Engineering Mathematics 1

[CHAPTER 2:PARTIAL FRACTION]

d. x3
(x  2)(x  3)

Ans: x 1 8 2)  27
5(x  5(x  3)

48

[CHAPTER 3: TRIGONOMETRY]

TRIGONOMETRY

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

i. Introduction of trigonometric
ii. Define sine, cosine, tangent, cosecant, secant and cotangent.
iii. Graph of sine, cosine and tangent.
iv. Positive and negative value of trigonometric.
v. Apply sine and cosine rules.

3.1 INTRODUCTION OF TRIGONOMETRY

3.1.1 ANGLES AND THEIR MEASURE

ANGLES IN UNIT DEGREES AND RADIANS

 Angles are measured in either degrees or radians (rad).

 The size of a radian is determined by the requirement that there are 2 radians in a circle.

 Thus 2 radians equals 360 degrees.

 This mean that 1 radian  180 and 1 degree   radians.
, 180

EXAMPLE 1 (b) 135



1. Express in radian measure:

(a) 54

SOLUTION

(a) 54  54   0.9425 rad
180

(b) 135  135   3  rad
180 4

49

[CHAPTER 3: TRIGONOMETRY]

2. Express each angle in degree measure:

(a)  rad (b) 5 rad
3 9
SOLUTION

(a)  rad    180  60
3 3 

(b) 5 rad  5  180  100
9 9 

LET’S PRACTICE 1

1. Express in radian measure:

a) 330
b)  160
c) 42

2. Express each angle in degree Ans: 5.76,  2.79 & 0.733
measure: Ans: 76.390 & 7560

a) 4 rad
3

b) 251

RIGHT ANGLE TRIANGLE

 In mathematics, the pythagorean theorem, also known as Pythagoras’ theorem, is a
fundamental relation in Euclidean geometry among the three sides of a right triangle.

 It states that the area of the square whose side is the hyotenuse is equal to the sum of the
areas of the squares on the other two sides.

 By the Pythgorean theorem that the sum of the squares of each of the smallest sides equals
the square of the hypotenuse.

 Trigonometric functions are how the relationships amongst the length of the sides of a right
triangle vary as the other angles are changed.

 Formula;

c  a2 b2

50

[CHAPTER 3: TRIGONOMETRY]
3.1.2 TRIGONOMETRIC RATIOS

 The ratios of the sides of a right triangle are called trigonometric ratios.
 Three common trigonometric ratios are the sin (sin), cosine (cos) and tangent (tan).
 The three others trigonometric ratios are the cosecant (csc), secant (sec) and

cotangent (cot).

Trigonometric ratios in right triangles

SOH-CAH-TOA : an easy way to remember trigonometric ratios

 The word SOHCAHTOA helps us remember the definitions of sine, cosine and
tangent.

Table 1:

ACRONYM PART VERBAL DESCRIPTION MATHEMATICAL
DEFINITION

SOH Sine is Opposite over Hypotenuse sin A  opposite
hypotenuse

CAH Cosine is Adjacent over Hypotenuse cos A  adjacent
hypotenuse

TOA Tangent is Opposite over Adjacent tan A  opposite
adjacent

51

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 2

Find sin A and tan A Figure 2:

SOLUTION

sin A  opposite tan A  opposite
hypotenuse adjacent

sin A  3 tan A  3
5 4

LET’S PRACTICE 2

1.

Find: 5 12 12
13 13 5
a) cosF
b) sin F
c) tan F

Ans: , &

52

[CHAPTER 3: TRIGONOMETRY]

2.

a) sin H d) cscG
b) cos H e) secG
c) tan H f) cotG

Ans: 8 , 15 , 8 , 17 , 17 & 15
17 17 15 8 15 8

3.

Find: d) cosecB
e) sec B
a) sin B f) cot B
b) cos B
c) tan B

Ans: 5 , 12 , 5 , 13 , 13 & 12
13 13 12 5 12 5

4.

Given BAC  60 0 . Find the
length of BC.

Ans: 22.52

53

[CHAPTER 3: TRIGONOMETRY]

3.2 GRAPH OF SINE, COSINE AND TANGENT

GRAPH OF SINE

y  sin x

x0  90 0 00 90 0 1800 270 0 360 0

x rad    0   3 2
2
y 0 22 0
1
1 0 1

GRAPH OF COSINE

y  cos x

x0  90 0 00 90 0 1800 270 0 360 0

x rad    0   3 2
2
y 1 22
0
0 1 0 1

54

[CHAPTER 3: TRIGONOMETRY]

GRAPH OF TANGENT

y  tan x

x 0  90 0  45 0 00 45 0 90 0 1800 270 0

x rad      0   3
2 4 42
y 0 2
 1 1
0

 An asymptote is an imaginary line that a curve get closer and closer to but never
touches.

55

[CHAPTER 3: TRIGONOMETRY]

3.3 POSITIVE AND NEGATIVE VALUE OF TRIGONOMETRIC

FOUR QUADRANTS

RELATED ANGLE

 Reference angles allow us to evaluate more complex angles and makes easier when
evaluating angles.

 A reference angle x and x ' , is the positive acute angle made by the terminal side of the

angle x and x-axis.
 The figure shows differing angles that lies in quadrants I, II, III and IV.
 Remember that in quadrant I, an angle and its related angle are the same measure.

  180 0  a  a

  180 0  a   360 0  a
56
where;
= Actual angle

a = Reelated angle

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 3

Find the value of for sin   0.5 where 0 0    360 0

SOLUTION Sin positive Quadrant 1:
(Quadrant 1 and 2)   300
sin  0.5
Quadrant 2:
  sin 1 0.5   180 0  30 0  150 0
  300

Therefore;
  30 0 & 150 0

LET’S PRACTICE 3

1. Find the value of  for the followings where 0 0    360 0
a. cos  0.8264

Ans: 34.270 & 325.730

b. sin  0.8660

c. cos ec  5.461 Ans: 600 & 1200

Ans: 10.550 & 169.450

57

[CHAPTER 3: TRIGONOMETRY]

d. sec  1.5642

Ans: 50.260 & 309.740

2. Find the value of  for the followings where 0 0    360 0 ;
a. cos  0.9061

Ans: 154.970 & 205.030

b. tan  1.364

c. sin  0.5246 Ans: 126.250 & 306.250
d. cot  2.53
Ans: 211.640 & 328.360

Ans: 158.430 & 338.430

58

[CHAPTER 3: TRIGONOMETRY]

3. Find the values of  in the range 0 0    360 0 for each of the following,

a. sin 1   0.5870
2

Ans: 71.88 & 288.12

b. cos 1   0.7384
2

Ans: 275.2

c. cos2  0.4756

Ans: 30.8 , 149.2 , 210.8 & 329.2

59

[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 Acos A

cos 2A  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

60

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 3

Solve sin x  2  3where 0 0  x  360 0

SOLUTION Sin positive Quadrant 1:
(Quadrant 1 and 2) x  90 0
sin x  2  3 Quadrant 2:
sin x  1 Reference Angle x  180 0  90 0  90 0
x  sin 1 1
x  900 Therefore;
x  900

EXAMPLE 4

Solve the equation 2 cos 2   3 cos   1  0 for the values of  in the range 0 0  x  360 0

SOLUTION

2 cos2   3cos 1  0

2cos 1cos 1  0

2 cos  1  0 cos  1  0 Cos positive
cos  1 (Quadrant 1 and 4)
cos   1 Cos positive   cos 1 1 Reference angle
2 (Quadrant 1 and 4)   00

  cos 1  1  Reference angle
 2 

  600 Quadrant 1:

  00

Quadrant 1: Quadrant 4:

  600   360 0  00  360 0

Quadrant 4:

  360 0  60 0  300 0

Therefore;

 00,600,3000 &3600

61

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 5

1. Solve each of the following equations for 0 0    360 0 .

a. 4sin  3cos

Ans: 36.87 & 216.87

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 the equation 1  2sin  4cos2  0 for values of  in the range 0 0    360 0 .

Ans: 48.59 , 131.41 , 210 & 330

62

[CHAPTER 3: TRIGONOMETRY]

3. Solve each of the following equations for 0 0  x  360 0 .
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
e. 3 cos 2 x  2 sin x cos x
Ans: 30 & 150

Ans: 56.31 & 236.31

63

[CHAPTER 3: TRIGONOMETRY]

4. Solve each of the following equations for values of  in the range 0 0    360 0 .

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. 3cos  2 cot 

Ans: 41.81 & 138.19

64

[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 5 SOLUTION:
1.
 ABC  180 0  75 0  40 0
Find the length of AC.
ABC  65 0

AC 8.2
sin 650  sin 400

AC 8.2sin 650  11.56cm
 sin 400

2. 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 300
43  75

A  sin 1  43sin 300   16.660
75

65

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 6
1

Find the length of AC Ans: 4.52cm
2.

Find  ABC Ans: 31.69
3.

Find  ABC and the length of AB

Ans: 11.82 & 10.02cm

66

[CHAPTER 3: TRIGONOMETRY]
4.

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

5. Solve the triangles with the given parts:

a) a  45.7cm , A  68.20 , B  47.00
b) a  4.608m , b  3.207m , A  18.230
c) b  742mm , B  530 , C  3.50

Ans:

a) C  64.80 , b  36cm , c  44.54cm
b) B  12.580 , C  149.190 , c  7.545m
c) A  123.50 , a  774.75mm , c  56.72mm

67

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 6 SOLUTION:
1. Given a = 16.4cm, b = 11.8cm and C  670 .
c2  a2  b2  2abcosC
Find the length of c.
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  540

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

 B  1800  50.290  540  75.710

68

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 7

1. Given a = 7cm, b = 5cm and C  600 . Find
the length of c.

Ans: 6.24cm

2. Given a = 7cm, b = 5cm and c = 3cm. Find B
.

Ans: 38.210

3. Given a = 15cm, b = 12cm and c = 14cm.
Find A .

Ans: 69.990
69

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 8
1. Given ABC  640 . Find:

a. A
b. length of c

Ans: 74.050 & 6.4cm

2. Given the length of JL= 6.2cm, KL = 4.6cm and
K  520 . Find:
a. the angle of I
b. length of JK

Ans: 92.220 & 7.86cm

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

Ans: 35.010 & 62.250
70

[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 7

Given AC = 15cm, BC = 9cm and ACB  640 . SOLUTION:
Find the area of triangle.
Area of ABC;

 1 159sin 640
2

 60.67cm2

LET’S PRACTICE 9
1. Given AC = 6cm, BC = 5 cm and ACB  490 . Find
the area of triangle ABC.

2. Given DF = 8cm, EF = 12cm and DFE  1260 . Ans: 11.32cm2
71

[CHAPTER 3: TRIGONOMETRY]
Find the area of triangle DEF.

Ans: 38.83cm2

3. Given AC = 11cm, BC = 14.5cm and B  480 .
Find the area of triangle.

Ans: 64.18cm2

4. Given ED = 7cm, DF = 9cm and EF = 6cm. Find
the area of triangle DEF.

Ans: 20.98cm2
72

[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. Types of complex number

4.1 INTRODUCTION OF COMPLEX NUMBER

Introduction including: -

 Complex Number in general form
 Definition of  1
 Real part
 Imaginary part

Real Numbers are numbers like 1, 10.158, -0.34, 2/5, 3 , or any number
that we think of.

Imaginary Numbers when squared give a negative result

 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

73

[CHAPTER 4: COMPLEX NUMBER]
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
Purely Imaginary
3 + 2i 3 2
5 5 0
−6i 0 −6

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

74

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 1

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

SOLUTION:
1.  4  4(1)  4i2  2i
2. 7i  i 2  7i 3  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. e  e2  2e2

Simplify:
1. 2i  3i

2. 4   2 4. c   k 2
Ans:  6 , 4  2i , e  e2i , c  k 2i

75

[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 1  4 Add real part
( 3  2i ) + (1  7i ) Add imaginary part

2  7i  9i Answer

4  9i

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

SOLUTION 3  4  7 Add real part
Add imaginary part
(3  5i) + (4  3i)
Answer
5  3i  2i

7  2i

76

[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

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

77

[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 1  6 Substract real part
Substract imaginary part
( 7  3i ) - (1 7i )
Answer
3  7i  4i

6  4i

b. Perform the subtraction

SOLUTION 5  3  2 Substract real part
Substract imaginary part
5  3i  3  2i
Answer
3  2i  i

2i

78

[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  11i
Ans: 7  i
c. 8i  6   5 1i
Ans: 11  9i
d. 2  6i   4  9i Ans: 6  3i
Ans: 36  52i
e. 30  4i   6  48i Ans:  74  27i
Ans:  25  12i
b.  70  6i  4  21i Ans: 104  36i

c.  32i  11  14  20i 79

d. 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

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

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

 (1)  (2i)  (1) Remember:- i 2  1

 0  2i

c. Solve the equation 3  2i1 7i
3  2i1  7i  [(31)  (2  7)]  [(3 7)  (2 1)i]

 11  23i

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[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 4
Solve the following operation of complex number.

a. 8i  6 5 1i

Ans:  22  46i

b. 8  2i1  3i

Ans: 2  26i

c.  7  9i5  2i

Ans:  53  30i

d. 2  6i 4  9i

Ans:  44  6i

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 .

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[CHAPTER 4: COMPLEX NUMBER]

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

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
16  20i  20i  25i c. Add like terms (and notice how on the
bottom 20i  20i cancels out.
 8  10i  12i 15
16  20i  20i  25 d. Lastly we should put the answer back into
a  bi form
  7  22i
41

  7  22 i
41 41

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[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 5

Solve the following operation of complex number.

a. 5 1i
8i  6

Ans:  22  46i
 100
1 3i
b. 8  2i

Ans: 2  26i
68

c. 5  2i
 7  9i

Ans:  17  59i
57

d.  4  9i
 2  6i

Ans:  46  42i
40

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[CHAPTER 4: COMPLEX NUMBER]
 There is the faster way though. In the previous example, what happened at the bottom:

(4  5i)(4  5i)  16  20i  20i  25i 2
The middle terms (20i  20i) cancel out! Also i 2  1 so we end up with this:

(4  5i)(4  5i)  42  52
Which is really quite a simple result. The general rule is:

(a  bi)(a  bi)  a 2  b2

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.

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[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 and

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 6

1. Show the following complex number on Argands Diagram.

a. x  yi b. 3 5i

85

c. 4  7i [CHAPTER 4: COMPLEX NUMBER]

d. 2  5i

LET’S PRACTICE 6

Represent the following complex number on Argand Diagram
a. 4  9i

Ans:

b.  3  6i

Ans:

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[CHAPTER 4: COMPLEX NUMBER]

c.  7  2i

Ans:

d. 8  5i

Ans:

4.4.2 MODULUS AND ARGUMENT
 Refer to argand diagram below:

b z(a,b)

O a+bi

a real

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[CHAPTER 4: COMPLEX NUMBER]

 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.

z = x2  y2

 Argument of z is angle  ; tan  y .
x
 Hence, the argument, donated by;

  tan 1 y 
 x 

EXAMPLE 7

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

1. z  3  4i 2. z  4  7i

Modulus, z = 32  42 Modulus, z = 42  72

= 9  16 = 16  49
= 25 = 65
=5 = 8.06

Argument, θ = tan1 4 Argument, θ = tan1 7
3 4
= 53.13º
= 60.25 

θ in Quadrant 4;

  360  60.25 = 299.75º

Argand diagram: Argand diagram:

θ
θ

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[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 7
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

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[CHAPTER 4: COMPLEX NUMBER]

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( )
Trigonometric
r cos  i sin 
Exponent
 can be in degree( )
rei

 must be in radian (rad)

EXAMPLE 8

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

a. Find modulus z, z
z  42  92
 16  81
 97 ;
z  9.85

b. Find argument of z, 

Argument  = tan1 9
4

  66.04

Hence, z in polar form,

z  = 9.8566.04

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[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 8
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

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[CHAPTER 4: COMPLEX NUMBER]
EXAMPLE 9

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

z  530→ in polar form ( r  )

= 5cos 30  i sin 30

LET’S PRACTICE 9
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

Ans: 15(cos 53.13  i sin 53.13)

3. y  40270

Ans: 40(cos 270  i sin 270)

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[CHAPTER 4: COMPLEX NUMBER]

4. x  21.3e 2.13i

Ans:, 21.3(cos122  i sin 122)

EXAMPLE 10

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

r  z (modulus, z )

z = 62  42

= 36  16

= 52
= 7.21

Argument,  = tan1 4
6

= 33.69

Since z is in Quadrant 4,

 = 360  33.69= 326.31 (convert to radian)

=5.695rad

 z  rei  7.21e5.695i

 = 33.69

Z(6,-4)

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[CHAPTER 4: COMPLEX NUMBER]

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

Ans: 11.7e5.742i

2. r  11  52i

Ans: 53.2e1.36i

3. p  32275

Ans: 32e 4.8i

4. x  38(cos86  i sin 86)

Ans: 38e1.5i

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[CHAPTER 4: COMPLEX NUMBER]
EXAMPLE 11

If z  245 ,express z in Cartesian form
Firstly, write z in trigonometric form:

z = 2 cos 45  i sin 45
= 2 0.7071  i0.7071

= 1.4142  1.4142i - Cartesian form, a  bi

LET’S PRACTICE 11

1. State the following complex number in the form of polar and exponent.
z  42(cos 56  i sin 56)

Ans:
23.49  34.82i

2. y  34e1.48i

Ans: 30.83  33.86i

3. y  28130

Ans:  18  21.45i

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[CHAPTER 4: COMPLEX NUMBER]

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 12

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

a. z1  z2

SOLUTION:

= 335 547
= 3 5(35  47)

= 1582

b. z2
z1

SOLUTION:

= 547
335

= 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:

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

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[CHAPTER 4: COMPLEX NUMBER]

b. a  c
b

SOLUTION:

 414  745
6  71.5

 4  714  (71.5)  45
6

 4.67130.5

c. bc
a

SOLUTION:

 6  75  745
414

= 6  7  75  45  14
4

 10.5  44

d. c 2
ab

SOLUTION:

 (745) 2
414  6  71.5

 72 2  45
4  6(14  71.5)

 49 (90  61)
24

 2.04151

97