Engineering Mathematics 1, Self Learning

3RD TOPIC : COMPLEX NUMBER

Example 3.2.2
Multiply the complex numbers below
c) (4 − ) 3 + 5
d) (3 − 5 )(4 − 2 )

Solution 3.2.2 (c)

(4 − ) 3 + 5 Expand the function

= 12 + 20 − 3 − 5 2 2 = −1

= 12 + 20 − 3 − 5 −1

= 12 + 20 − 3 + 5 Solution 3.2.2(d)
= 17 + 17

(3 − 5 )(4 − 2 ) Expand the
function

= 12 − 6 − 20 + 10 2 2 = −1

= 12 − 6 − 20 + 10(−1)

= 12 − 6 − 20 − 10

= 2 − 26

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3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.2.2

Simplify the following complex number:
a) 4 1 − 3
b) 2 2 + 5
c) 4 + 3 2 + 7
d) (7 − )(2 + 6 )
e) 3 − 3 4 − 4
f) 5 − 6 (4 + 2 )
g) 2 3 + 4 2 −
h) (9 − 2)(4 + 4 )
i) (4 + 5 )2

Answer :
a) 4 + 12
b) 4 + 10
c) −13 + 34
d) 20 + 30
e) −24
f) 17 − 24
g) 20 + 10
h) 28 − 44
i) −9 + 40

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3RD TOPIC : COMPLEX NUMBER

31.2.4OPPAERRTAITAILONFROAFCCTOIOMNPLEX NUMBER

3.2.3 Division of Complex Number

When it comes to division of complex number, make sure we solve it until
the denominator become the real number.

How to do it?
1. Put the division into fraction form
2. Multiply the numerator and denominator with conjugate of the

denominator
3. Simplify the expression

What is Conjugate?
The different sign of the imaginary part

Complex Number Conjugate
+ −
− +

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3RD TOPIC : COMPLEX NUMBER

31.2.4OPPAERRTAITAILONFROAFCCTOIOMNPLEX NUMBER

Steps in solving division of complex numbers
If given = + and = + , hence
Put in fraction form
+
= +
Multiply the fraction with
conjugate of denominator
+ −
= + × −
Expand numerator and
denominator
− + − 2
= 2 − + − 2 2
2 = −1
− + − (−1)
= 2 − 2(−1)
+ − +
= 2 + 2

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3RD TOPIC : COMPLEX NUMBER

Example 3.2.3

Solve the division of complex numbers below
2+3
a) 14−+2
b) 2−3

Solution 3.2.3 (a)

2 + 3 Solution 3.2.3 (b)
4 +
1 − 2
2 + 3 4 − 2 − 3
= 4 + × 4 −
1 − 2 2 + 3
8 − 2 + 12 − 3 2 = 2 − 3 × 2 + 3
= 16 − 4 + 4 − 2

8 + 10 − 3(−1) 2 + 3 − 4 − 6 2
= 16 − (−1) = 4 + 6 − 6 − 9 2

11 + 10 2 − − 6(−1)
= 16 + 1 = 4 − 9(−1)

= 11 + 10
17 17
2 + 6 −
= 4+9

8 −
= 13

= 8 − 1
13 13

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3RD TOPIC : COMPLEX NUMBER

Example 3.2.3

Solve the division of complex numbers below
4−4
c) 5 +3

d) 1+2
4

Solution 3.2.3 (c)

4 − 4
5 + 3

4 − 4 −5 + 3
= 5 + 3 × −5 + 3

−20 + 12 + 20 2 − 12 Solution 3.2.3 (d)
= −25 2 + 15 − 15 + 9
1 + 2
−32 + 12 + 20(−1) 4
= −25 −1 + 9
1 + 2 −4
−32 + 12 − 20 = 4 × −4
= 25 + 9
−4 − 8 2
−32 − 8 = −16 2
= 34
−4 − 8(−1)
16 4 = −16(−1)
17 17
= − − −4 + 8
= 16

= − 1 + 1
4 2

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3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.2.3

Simplify the following complex number:

a) 4−3 Answer :
4+2
a) 1 −
2
2+5
b) 2+3 b) 19 + 4
13 13

c) 2+6 c) 4 + 22
7− 25 25

d) 3−3 d) 3
4−4 4

e) 3+4 e) 2
2− 5

f) 9−2 f) − 9 − 1
4 4 2

4+5 g) 13 + 3
4 −2 8 4
g)

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3RD TOPIC : COMPLEX NUMBER

31.3.4APRAGRATNIDALDIAGRAM

3.3.1 Draw argand Diagram
Argand diagram is a graphical presentation of complex number.
If the complex number is = + , then a is a real part and bi is imaginary
part.
To present the complex number in Argand Diagram is just like presenting in
Cartesian Plane.
Therefore, in Argand Diagram, a will be in horizontal axis (x-axis) and bi will
be in vertical axis (y-axis).

Example:
If we have z= 3 + 4 , with 3 is real part and 4i is imaginary part,

y Imaginary axis

z=3+4i
4

Real axis

x
3

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3RD TOPIC : COMPLEX NUMBER

31.3.4APRAGRATNIDALDIAGRAM

3.3.2 Modulus and argument of complex number
In each Argand Diagram must have the value of Modulus and Argument of
complex number.

What is Modulus?
Modulus is the length from the origin to the point of complex number as we
call it line segment

How to calculate modulus:

= 2 + 2

Where x is real number, and y is imaginary number

What is Argument?

Argument is the angle generated from the positive x-axis to the line segment.

Just like the example in the page before, z= 3 + 4

How to calculate argument:

arg = −1


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3RD TOPIC : COMPLEX NUMBER

31.3.4APRAGRATNIDALDIAGRAM

y

Modulus is the

z=3+4i length of this

4 segment

Argumant, is the
3 angle generated

x

By referring to the diagram above, here’s how to find the value of modulus
and argument:

Modulus,

= 32 + 42
= 9 + 16
= 25
= 5unit

Argument, 4
3
= arg = −1

= 53.130

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3RD TOPIC : COMPLEX NUMBER

31.3.4APRAGRATNIDALDIAGRAM

Here’s we show you the argument principles for you to refer the value of
argument in different Quadrant

y y
z z



x x

1st quadrant 2nd quadrant
= arg = 180 − arg

y y



x x
z
z
4th quadrant
3rd quadrant = 360 − arg
= 180 + arg

is the argument, arg is the value from calculator

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3RD TOPIC : COMPLEX NUMBER

Example 3.3

1. Sketch the argand diagram for = 2 + 5 , hence calculate the modulus and
argument

Solution 3.3 x

1. z = 2 + 5i

y
= 2 + 5

5



2

Modulus = 22 + 52
= 4 + 25
= 29unit

Argument, 5
2
arg = −1 1st quadrant
= arg
= 68.2°

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3RD TOPIC : COMPLEX NUMBER

Example 3.3

2. Sketch the argand diagram for = −4, hence calculate the modulus and
argument

Solution 3.3
2. = − 4

y

= − 4 1
−4
x

Modulus = 12 + (−4)2
= 1 + 16
= 17unit

Argument, arg = −1 1 2nd quadrant
−4 = 180 − arg
= −14.04°
= 180 − 14.04
= 165.96°

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3RD TOPIC : COMPLEX NUMBER

Example 3.3

3. Sketch the argand diagram for = 5 − 3 , hence calculate the modulus and
argument

Solution 3.3 x
3. = 5 − 3 5

y = 5 − 3



−3

Modulus = 52 + (−3)2
= 25 + 9
= 34unit

Argument, arg = −1 −3 4th quadrant
5 = 360 − arg
= −30.96°
= 360 −
30.96

= 329.04°

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3RD TOPIC : COMPLEX NUMBER

Example 3.3

4. Sketch the argand diagram for = 5 + 3 − (8 + 6 ), hence calculate the
modulus and argument

Solution 3.3

4. = 5 + 3 − (8 + 6 ) y
= 5 + 3 − 8 − 6

= −3 − 3
x
−3

= −3 − 3 −3

Modulus = (−3)2+(−3)2
= 9+9
= 18unit

Argument, arg = −1 −3 3rd quadrant
−3 = 180 + arg
= 45°
= 180 +
45
= 225°

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3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.3

Sketch the argand diagram for the following, hence
calculate the modulus and argument
a) 1 − 3
b) 2 + 5
c) −4 − 3
d) 7 −
e) 3 − 3 + 4 − 4
f) 6 − (4 + 2 )
g) 4 − 2 −

Answer :

a) = 10 ; θ = 288.43
b) = 29 ; = 68.20°
c) = 5 ; θ = 216.87°
d) = 50 ; θ = 351.87°
e) = 98 ; θ = 315°
f) = 32 ; θ = 135°
g) = 5 ; = 26.57°

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3RD TOPIC : COMPLEX NUMBER

31.4.4FOPARRMTSIAOLF COMPLEX NUMBER

3.4.1 Other Forms of Complex Number

We normally know the complex number in terms of = + . Actually
there are other forms of complex number.

Forms of Complex Number Formula

Trigonometric Form (cos + sin )

Polar Form ∠
Exponential Form
Where must in radian

How it’s been generated into other form?

y = +




x

The coordinate of the Argand diagram is (a, b), then we convert the value of
a and b using the angle, and the length, :
+ = +
Therefore + = ( + )

And we can also write it in angle notation that we called it as polar form:
∠

The symbol of ∠ in polar form is called CIS.
Exponential form being generated by using Euler formula

= +

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3RD TOPIC : COMPLEX NUMBER

Example 3.4.1
1. Express = 5 − 3 in trigonometric form and polar form

Solution 3.4.1

1. = 5 − 3

To change from cartesian form to other form, we must find the
value of modulus and argument.

= 52 + (−3)2
= 25 + 9
= 34units

arg = −1 −3 4th quadrant
5 = 360 − arg

= −30.96°
= 360 − 30.96
= 329.04

Trigonometric form; = cos + sin
= 34 cos 329.04° + sin 329.04°

Polar form; = ∠
= 34∠329.04°

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116

3RD TOPIC : COMPLEX NUMBER

Example 3.4.1
2. Express = −2 − 6 in trigonometric form and exponential form

Solution 3.4.1

2. = −2 − 6

To change from cartesian form to other form, we must find the
value of modulus and argument.

= (−6)2+(−2)2
= 36 + 4
= 40units

arg = −1 −6
−2

= 71.56°
= 180 + 71.56

= 251.57

= 251.57 ×
180
= 4.39

Trigonometric form; = cos + sin
= 40 cos 251.57° + sin 251.57°

Exponential form; = The value of in
= 40 4.39 exponential form must

in radian

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3RD TOPIC : COMPLEX NUMBER

Example 3.4.1

3. Express = 5(cos 125° + sin 125°) in polar form, exponential form and
cartesian form

Solution 3.4.1

3. = 5(cos 125° + sin 125°)

= 5
125°
= 125 ×
= 180

= 2.18

Polar form; = ∠
= 5∠125°

Exponential form; = The value of in
= 5 2.18 exponential form must

in radian

Cartesian form; = 5(cos 125° + sin 125°)
= 5 −0.57 + 0.82i
= −2.85 + 4.1
Just use the calculator to
find cos 125° and
sin 125°

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3RD TOPIC : COMPLEX NUMBER

Example 3.4.1
4. Express = 4 2.5 in trigonometric form and cartesian form

Solution 3.4.1

3. = 4 2.5

=4 180
= 2.5
180 From radian to degree; ×

= 2.5 ×
180
= 143.24° From degree to radian, ×

Trigonometric form; = cos + sin
= 4 cos 143.24° + sin 143.24°

Cartesian form; = 4 cos 143.24° + sin 143.24°
= 4 −0.80 + 0.60
= −3.2 + 2.4
Just use the calculator to
find cos 143.24° and
sin 143.24°

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119

3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.4.1

Express the following complex number according to the form stated

a) 1 + 3 (Polar Form and Trigonometric Form)

b) 2 − 5 (Polar Form and Exponential Form)

c) 3.5 cos 130° + sin 130° (Exponential Form and Cartesian Form)

d) 4 2.5 (Trigonometric Form and Cartesian Form)

e) 1.6∠150° (Exponential Form and Cartesian Form)

Answer :

a) Polar form; = 10∠71.5°
Trigonometric form; = 10(cos 71.5° + sin 71.5°)

b) Polar Form; = 29∠291.80°
Exponential Form; 29 5.09

c) Exponential Form; = 3.5 2.27
Cartesian Form; = −2.24 + 2.70

d) Trigonometric Form; = 4 cos 143.24° + sin 143.24°
Cartesian Form; = −3.2 + 2.4

e) Exponential Form; = 21.6 2.61
Cartesian Form; = −1.39 + 0.8

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3RD TOPIC : COMPLEX NUMBER

31.4.4FOPARRMTSIAOLF COMPLEX NUMBER

3.4.2 Multiplication and division of Complex Number in Polar and
Trigonometric form

In multiplication and division of complex number for polar and
trigonometric form, note that,

When we multiply,

We will multiply the modulus, and add the argument

When we divide,

We will divide the modulus, and minus the argument

Multiplication and division Polar form
If given = ∠ and = ∠

× = ( × )∠( + )

= ∠( − )


Multiplication and division Trigonometric form

If given = + and = ( + )

× = ( × ) ( + ) + ( + )

= ( − ) + ( − )


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3RD TOPIC : COMPLEX NUMBER

Example 3.4.2

1. If = 4∠120° and = 2∠85° find,
a) ×

b)

Solution 3.4.2 (1)

a) × = 4 × 2 ∠ 120 + 85
= 8∠205°

b) = 4 ∠ 120 − 85
2
= 2∠35°

2. If = 3(cos 135° + sin 135°) and = 1.5(cos 60° + sin 60°) find,
a) ×

b)

Solution 3.4.2 (2)

a) × = 3 × 1.5 cos(135 + 60 ) + sin(135 + 60 )
= 4.5 cos 195° + sin 195°

b) = 3 cos(135 − 60 ) + sin(135 − 60 )
1.5
= 2(cos 75° + sin 75°)

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3RD TOPIC : COMPLEX NUMBER

Example 3.4.2
3. If = 5∠150° and = 3(cos 77° + sin 77°) find, × and give your

answer in Polar form

Solution 3.4.2 (3)
Change = 3(cos 77° + sin 77°) in polar form
= 3∠77°
× = 5 × 3∠150 + 77

= 15∠227° (answer in polar form)

4. If = 4(cos 30° + sin 30°) and = 1.2∠56° find and give your answer in

trigonometric form

Solution 3.4.2 (4)

Change = 1.2∠56° into trigonometric form
= 1.2(cos 56° + sin 56°)

==134.2.33cocsos30− 2−6 5 6+ + sin 30 − 56
sin(−26 )

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3RD TOPIC : COMPLEX NUMBER

Example 3.4.2

3. If = 5 2 and = 3(cos 77° + sin 77°) find, × and give your answer
in Trigonometric form

Solution 3.4.2 (3)

Step 1 :State the modulus and argument of z,

= 5 180

= 2 ×

= 90°

Step 2: Change z into trigonometric form
= 5 cos 90° + sin 90°

Step 3
× = (5 × 3)(cos 90 + 77 + sin 90 + 77 )
= 15(cos 167° + sin 167°)

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3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.4.2

Find × and for the following and give the answer as stated:


a) = 4 c 100° + sin 100° and = 2.3(cos 55° + sin 55°)
(give answer in Trigonometric Form)

b) = 1.8∠325° and = 5∠180°
(give answer in Polar Form)

c) = 3.5 c 150° + sin 150° and = 2∠50°
(give answer in Polar Form)

d) = 2(cos 90° + sin 90°) and = 4∠125°
(give answer in Trigonometric Form)

e) = 3.8 1.5 and = 1.9∠105°
(give answer in Polar Form)

Answer :

a) × = 9.2 cos 155° + sin 155° ; = 1.74 cos 45° + sin 45°

b) × = 9∠475°; = 0.36∠175°


c) × = 7∠200°; = 1.75∠100°


d) × = 8(cos 215° + sin 215°); = 1.5 cos(−35 ) + sin(−35 )


e) × = 7.22∠190.94°; = 2∠(−105°)


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3RD TOPIC : COMPLEX NUMBER

Let’s play games…………..

BasicComplex
Numbers

Operation of
Complex Numbers

Modulus, Argument
& Forms of Complex

Numbers

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REFERENCES

1. John Bird ed.(2014), Higher Engineering Mathematics, 7th
Edition, Bell and Bain Ltd. Glasgow.
ISBN : 978-0-315-85882-1

2. Ruhana Mat Kia et. al. (2017), Engineering Mathematics 1
for Polytechnics, 2nd Edition, Mathematics, Science and
Computer Department, Politeknik Port Dickson, Negeri
Sembilan.
ISBN : 978-967-5677-49-6

3. Tho Lai Hoong, Thum Lai Chun (2019), Kunci Emas
Formula A+ SPM Tingkatan 4 & 5 KBSM Matematik
Tambahan Tip Pemeriksa, Sasbadi Sdn. Bhd., Selangor.
ISBN : 978-983-77-1409-0

4. Chan Tze Hin(2019), Kunci Emas Formula A+ SPM
Tingkatan 4 & 5 KBSM Matematik Tambahan Revisi,
Sasbadi Sdn. Bhd., Selangor.
ISBN : 978-983-77-1407-6

5. Mathcentre, 2009, The Double Angle Formulae, accessed
7 September 2021,
<https://www.mathcentre.ac.uk/resources/uploaded/m
c-ty-doubleangle-2009-1.pdf>

6. A.J. Reynolds, 2010, Section 4 : Sine and Cosine Rule,
accessed 8 September 2021
<https://www.cimt.org.uk/projects/mepres/step-
up/sect4/index.htm>

7. Algebraic Rules, Retrieved 26 August 2021, from
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4176dcd9dc6d4ab9caea20.png>

8. Basic of Algebra (2005-2021) Retrieved 22 August 2021,
from
<https://www.wyzant.com/resources/lessons/math/alg
ebra/equation-basics/>

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9. Transposition of formulae , Retrieved 30 August 2021,
from <https://cpb-eu-
w2.wpmucdn.com/mypad.northampton.ac.uk/dist/d/63
34/files/2018/10/Algebra-Formulae-Transposition-
1chm7ai.pdf>

10. Math is Fun, 2020, Definition of Variables, accessed 13
September 2021,
https://www.mathsisfun.com/definitions/variable.html

11. Math is Fun, 2018, Definition of Coefficient, accessed 13
September 2021,
https://www.mathsisfun.com/definitions/coefficient.ht
ml

12. Pierce, Rod. (3 Aug 2019). "Complex Numbers". Math Is
Fun. Retrieved 26 Sep 2021 from
http://www.mathsisfun.com/numbers/complex-
numbers.html

13. Complex Number, Retrieved 26 August 2021, from <
https://www.cuemath.com/numbers/complex-
numbers/>

14. Murray Bourne, Polar Form of Complex Numbers. .
Intmath.com. Retrieved 26 Sep 2021 from
<https://www.intmath.com/complex-numbers/4-polar-
form.php>

15. Dawkins, p., 2021. Complex Number Primer. [online]
Tutorial.math.lamar.edu. Available at:
<https://tutorial.math.lamar.edu/Extras/ComplexPrime
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16. Courses.lumenlearning.com. 2021. Multiply and divide
complex numbers | College Algebra. [online] Available at:
<https://courses.lumenlearning.com/ivytech-
collegealgebra/chapter/multiply-and-divide-complex-
numbers/> [Accessed 26 September 2021].

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ATTRIBUTION

1. Figure 2.1 (Chapter 2 : Trigonometry) is an adaption of
“Image : Graphing y = tan x” from the Engineering
Mathematics 1 for polytechnics Second Edition (2017).
Image have been modified.

2. Figure 2.4 (Chapter 2 : Trigonometry) is an adaption of
“Image : Graphing f(x) = cos (x)” from the Engineering
Mathematics 1 for polytechnics Second Edition (2017).
Image have been modified.

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