6i
5
d.  7  5i  5  4i

e. 2  3i  7  6i

f.  24  10i  12  14i

g..  30  52i  17 16i

h. 72  14i  21  43i

[CHAPTER 4: COMPLEX NUMBER]
4.2.2 SUBSTRACTION
 To subtract two or more complex numbers, we do same steps as for addition.
 Subtract each part separately.

Complex Number 2 Imaginary part

a  bi  c  di  a  c  b  d i

Complex Number 1 Real part

EXAMPLE 3

a. Subtract 7  3i and 1 7i

SOLUTION:
( 7  3i ) - (1 7i )

REAL PART 7 1  6
3  7i  4i
IMAGINARY PART
6  4i
ANSWER

b. Perform the subtraction (5 + 3 ) − (3 + 2 )

SOLUTION:

5  3i  3  2i 5  3  2
3  2i  i
REAL PART
IMAGINARY PART 2i

ANSWER

6

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 3

Solve the following operation of complex number.

a.  7  9i  5  2i

b. 8  2i  1 3i Ans: −12 + 7i
Ans: 7  i
c. 8i  6   5 1i
Ans: 11  9i
d. 2  6i   4  9i Ans: 6  3i
Ans: 36 + 44i
e. 30  4i   6  48i Ans:  74  27i
Ans:  25  12i
f.  70  6i  4  21i Ans: 104  36i

g.  32i  11  14  20i 7

h. 91  27i  13  9i

[CHAPTER 4: COMPLEX NUMBER]

4.2.3 MULTIPLICATION

 To multiply complex numbers, each part of the first complex number gets multiplied
by each part of the second complex number.

 Just use "FOIL", which stands for "Firsts, Outers, Inners, Lasts"

a  bic  di  ac  adi  bci  bdi2

(a  bi)(c  di)  (ac  bd)  (ad  bc)i

EXAMPLE 4 Remember:- i 2  1
Remember:- i 2  1
a. Solve the equation 3  2i1 7i

SOLUTION:

3  2i1  7i  (31)  (3 7i)  (2i 1)  (2i  7i)

 (3)  (21i)  (2i)  (14i 2 )
 (3)  (21i)  (2i)  (14)(1)
 (3)  (21i)  (2i)  (14)
 11 23i

b. Solve the equation 1  i2

SOLUTION:

1  i1  i  (11)  (1 i)  (1 i)  (i 2 )

 (1)  (2i)  (1)
= 2

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[CHAPTER 4: COMPLEX NUMBER]

c. Solve the equation −9(4 − 2 ) Remember:- i 2  1
SOLUTION:
−9(4 − 2 ) = (−9 × 4) + [(−9) × (−2 )]
= −38 + 18

d. Solve the equation 3 (2 + 6 )
SOLUTION:
3 (2 + 6 ) = (3 × 2) + (3 × 6 )
= 6 + 18
= 6 + 18(−1)
= −18 + 6

LET’S PRACTICE 4 Ans:  22  46i
Solve the following operation of complex number.

a. 8i  6 5 1i

b. 8  2i1  3i

c.  7  9i5  2i Ans: 2  26i
d. 2  6i 4  9i Ans:  53  30i

Ans:  44  6i

9

[CHAPTER 4: COMPLEX NUMBER]

4.2.4 DIVISION

 To perform the division of complex numbers, we use the conjugate.
 For example,

a  bi , the conjugate of the bottom number will be used.
c  di
 Bottom number is c  di , so the conjugate will be c  di . The sign for imaginary part

will be changed or the conjugate can be written as c  di  c  di .

 Then the division of a  bi when using conjugate is written as a  bi  c  di
c  di c  di c  di

 Hence, a  bi  c  di = a  bi c  di
c  di c  di c  di c  di

 =
ac  adi  bci  bdi2 ; i 2  1
cc  cdi  cdi  ddi2

= ac  bd   bc  ad i

c2  d 2

EXAMPLE 5

1. Do the division 2  3i
4  5i

SOLUTION: EXPLANATION:
a. Multiply top and bottom by the conjugate of
2  3i  4  5i
4  5i 4  5i 4  5i
b. Remember that i 2  1
 8  10i  12i  15i 2 2 c. Add like terms and notice how on the
16  20i  20i  25i
bottom 20i  20i cancels out.
 8  10i  12i 15 d. Lastly we should put the answer back into
16  20i  20i  25
a  bi form
  7  22i
41

  7  22 i
41 41

10

[CHAPTER 4: COMPLEX NUMBER]

2. Write 4  3i in the standard form + .   7  22 i
2  5i 41 41

SOLUTION:

4 + 3 4 + 3 2 − 5
2 + 5 = 2 + 5 × 2 − 5

8 − 20 + 6 − 15
= 4 − 10 + 10 − 25

8 − 14 − 15(−1)
= 4 − 25(−1)

8 + 15 − 14
= 4 + 25

23 − 14
= 29

23 14
= 29 − 29

LET’S PRACTICE 5

Solve the following operation of complex number.

a. ( )
()

  7  22 i
41 41

Ans:

b. 1 3i
8  2i

Ans: 2  26i
68
5  2i
c.  7  9i

Ans:

11

[CHAPTER 4: COMPLEX NUMBER]

d.  4  9i
 2  6i

Ans:  46  42i
40

4.3 CONJUGATE OF A COMPLEX NUMBERS
 A conjugate is where we change the sign in the middle. Refer the diagram below.

 In other word, it is obtained by changing the sign of the imaginary part.
 A conjugate is often written with a bar over it.

Conjugate solving Explanation
5  3i = 5  3i change the sign of the imaginary part

 Let z  3  4i ;
then the conjugate of z represented by
z  3  4i .

 If z  5  2i , then its conjugate z  5  2i .
 The conjugate is used to solve the division of complex numbers.
 It is important to ease the complex number with denominator.
 The trick is to multiply both top and bottom by the conjugate of the bottom.

4.3.1 THE PROPERTIES OF THE COMPLEX CONJUGATE

 For any complex numbers , , , the algebraic properties of the conjugate
operation:

+ = + − = −
= ∙
≠ 0
= ,

12

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 6

Given that = 2 − i and = 3 + 3i. Write each expression in the standard form a + bi
a. −
b. +

SOLUTION:
a. = 3 − 3
Therefore; − = (2 − ) − (3 − 3 )
− = 2 − − 3 + 3
− = − +

b. + = 2 − + 3 + 3
+ = 5 + 2
Therefore; + = −

LET’S PRACTICE 6

Given that z = 8 + 3i and w = 3 − 4i. Write each expression in the standard form a + bi.
a. + ̅

Ans: 16

b. −

Ans: −8

c. ̅ +

Ans: 11 +

13

[CHAPTER 4: COMPLEX NUMBER]

4.4 GRAPHICAL REPRESENTATION OF A COMPLEX NUMBER THROUGH ARGAND
DIAGRAM

 Complex number can be represented by argand diagram.
 If a complex number,

z  a  bi ; a → real part
b → imaginary part.

4.4.1 ARGAND’S DIAGRAM TO REPRESENT A COMPLEX NUMBER

 Based on the following diagram,

z  a  bi is the reflection image of z  a  bi

Where;

a  bi  (a  bi)  2a

and

(a  bi)  (a  bi)  2bi

EXAMPLE 7

1. Show the following complex number on Argands Diagram.

a. x  yi b. 3  5i

c. 4  7i d. 2  5i

14

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 7

Represent the following complex number on Argand Diagram

a. 4  9i

Ans:

b.  3  6i

Ans:

c.  7  2i

Ans:

d. 8  5i

Ans:

4.4.2 MODULUS AND ARGUMENT

 Refer to argand diagram below:  Modulus of the complex number z, donated by
z where it is a positive value which is equal to

the length of the segment OZ.

| | = +

 Argument of z is angle  ; tan   y .
x

Hence, the argument, donated by;


= tan

15

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 8

Find the modulus and the argument for the following complex numbers. Then show them on
argand diagram.

a. z  3  4i b. z  4  7i

Modulus, z = 32  42 Modulus, z = 42  72
= 9  16
= 25 = 16  49
=5 = 65
= 8.06
Argument, θ = tan1 4
3 Argument, θ = tan1 7
4
= 53.13º
= 60.25 
Argand diagram:
θ in Quadrant 4;

  360  60.25 = 299.75º

Argand diagram:

θ
θ

16

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 8

Find the modulus and the argument for the following complex numbers.
1. z  8  4i

Ans: z  8.94,   333.43

2. z  6  9i

Ans: z  10.8 ,   56.3

3. z  10i  12

Ans: z  15.62 ,   320.19

3. z  10i  12

Ans: z  15.62 ,   320.19

4.5 COMPLEX NUMBER IN OTHER FORM

Complex number can be expressed as shown in the following table.

Complex Number The form of

r  a  bi Cartesian

r  Polar
 can be in degree( )

r cos  i sin  Trigonometric

 can be in degree( )

rei Exponent
 must be in radian (rad)

17

[CHAPTER 4: COMPLEX NUMBER]
4.5.1 COMPLEX NUMBER IN POLAR FORM

EXAMPLE 9

Given a complex number z  4  9i . State z in polar form, z 

SOLUTION: 42  92
Modulus; z 

z  9.85

Argument; = tan
9

= tan 4
= 66.04°

In Polar Form; | |∠ = . ∠ . °

LET’S PRACTICE 9

Express for the following complex numbers in polar form
1. y  9  5i

Ans: 10.333.1

2. p  12  7i

Ans:13.8930.26

3. r  9(cos120  i sin120)

4. x  6.4e1.12i Ans: 9120

Ans: 6.464.17

18

[CHAPTER 4: COMPLEX NUMBER]

4.5.2 COMPLEX NUMBER IN TRIGONOMETRIC FORM

EXAMPLE 10

If z  530, write z in trigonometric form { r cos  i sin  }

SOLUTION:
z  530→ in polar form ( r  )

 z  5 cos30  i sin 30

LET’S PRACTICE 10
Express for the following complex numbers in trigonometric form
1. k  7 10i

Ans: 12.2(cos 305  i sin 305)

2. p  9  12i

3. y  40270 Ans: 15(cos 53.13  i sin 53.13)
4. x  21.3e 2.13i Ans: 40(cos 270  i sin 270)

Ans:, 21.3(cos122  i sin122)

19

[CHAPTER 4: COMPLEX NUMBER]
4.5.3 COMPLEX NUMBER IN EXPONENTIAL FORM

EXAMPLE 11
Given z  6  4i . State z in the form of rei

SOLUTION:
Modulus; | | = 6 + (−4) = √52

| | = 7.21

Argument; = tan

4
= tan 6
= 33.69°

Since is in Quadrant 4;

= 360° − 33.69° = 326.31°

= 326.31° × = 5.695
180°

In exponential form; = = . .

LET’S PRACTICE 11
Express for the following complex numbers in exponential form.
1. a  6i  10

Ans: 11.7e5.742i

20

[CHAPTER 4: COMPLEX NUMBER]

2. r 11 52i

3. p  32275 Ans: 53.2e1.36i
Ans: 32e 4.8i
4. x  38(cos86  i sin 86) Ans: 38e1.5i

4.5.4 COMPLEX NUMBER IN CARTESIAN FORM

EXAMPLE 12

If z  245, express z in Cartesian form.

SOLUTION:
Write z in trigonometric form:

z = 2 cos 45  i sin 45

In Cartesian form:
= 2(0.7071 + 0.7071 )
= . + .

21

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 12

Express for the following complex numbers in Cartesian form.
a. z  42(cos 56  i sin 56)

Ans: 23.49  34.82i

b. y  34e1.48i

Ans: 30.83  33.86i

c. y  28130

Ans: 18  21.45i

4.6 MULTIPLICATION AND DIVISION OF COMPLEX NUMBERS IN POLAR
FORM

 Complex numbers in polar form are especially easy to multiply and divide. The rules
are:
i. Multiplication rule: To form the product, multiply the magnitudes and add
the angles.
ii. Division rule: To form the quotient, divide the magnitudes and subtract the
angles.

EXAMPLE 13

1. Given z1  335 and z2  547 . Calculate:
a. ×

SOLUTION:
× = (3∠35°)(5∠47°)
× = (3 × 5)∠(35° + 47°)
× = 15∠82°

22

[CHAPTER 4: COMPLEX NUMBER]
b.

SOLUTION:

= ∠ °°b
∠

= 5 ∠(47° − 35°)
3

= 1.67∠12°


2. Given a  414 , b  6  71.5 , c  745 . Solve:
a. a  b  c

SOLUTION:

a  b  c  414 6  71.5 745
a  b  c  (4  6  7)(14  71.5  45)
a  b  c  168 12.5

b. a  c
b

SOLUTION:

a  c  414  745
b 6  71.5

a  c  4  714  (71.5)  45
b 6

a  c  4.67130.5
b

23

[CHAPTER 4: COMPLEX NUMBER]

c. bc
a

SOLUTION:

b  c  6  75  745
a 414

b  c  6  7  75  45  14
a 4

b c  10.5  44
a

d. c2
ab

SOLUTION:

c2  (745) 2
ab 414  6  71.5

c2  4 72 2  45
ab 6(14  71.5)

c2  49 (90  61)
ab 24

c2  2.04151
ab

LET’S PRACTICE 13

1. Solve the following expression in an exponential form.
25(cos180  i sin180)  8(cos12  i sin 12)
20(cos 50  i sin 50)

Ans: 10e 2.48i

24

[CHAPTER 4: COMPLEX NUMBER]

2. Given that Z1  8(cos34  i sin 34) and Z2  6040 . Solve Z2 in trigonometry form
Z1

Ans: 7.5(cos 6  i sin 6)

3. Given z1  3  2i and z2  3(cos30  i sin 30)
a. Calculate z1  z2 and express the answer in the form of a  bi and exponential
form.

Ans: 4.8  9.7i & 10.8e1.11i

b. Calculate z1 and express the answer in the form of a  bi and exponential form.
z2

Ans: 1.2  0.077i & 1.2e0.064i
4. Given z1  8145 , z2  2460

a. Calculate z1  z2 and express the answer in the form of a  bi

Ans:  174  81.14i
b. State 2z2 in trigonometric form. Illustrate the answer on Argand diagram.

z1

 Ans: 6 cos 850  i sin 850

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