Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
College: Engineering
Campus: Bambang
DEGREE PROGRAM BSEE II
SPECIALIZATION
Electrical
YEAR LEVEL
2nd Year
COURSE NO.
COURSE TITLE
TIME FRAME
ELCE03
ENGINEERING MATHEMATICS FOR EE
6 hours
WK NO. 1
IM NO.
I.
UNIT TITLE/CHAPTER TITLE
Complex Algebra
II.
LESSON CONTENT
1. Complex Number
2. Operations of Complex Numbers
3. Modulus
4. Axiomatic Foundations
5. Graphical Representation of Complex Numbers
6. Forms of Complex Numbers
III.
LESSON OVERVIEW
This lesson is an introduction to the complex number system and the algebra of complex numbers.
IV.
DESIRED LEARNING OUTCOMES
At the end of the lesson, the students should be able to:
1.
2.
3.
4.
V.
define an imaginary number; a complex number;
distinguish the difference between a real number, an imaginary number, and a complex number;
convert complex numbers from one form to another; and,
perform complex algebra.
LESSON CONTENT
REAL NUMBER SYSTEM
The real number system consists of:
✓ Natural Numbers (or counting numbers) – these are positive integers such as 1, 2, 3, 4, 5, …
✓ Negative Integers & Zero (i.e., …, -5, -4, -3, -2, -1, 0)
✓ Rational Numbers (or fractions) – these are numbers that can not be reduced to a whole number
(i.e., 2/3, ¾, 6/13)
✓ Irrational Numbers (i.e., √2, 𝜋, 𝑒)
GRAPHICAL REPRESENTATION
-
origin: point corresponding to zero
negative: all numbers to the left of zero
positive: all numbers to the right of zero
COMPLEX NUMBER SYSTEM
- Origin: A mathematician wondered what the roots to 𝑥 2 + 1 = 0 are.
- This gave birth to the imaginary number 𝑖 = √−1.
▪
▪
▪
2
where 𝑖 2 = (√−1) = −1
𝑖 3 = 𝑖 2 ⋅ 𝑖 = (−1) ⋅ 𝑖 = −𝑖
𝑖 4 = (𝑖 2 )2 = (−1)2 = +1
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Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
-
In general, for 𝑖 𝑛 , divide 𝑛 by 4 and determine its remainder…
▪ if remainder is 0, 𝑖 𝑛 = +1
▪ if remainder is 1, 𝑖 𝑛 = 𝑖
▪ if remainder is 2, 𝑖 𝑛 = −1
▪ if remainder is 3, 𝑖 𝑛 = −𝑖
Example 1.1 Evaluate 𝑖 385 .
Solution:
𝑖 𝑛 = 𝑖 385
𝑛 = 385
𝑛
385
=
= 96 r. 1
4
4
∴ 𝑖 385 = 𝑖 4(96) ⋅ 𝑖1 = (+1)96 ⋅ 𝑖 = 𝒊
OPERATIONS OF COMPLEX NUMBERS
ADDITION
If 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖, the sum
𝑧1 + 𝑧2 = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖
(3
(−7
Example 1.2
+ 2𝑖) +
− 𝑖)
Solution:
Simply add the real parts 3 and −7, and then add the imaginary parts 2 and −1, separately as if
the imaginary number 𝑖 is a variable.
(3 − 7) + (2 − 1)𝑖
−𝟒 + 𝒊
SUBTRACTION
If 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖,
𝑧1 − 𝑧2 = (𝑎 − 𝑐) + (𝑏 − 𝑑)𝑖Example 1.3
(8 − 6𝑖) − (2𝑖 − 7)
Solution:
Just like in addition, perform the subtraction of the two given complex numbers as if the imaginary
number 𝑖 is a variable.
(8 + 7) + (−6 − 2)𝑖
𝟏𝟓 − 𝟖𝒊
MULTIPLICATION
If 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖,
𝑧1 ⋅ 𝑧2 = (𝑎 + 𝑏𝑖) ⋅ (𝑐 + 𝑑𝑖)
= (𝑎𝑐) + (𝑎𝑑𝑖) + (𝑏𝑐𝑖) + (𝑏𝑑𝑖 2 )
= 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑏𝑐𝑖 + 𝑏𝑑(−1)
= (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖
Example 1.4 (2 − 3𝑖)(4 + 2𝑖)
Solution:
(2 − 3𝑖)(4 + 2𝑖) = (2)(4) + (2)(2)𝑖 + (−3)(4)𝑖 + (−3)(2)(𝑖 2 )
= 8 + 4𝑖 − 12𝑖 − 6(−1)
= (8 + 6) + (4 − 12)𝑖
= 𝟏𝟒 − 𝟖𝒊
DIVISION
If 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖,
𝑧1 (𝑎 + 𝑏𝑖)
=
𝑧2 (𝑐 + 𝑑𝑖)
𝑎 + 𝑏𝑖 𝑐 − 𝑑𝑖
=
⋅
𝑐 + 𝑑𝑖 𝑐 − 𝑑𝑖
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INSTRUCTIONAL MODULE
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(𝑎𝑐) − (𝑎𝑑)𝑖 + (𝑏𝑐)𝑖 − (𝑏𝑑)𝑖 2
=
(𝑐)2 − (𝑐𝑑)𝑖 + (𝑐𝑑)𝑖 − (𝑑)2 𝑖 2
𝑎𝑐 − 𝑎𝑑𝑖 + 𝑏𝑐𝑖 − 𝑏𝑑(−1)
=
𝑐 2 − 𝑑 2 (−1)
(𝑎𝑐 + 𝑏𝑑) + (𝑏𝑐 − 𝑎𝑑)𝑖
=
𝑐2 + 𝑑2
NOTE: 𝑐 − 𝑑𝑖 is the complex conjugate of 𝑧2 = 𝑐 + 𝑑𝑖. (𝑧̅2 = ̅̅̅̅̅̅̅̅
𝑐 + 𝑑𝑖 = 𝑐 − 𝑑𝑖)
REMARKS: To get the complex conjugate of any complex number, simply change the sign of the
imaginary part from positive to negative, or vise versa.
Example 1.5 (3 − 2𝑖)/(−1 + 𝑖)
Solution:
3 − 2𝑖
3 − 2𝑖 −1 − 𝑖
=
⋅
−1 + 𝑖 −1 + 𝑖 −1 − 𝑖
(3)(−1) + (3)(−𝑖) + (−2𝑖)(−1) + (−2𝑖)(−𝑖)
=
(−1)(−1) + (−1)(−𝑖) + (𝑖)(−1) + (𝑖)(−𝑖)
−3 − 3𝑖 + 2𝑖 + 2𝑖 2
=
1 + 𝑖 − 𝑖 − 𝑖2
−3 − 3𝑖 + 2𝑖 + 2(−1)
=
1 − (−1)
(−3 − 2) + (−3 + 2)𝑖
=
2
−5 − 𝑖
=
2
−𝟓 𝟏
=
− 𝒊
𝟐
𝟐
ABSOLUTE VALUE
The absolute value or modulus of a complex number 𝑎 + 𝑏𝑖 is defined as |𝑎 + 𝑏𝑖| = √𝑎2 + 𝑏 2 .
Example 1.6 Evaluate |3 + 4𝑖|.
Solution:
|3 + 4𝑖| = √32 + 42 = √25 = 𝟓
If 𝑧1, 𝑧2 , 𝑧3 , …, 𝑧𝑚 are complex numbers, the following properties hold.
|𝑧1 𝑧2 … 𝑧𝑚 | = |𝑧1 | |𝑧2 | … |𝑧𝑚 |
1) |𝑧1 𝑧2 | = |𝑧1 | |𝑧2 | or
𝑧
|𝑧 |
𝑧2
2
2) | 1 | = |𝑧1 | if
𝑧2 ≠ 0
3) |𝑧1 + 𝑧2 | ≤ |𝑧1 | + |𝑧2 |
4) |𝑧1 + 𝑧2 | ≥ |𝑧1 | − |𝑧2 |
or
and
|𝑧1 + 𝑧2 + ⋯ + 𝑧𝑚 | ≤ |𝑧1 | + |𝑧2 | + ⋯ + |𝑧𝑚 |
|𝑧1 − 𝑧2 | ≥ |𝑧1 | − |𝑧2 |
AXIOMATIC FOUNDATIONS OF THE COMPLEX NUMBER SYSTEM
Let 𝑧1 = 𝑎 + 𝑏𝑖, 𝑧2 = 𝑐 + 𝑑𝑖, 𝑧3 = 𝑒 + 𝑓𝑖 belong to a set 𝑆 of complex numbers.
1. Equality 𝑎 + 𝑏𝑖 = 𝑐 + 𝑑𝑖 if and only if 𝑎 = 𝑐 & 𝑏 = 𝑑
2. Closure Law
𝑧1 + 𝑧2 & 𝑧1 𝑧2 belong to 𝑆
3. Commutative Law of Addition 𝑧1 + 𝑧2 = 𝑧2 + 𝑧1
4. Associative Law of Addition 𝑧1 + (𝑧2 + 𝑧3 ) = (𝑧1 + 𝑧2 ) + 𝑧3
5. Commutative Law of Multiplication
𝑧1 𝑧2 = 𝑧2 𝑧1
6. Associative Law of Multiplication
𝑧1 (𝑧2 𝑧3 ) = (𝑧1 𝑧2 )𝑧3
7. Distributive Law 𝑧1 (𝑧2 + 𝑧3 ) = 𝑧1 𝑧2 + 𝑧1 𝑧3
8a. Additive Identity 𝑧1 + 0 = 0 + 𝑧1 = 𝑧1 (NOTE: 0 is called the identity with respect to addition.)
8b. Multiplicative Identity 1 ⋅ 𝑧1 = 𝑧1 ⋅ 1 = 𝑧1
(NOTE: 1 is called the identity with respect to
multiplication.)
9. Additive Inverse For any complex number 𝑧1 there is a unique number 𝑧 in 𝑆 such that 𝑧 + 𝑧1 = 0;
𝑧 is called the inverse of 𝑧1 with respect to addition and is denoted by −𝑧1.
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NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
10. Multiplicative Inverse For any 𝑧1 ≠ 0 there is a unique number 𝑧 in 𝑆 such that 𝑧1 𝑧 = 𝑧𝑧1 = 1; 𝑧 is
called the inverse of 𝑧1 with respect to multiplication and is denoted by 𝑧1−1 or 1/𝑧1 .
GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS
If real scales are chosen on two mutually perpendicular axes 𝑋′ 𝑂𝑋 and 𝑌 ′ 𝑂𝑌 (called the 𝑥 and 𝑦 axes
respectively) as shown in the figure, we can locate any point in the plane determined by these lines by
the ordered pair of real numbers (𝑥, 𝑦) called rectangular coordinates of the point. Examples of the
location of such points indicated by 𝑃, 𝑄, 𝑅, 𝑆 and 𝑇 in the figure.
Since a complex number 𝑥 + 𝑖𝑦 can be considered as an ordered pair of real numbers, we can represent
such numbers by points in an 𝑥𝑦 plane called the complex plane or Argand diagram. The complex
number represented by 𝑃, for example, could then be read as either (3,4) or 3 + 4𝑖. To each complex
number there corresponds one and only one point in the plane, and conversely to each point in the plane
there corresponds one and only one complex number. Because of this we often refer to the complex
number 𝑧 as the point 𝑧. Sometimes we refer to the 𝑥 and 𝑦 axes as the real and imaginary axes
respectively and to the complex plane as the 𝑧 plane. The distance between two points 𝑧1 = 𝑥1 + 𝑖𝑦1 ,
and 𝑧2 = 𝑥2 + 𝑖𝑦2 in the complex plane is given by |𝑧1 − 𝑧2 | = √(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2 .
FORMS OF COMPLEX NUMBERS
✓ Rectangular Form: 𝑧 = 𝑎 + 𝑏𝑖
✓ Polar Form: 𝑧 = 𝑟∠𝜃
𝑟
✓ Exponential Form: 𝑧 = 𝑟𝑒 𝑖𝜃
✓ Trigonometric Form: 𝑧 = 𝑟 cis 𝜃 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
RECTANGULAR FORM
𝑧 = 𝑎 + 𝑏𝑖
POLAR FORM
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Bambang, Nueva Vizcaya
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𝑧 = 𝑟∠𝜃
Conversion from Rectangular to Polar
Conversion from Polar to Rectangular
𝑟 = √𝑎 2 + 𝑏 2
𝑏
𝜃 = tan−1
𝑎
𝑎 = 𝑟 cos 𝜃
𝑏 = 𝑟 sin 𝜃
EXPONENTIAL FORM
𝑧 = 𝑟𝑒 𝑖𝜃
𝑟
TRIGONOMETRIC FORM
𝑧 = 𝑟 cis 𝜃 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
NOTE:
= cos 𝜃 + 𝑖 sin 𝜃 is known as Euler’s Formula.
Example 1.7 Convert each of the given complex numbers into their three other forms:
1.7.1 𝑧 = 3 + 2𝑖
1.7.2 𝑧 = −3𝑖
1.7.3 𝑧 = 6 cis 240°
1.7.4 𝑧 = 2𝑒 −𝑖𝜋/4
𝑒 𝑖𝜃
Solution:
1.7.1 𝑧 = 3 + 2𝑖
The complex number 𝑧 = 3 + 2𝑖 is given in rectangular form as
𝑧 = 𝑎 + 𝑏𝑖 = 3 + 2𝑖
where 𝑎 = 3 and 𝑏 = 2.
Since 𝑎 and 𝑏 are both positive, as shown
the angle 𝜃 is found on the first quadrant of the complex plane (or Argand diagram) and is
found as
𝑏
2
𝜃 = tan−1 = tan−1 = 𝟑𝟑. 𝟔𝟗°
𝑎
3
The vector sum of 𝑎 and 𝑏 (or the hypotenuse of the right triangle formed together with 𝑎 and
𝑏) is found as
𝑟 = √𝑎2 + 𝑏 2 = √32 + 22 = √9 + 4 = √𝟏𝟑
Therefore, the polar form of 𝑧 = 3 + 2𝑖 is
𝑧 = 𝑟∠𝜃 = √𝟏𝟑∠𝟑𝟑. 𝟔𝟗°
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NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
For the exponential form of a complex number, take note of 𝜃 𝑟 . This means that the angle
affixed to the imaginary number must be in radians. We must therefore convert the angle 𝜃
from degrees to radians, and this can easily be achieved by simply multiplying the degree𝜋
angle by
. Thus, the exponential form of 𝑧 = 3 + 2𝑖 is
180°
𝑟
𝜋
𝑖(33.69°⋅
𝜋
)
180° = √𝟏𝟑𝒆𝟎.𝟓𝟖𝟖𝟎𝒊
𝑧 = 𝑟𝑒 𝑖𝜃 = 𝑟𝑒 𝑖𝜃⋅180° = √13𝑒
Finally, the trigonometric form of 𝑧 = 3 + 2𝑖 is
𝑧 = 𝑟 cis 𝜃 = √𝟏𝟑 𝐜𝐢𝐬 𝟑𝟑. 𝟔𝟗°
1.7.2
𝑧 = −3𝑖
The complex number 𝑧 = −3𝑖 is also given in rectangular form as
𝑧 = 𝑎 + 𝑏𝑖 = −3𝑖
where 𝑎 = 0 and 𝑏 = −3.
Since 𝑎 (the real part of the complex number) is zero, the complex vector is vertical with
direction pointing downwards (negative) as shown.
The angle 𝜃 is measured from the positive R-axis and can be solved as
−3
𝜃 = tan−1
= 270°
0
(NOTE: Trying this in a scientific calculator will result in “Math ERROR”. This is because the
−3
calculator is unable to solve for the
part of the equation. But as can be seen from the
0
Argand diagram, the terminal side of the angle has travelled exactly three quadrants from the
positive R-axis in counterclockwise direction. Since one quadrant is equivalent to 90°, three
quadrants is therefore equivalent to 3 × 90° = 270°.)
𝑟 can be solved as
𝑟 = √𝑎2 + 𝑏 2 = √02 + (−3)2 = √9 = 3
Therefore, the polar form of 𝑧 = −3𝑖 is
𝑧 = 𝑟∠𝜃 = 𝟑∠𝟐𝟕𝟎°
Furthermore, the exponential form of 𝑧 = −3𝑖 is
𝑟
𝑖(270°⋅
𝜋
)
180° = 𝟑𝒆𝒊 𝟑𝝅/𝟐
𝑧 = 𝑟𝑒 𝑖𝜃 = 3𝑒
Finally, the trigonometric form of 𝑧 = −3𝑖 is
𝑧 = 𝑟 cis 𝜃 = 𝟑 𝐜𝐢𝐬 𝟐𝟕𝟎°
1.7.3
𝑧 = 6 cis 240°
The complex number 𝑧 = 6 cis 240° is given in trigonometric form as
𝑧 = 𝑟 cis 𝜃 = 6 cis 240°
where 𝑟 = 6 and 𝜃 = 240°.
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Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
The polar form of 𝑧 = 6 cis 240° is
𝑧 = 𝑟∠𝜃 = 𝟔∠𝟐𝟒𝟎°
The exponential form of 𝑧 = 6 cis 240° is
𝑟
𝑖(240°⋅
𝜋
)
180° = 6𝑒 𝑖 4𝜋/3
𝑧 = 𝑟𝑒 𝑖𝜃 = 6𝑒
As for the rectangular form, 𝑎 and 𝑏 can be solved as
𝑎 = 𝑟 cos 𝜃 = 6 cos 240° = −3
𝑏 = 𝑟 sin 𝜃 = 6 sin 240° = −3√3
respectively. Thus, the rectangular form of 𝑧 = 6 cis 240° is
𝑧 = 𝑎 + 𝑏𝑖 = −𝟑 − 𝟑√𝟑 𝒊
1.7.4
𝑧 = 2𝑒 −𝑖𝜋/4
The complex number 𝑧 = 2𝑒 −𝑖𝜋/4 is given in exponential form as
𝑟
𝑧 = 𝑟𝑒 𝑖𝜃 = 2𝑒 −𝑖𝜋/4
−𝜋
where 𝑟 = 2 and 𝜃 𝑟 = .
4
(NOTE: The angle 𝜃 𝑟 =
−𝜋
4
is measured from the positive R-axis in a clockwise direction,
𝜋
4
hence the negative angle. Though this is not incorrect, this is not equivalent to . Positive
𝜋
𝜋
4
is located on the first quadrant of an Argand diagram while negative is located on the fourth
4
quadrant.
Measured in a counterclockwise direction, this angle is equivalent to
7𝜋
7𝜋 180°
𝜃 𝑟 = or 𝜃 = ⋅
= 315°.)
4
4
𝜋
The polar form of 𝑧 = 2𝑒 −𝑖𝜋/4 is
𝑧 = 𝑟∠𝜃 = 𝟐∠𝟑𝟏𝟓°
The trigonometric form of 𝑧 =
is
𝑧 = 𝑟 cis 𝜃 = 𝟐 𝐜𝐢𝐬 𝟑𝟏𝟓°
And the rectangular form of 𝑧 = 2𝑒 −𝑖𝜋/4 is
𝑧 = 𝑎 + 𝑏𝑖 = (𝑟 cos 𝜃) + (𝑟 sin 𝜃)𝑖 = (2 cos 315°) + (2 sin 315°)𝑖 = √𝟐 − √𝟐 𝒊
2𝑒 −𝑖𝜋/4
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IM No.: ELCE03-2S-2021-2022
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VI.
LEARNING ACTIVITIES
NAME: _____________________________________
COURSE-YR: ________________________
SCORE: _____________________
DATE: _______________________
Activity 1.1
Plot each of the given complex numbers in an Argand diagram and label all the necessary parts. Also,
convert each to its three other forms.
1. 𝑧 = −1 + 2𝑖
2. 𝑧 = −√6 − √2 𝑖
3. 𝑧 = 14.7∠45°49′
4. 𝑧 = −3 cis 310°
5. 𝑧 = 5𝑒 𝑖𝜋
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Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
VII.
EVALUATION (Note: Not to be included in the student’s copy of the IM)
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“In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293,
the copy righted works included in this material may be reproduced for
educational purposes only and not for commercial distribution”
Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
NAME: _____________________________________
COURSE-YR: ________________________
IM No.: ELCE03-2S-2021-2022
SCORE: _____________________
DATE: _______________________
Quiz 1.1
Given: 𝑧1 = 1 + 2𝑖
𝑧2 = 2 − 3𝑖
1.
2.
3.
4.
5.
(10 pts.)
(10 pts.)
(10 pts.)
|3𝑧1 − 4𝑧2 | =?
𝑧13 − 3𝑧12 + 4𝑧1 − 8 =?
Simplify:
(15 pts.)
(15 pts.)
3𝑖 30 − 𝑖19
2𝑖 − 1
Convert item 2 into its three other forms.
Convert item 3 into its three other forms.
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“In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293,
the copy righted works included in this material may be reproduced for
educational purposes only and not for commercial distribution”
Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
VIII.
ASSIGNMENT
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“In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293,
the copy righted works included in this material may be reproduced for
educational purposes only and not for commercial distribution”
Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
NAME: _____________________________________
COURSE-YR: ________________________
IM No.: ELCE03-2S-2021-2022
SCORE: _____________________
DATE: _______________________
Assignment 1.1
Simplify each of the following:
1. (5 + 3𝑖) + {(−1 + 2𝑖) − (7 − 5𝑖)}
2. (8 − 6𝑖) − (−7 − 𝑖)
3. (4 + 2𝑖)(2 − 3𝑖)(2 − 𝑖)
4. (−1 + 2𝑖){(7 − 5𝑖) + (−3 + 4𝑖)}
5+5𝑖
20
5.
+
3−4𝑖
4+3𝑖
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NVSU-FR-ICD-05-00 (081220)
“In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293,
the copy righted works included in this material may be reproduced for
educational purposes only and not for commercial distribution”
Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bambang, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.: ELCE03-2S-2021-2022
IX.
A)
REFERENCES
Book/Printed Resource
Bird, J. (2020). Higher Engineering Mathematics, 6th Edition. Oxford, UK: Elsevier's Science and
Technology Rights Department.
Dulay, V. N. (2001). Reviewer/Text in Advanced Engineering Mathematics. Manila: Rex Book Store.
Kreszig, E. (2006). Advanced Engineering Mathematics 10th Edition. New York: John Wiley and Sons,
Incorporated.
Spiegel, M. R. (1965). Schaum"s Outline of Theory and Problems of Laplace Transforms. USA: McGraw
Hill, Incorporated.
B)
ONLINE RESOURCES
http://www.soton.ac.uk/~cjg/eng1/modules/modules.html
http://www.mit.opencourseware.com
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“In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293,
the copy righted works included in this material may be reproduced for
educational purposes only and not for commercial distribution”