ENGINEERING MATHEMATICS 1

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 5

Find the value of  for sin = 0.5870, where 0° ≤ ≤ 360°

SOLUTION:

sin   0.5870 Sin is positive value QUADRANT 1:
2 (Quadrant 1 and 2)

2 = 35.94°
2 = sin 0.5870
= 2(35.94°) = 71.89°

= 35.94° QUADRANT 2:
2

2 = 180° − 35.94° = 144.06°

= 2(144.06°) = 288.12°

Therefore;
= . ° & . °

LET’S PRACTICE 5

Find the values of  in the range 0° ≤ ≤ 360° for each of the following,

a. cos 1    0 .7384
2

Ans: 275.2

b. cos2  0.4756

Ans: 30.8 , 149.2 , 210.8 & 329.2

45

[CHAPTER 3: TRIGONOMETRY]
3.4 TRIGONOMETRIC EQUATIONS AND IDENTITIES

1. Basic Identities

Sin  Tan
Cos

2. Trigonometric Identities

cos2   sin2   1
1 tan2   sec2 
cot2  1  cosec2

3. Angle Sum and Difference Identities

sin     sin cos  cos sin 

cos     cos cos  sin sin 

tan     tan  tan 
1 tan tan 

4. Double Angle Identities

sin 2A  2sin A cos A

cos 2 A  cos2 A  sin 2 A

cos 2A  1  2sin 2 A

cos 2A  2 cos2 A  1

tan 2 A  1 2 tan A
 tan 2A

46

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 6

1. Solve + 2 = 3, where 0° ≤ ≤ 360°

SOLUTION:

sin   2  3 Sine positive Quadrant 1:
sin   1 (Quadrant 1 and 2) = 90°
  sin 1 1
  90 Reference Angle Quadrant 2:
x = 180° − 90° = 90°

Therefore;
= °

2. Solve 5 − 2 = 0, where 0° ≤ ≤ 360°.

SOLUTION:

5 − 2 = 0 Quadrant 1:
= 21.80°
5 2
− = 0 Quadrant 3:
= 180° + 21.80° = 201.80°
5 − 2 = 0 Tangent is positive
(Quadrant 1 and 3)
2
= 5 2
5
= = 21.80°

Therefore;
= . ° & . °

LET’S PRACTICE 6

1. Solve each of the following equations for 0° ≤ ≤ 360°.
a. 4sin  3cos

Ans: 36.87 & 216.87

47

[CHAPTER 3: TRIGONOMETRY]

b. sin  cos  0

Ans: 135 & 315

c. 4 tan  3sec

Ans: 48.59 & 131.4

d. 2 tan  cos ec  3

Ans: 48.19 & 311.81

2. Solve each of the following equations for 0° ≤ ≤ 360°.
a. 2 cos 2 x  cos x

Ans: 60 , 90 , 270 & 300

b. 5sin x cos x  sin x

Ans: 78.46 & 281.54

c. sin 2 x  sin x  2  0

d. 2 tan 2 x  sec x tan x Ans: 270

Ans: 30 & 150

48

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 7

Solve the equation 2 cos 2   3 cos   1  0 for the values of  in the range 0° ≤ ≤ 360°.

SOLUTION:

2 cos2   3cos  1  0

2cos 1cos 1  0

2 cos   1  0

cos   1 Cos positive Quadrant 1:
2 (Quadrant 1 and 4) θ = 60°
Quadrant 4:
  cos 1  1  Reference angle θ = 360° − 60° = 300°
 2 

  60 

cos 1  0 Cos positive Quadrant 1:
cos  1 (Quadrant 1 and 4) = 0°
  cos 1 1
  0 Reference angle Quadrant 4:
= 360° − 0° = 360°

Therefore;
= 0°, 60°, 300° & 360°

49

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 7

1. Solve the equation 1  2sin  4 cos 2  0 for values of  in the range of
0° ≤ ≤ 360°.

Ans: 48.59 , 131.41 , 210 & 330
2. Solve each of the following equations for values of  in the range 0° ≤ ≤ 360°.

a. 8 sin 2   2 cos   5  0

Ans: 41.41 , 120 , 240 & 318.59

b. 7 sin 2   cos 2   5 sin 

Ans: 19.47 , 30 , 150 & 160.53

c. 3 cos  2 cot 

Ans: 41.81 & 138.19

50

[CHAPTER 3: TRIGONOMETRY]
3.5 SINE AND COSINE RULES

3.5.1 SINE RULE

Sine Rule:

a A  b  c
sin sin B sin C

sin A  sin B  sin C
a b c

EXAMPLE 8

a. Find the length of AC. SOLUTION:

 ABC  180   75   40 

ABC  65 

AC 8.2
sin 65  sin 40

AC  8.2 sin 65  11.56cm
sin 40

b. Solve the triangle with the following given dimensions:

a  43cm , b  75 cm and B  30 0

SOLUTION:

sin A  sin B
a b

sin A sin 30
43  75

A  sin 1  43sin 30   16.66
75

51

[CHAPTER 3: TRIGONOMETRY]

C  180 0  30 0  16 .66 0  133 .34 

c  b
sin C sin B

c 75
sin133.34  sin 30

c  75 sin 133.34   109.09cm
sin 30

LET’S PRACTICE 8
1

Find the length of AC Ans: 4.52cm
2.

Find  ABC

Ans: 31.69

52

[CHAPTER 3: TRIGONOMETRY]
3.

Find  ABC and the length of AB Ans: 11.82 & 10.02cm
4.

Find the length of AC and BC. Ans: 5.19cm & 6.35cm

5. Solve the triangles with the given parts:

a) = 45.7 , = 68.20°, = 47°
b) = 4.608 , = 3.207 , A = 18.23°
c) = 742 , B = 53°, C = 3.5°

a) C = 64.8°, = 36 , = 44.54
Ans: b) B = 12.58°, C = 149.19°, c = 7.545

c) A = 123.5°, = 774.75 , = 56.72

53

3.5.2 COSINE RULE [CHAPTER 3: TRIGONOMETRY]

Cosine Rule:
a2  b2  c2  2bccos A
b2  a2  c2  2accos B
c2  a2  b2  2abcosC

EXAMPLE 9

1. Given = 16.4 , = 11.8 and SOLUTION:
∠C = 67°. Find the length of .
c2  a2  b2  2abcosC

c2  16.42 11.82  216.411.8cos670

c2  256.97
c  256.97
c  16.03cm

2. Solve the triangle given that a  6.00cm , b  7.56cm and ∠C = 54°.

SOLUTION:

c2  a2  b2  2abcosC

c2  62  7.562  267.56cos540

c  6.31cm

For A , use the law of sines:

sin A  sin C
a c

sin A sin540
6  6.31

sin A  6 sin 540
6.31

A  50.290

= 180° − (50.29° + 54°) = . °

54

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 9

1. Given = 7 , = 5 and ∠C =
60°. Find the length of .

Ans: 6.24cm

2. Given = 7 , = 5 and = 3 .
Find B .

3. Given = 15 , = 12 and = Ans: 38.21°
14 . Ans: 69.99°
Find A .

55

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 10

1. Given ∠ABC = 64°. Find:
a. A
b. length of

Ans: 74.05°, 6.4

2. Given the length of = 6.2 , =
4.6 and ∠K = 52°. Find:
a. the angle of ∠L
b. length of

Ans: 92.22° , 7.86

3. Given the length of = 9.1 , =
5.9 and = 10.2 . Find:
a. P
b. Q

Ans: 35.01°, 62.25°

56

[CHAPTER 3: TRIGONOMETRY]

3.6 AREA OF TRIANGLE

Area of triangle ABC:

 1 ab sin C
2

 1 bc sin A
2

 1 ac sin B
2

EXAMPLE 10

Given = 15 , = 9 and SOLUTION:
∠ACB = 64°. Find the area of triangle.
Area of ABC;

 1 159sin 640
2

 60.67cm2

2. Given = 6 , = 5 and SOLUTION:
∠ABC = 49°. Find the area of triangle
. 56
sin = sin 49°

5 sin 49°
sin = 6 = 0.6289
A = sin 0.6289 = 38.97°

C = 180° − 49° − 38.97° = 92.03°

Area of ABC;

= 1 (5)(6) sin 92.03° = 14.99
2

57

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 11

1. Given = 8 , = 12
and ∠DFE = 126°. Find the area
of triangle .

2. Given = 11 , = 14.5 Ans: 38.83cm2
and ∠B = 48°. Find the area of Ans: 64.18cm2
triangle.

3. Given = 7 , = 9 and
= 6 . Find the area of
triangle .

Ans: 20.98cm2

58

[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

59

[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.

60

[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

61

[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

62

[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
63
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

64

[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 65

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

66

[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

Solve the following operation of complex number.

a. 8i  6 5 1i

Ans:  22  46i

b. 8  2i1  3i

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

Ans:  44  6i

67

[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

68

[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. ( )

()

Ans:

b. 1 3i
8  2i

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

Ans:

d.  4  9i
 2  6i

Ans:  46  42i
40

69

[CHAPTER 4: COMPLEX NUMBER]
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
= ,

70

[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 +

71

[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

72

[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

73

[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

74

[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)

75

[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

76

[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)

77

[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

78

[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 )
= . + .

79

[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°

80

[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

81

[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

82

[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

83

[CHAPTER 5: MATRIX]

MATRIX

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

i. matrix definition
ii. dimension or order of matrix
iii. types of matrix
iv. operation of matrix (addition, subtraction, multiplication and division)
v. simultaneous equation using matrix (inverse matrix method and

Cramer’s rule)

5.1 INTRODUCTION OF MATRIX

 A matrix is an ordered rectangular array of numbers or functions.
 The numbers or functions are called the elements or the entries of the matrices.
 The matrices are denoting by capital letters.

A matrix is a rectangular arrangement of numbers in
rows and columns.

Figure 1

Referring on Figure 1 above;
 Its dimension are 2  3
 2 rows and 3 columns
 The entries of the matrices are 2,5,10,4,19,4

84

[CHAPTER 5: MATRIX]

MATRIX NOTATION

 In order to identify an entry in a matrix, we simply write a subscript of the respective
entry’s row followed by the column.

 In general, the matrices can be denoted by as in Figure 2.

Figure 2

- Each aij is called an element of the matrices (or an entry of the matrices).
- This denotes the element in row i and column j.
- The entries of the matrices are organized in horizontal rows and vertical

columns.

DIMENSION OR ORDER OF A MATRIX

 The number of ROWS in matrix is represented with ‘m, and the number of
COLUMNS is represented with ‘n’.

 Hence the matrices can be called (m x n) order matrices.
 The numbers m and n are called dimensions/ order/ size of the matrices.
 Example:

MATRICES SIZE/ ORDER

A  1 1 0 4 1 4

 2 1 - A row & 4 columns
B   3 1
3  2
 0 4
- 3 rows & 2 columns

85

[CHAPTER 5: MATRIX]

EXAMPLE 1

34  5 31
A  22 15 
55 

52 13  9

i. What are the dimensions of a matrix above?

ii. Identify entry A23

SOLUTION:

i. Dimension of a matrix; 33

ii. 55

LET’S PRACTICE 1

 2  5 14 23  9
1. A   4 19 
4 34 9 

 41 5 30 1  6

i. What are dimension of A?

ii. Identify entry A34

iii. Identify entry A12

Ans: 3 5 , 1,  5

12 7 21 31 11
 27 19
2. V   45 2 14 71 26
15 36
 3 13 55 34 15

 4

i. What are the dimensions of V above?
ii. Identify the entry V14
iii. What is the matrices notations to donate the entry 15

Ans: 4  5, 31, V32 and V45

86

[CHAPTER 5: MATRIX]

TYPES OF MATRIX

a. RECTANGULAR MATRIX

 A matrix in which number of rows is not 1 2 3
equal to number of columns. 3 1
A  4 2 2
3
1 2 3

b. ROW MATRIX c. COLUMN MATRIX d. SQUARE MATRIX

 A matrix with single  A matrix with  A matrix with equal

row and any number single column and number of rows and

of columns. any number of columns.

A  1 2 3 4 5 rows.  m  n

1 1 2 1
2 A  2 1 2
A  3
1 2 1
4

e. DIAGONAL MATRIX f. SCALAR MATRIX g. ZERO MATRIX

 Is a square matrix in  A diagonal matrix  A matrix in which every

which all the in which all the element is zero.

elements except diagonal elements 0 0 0
those on the leading are equal. A  0 0 0
diagonal are zero.
4 0 0 0 0 0
2 0 0 A  0 4 0
A  0 5 0
0 0 4
0 0 7

h. IDENTITY MATRIX /UNIT MATRIX i. TRANPOSE MATRIX

 When the diagonal elements  A matrix which is formed by turning
are one and nondiagonal all the rows of a given matrix into
elements are zero. columns and vice versa.

 A unit matrix is always a  The transpose of matrix A is written
square matrix.
1 0 0 as AT
A  0 1 0
0 0 1 1 2 3T  1 4
4 5 6 2 5
3 6

87

[CHAPTER 5: MATRIX]
5.2 OPERATION OF MATRIX

 Addition
 Subtraction
 Multiplication

- By scalar
- By matrices

5.2.1 ADDITION

 The two matrices must be the same size.
 The rows must match in size
 The columns must match in size

 Add the numbers in the matching positions.

EXAMPLE 2

These are the calculations:

3 8  4 0  7 8 347 80 8
4 6 1  9 5  3 41 5
6   9  3

5.2.2 SUBTRACTING

 The two matrices must be the same size.
 The rows must match in size
 The columns must match in size

 Subtract the numbers in the matching positions.

EXAMPLE 3

These are the calculations:

3 8  4 0   1 8 3  4  1 80  8
4 6 1  9  15 4 1 3
 3 6   9  15

88

[CHAPTER 5: MATRIX]

LET’S PRACTICE 2

1. Find 2  3  1  5
 4    2
2   3

Ans: 1  8
 1 
0 

2. If A  2  3 and B   1  5 . Find A B.
 4 2   3  2

Ans: 3 2
 7 4

3. If A 3 5 4 and B  1 4 2 . Find A  B
 1 4 6  5 2 3

Ans: 2 1 6
 6 2 9

4. If A 3 5 4 and B  1 4 2 . Find A  B
 1 4 6  5 2 3

Ans: 4 9 2
4 6 3

89

[CHAPTER 5: MATRIX]
5.2.3 MULTIPLICATION
MULTIPLICATION BY SCALAR
 To multiply a matrix by a single number is easy.

EXAMPLE 4

These are the calculations:

2  4 0  8 0 24 8 20  0
1  9 2 18 21  2
2  9  18

MULTIPLYING A MATRIX BY ANOTHER MATRIX
 To multiply a matrix by another matrix, we need to do the “dot product”.

EXAMPLE 5

The “dot product” is multiply matching members, then sum up:
 First row, first column:

1,2,3 7,9,11  58 1(7) + 2(9) + 3(11) = 58

90

[CHAPTER 5: MATRIX]

 First row, second columns: 1(8) + 2(10) + 3(12) = 64
4(7) + 5(9) + 6(11) = 139
1,2,3  8,10,12  64 4(8) + 5(10) + 6(12) = 154

 Second rows, first column:

4,5,6  7,9,11  139

 Second rows, second columns:

4,5,6  8,10,12  154

 THE ANSWER:

1 2 3 7 8  58 64 
4 5 6  10 139 154
  9 12 

11

REMEMBER

 In arithmetic:
3  5  5  3 [Commutative Law]

 But this is NOT generally true for matrices (matrix multiplication is not
commutative).
AB  BA

91

[CHAPTER 5: MATRIX]

EXAMPLE 6

Given A 1 2 and B  2 0 . Prove that AB  BA .
3 4 1 2

SOLUTION:

A B  1 2  2 0
3 4 1 2

A B  4 4
10 8
The answer is
B  A  2 0  1 2 DIFFERENT
1 2 3 4
[PROVED]

B  A  2 4
7 10

LET’S PRACTICE 3

1. Given C  1 2 0 . Find 2C
0 1  3

Ans: 2 4 0
0 2  6

2. If A 2 0 1 Find 5A  (3A)
1 3 2
.

Ans: 4 0 2
 2 6 4

3. Given A 3 1 and B  2 3 . Find AB
4 2 1 5

Ans: 7 14
10 22

92

[CHAPTER 5: MATRIX]

X  3 1 Y   2 4
2 5   3 1 . Find XY .
4. If and

Ans:  3 11
 19 13

3 1 2 2 0
 2 4 0  4 . Find AB
5. If A and B   1 2

 3

Ans: 1 0
 8 16

 2 0 1  5 1  2
   1 
6. If A   3 5 2  and B  0 4  . Find AB

 4 1 4   2  3 3 

 8 5  7
 
Ans:  14 3 20 

 13  16 24 

 3 2 5 2  1 0
  3 2 . Find AB
7. If A  0 1 6  and B  5

 4 2  1 1 4 2

 5 7 6
 19 10
Ans:  3

 3 10 2 

93

[CHAPTER 5: MATRIX]

5.3 DETERMINANT

DETERMINANT OF A MATRIX
 Determinant of a matrix is a special number that can be calculated from a square
matrix.
 The symbol for determinant is two vertical lines either side.
 The determinant of a matrix may be negative or positive.
 Example:

A - means the determinant of the matrix A

CALCULATING THE DETERMINANT

 First of all the matrices must be square.
 have the same number of rows as columns.

i. FOR A 2  2 MATRIX

 For a 2 2 matrix (2 rows and 2 columns)

 Formula:

A  a b The determinant is;
c d  A  ad  bc

“BUTTERFLY” rule for matrix 2 2

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