engineering mathematics mohamed altemaly

8/19/2019 Engineering Mathematics Mohamed Altemaly

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Elminia University

Faculty of Engineering

ngineering Mathematics

Part 1

Dr. li Mohamed Eltamaly


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Preface

Most complex scientific and engineering models of the real world are

Differential equations. Here are some that I know of: heat flow,

electrostatic potential, waves (radio, light, sound, water), metal beam

bending, quantum mechanics, hydrogen bombs, electrons in telegraph

wires, optics, classical mechanics, general relativity, distributions of

organisms, ice sheets, tsunamis, air flow, ocean currents, weather,

auroras, blood flow, plate tectonics, supernovas. For this reason weintroduce this notes for students in faculty of engineer.

By the end of the course the student should:

•

be familiar with the concept of a complex number and be able

perform algebraic operations on complex numbers, both with

numeric and symbolic entries, solve simple equations with

complex roots, and in particular describe geometrically the

roots of unity;

• be familiar with the concept of a matrix and be able to perform

algebraic operations on matrices, both with numeric and

symbolic entries, be able to define a determinant and calculate

one both directly and by using row and column operations,

understand the definition and use of the inverse of a non-


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singular matrix, and be able to solve simple systems both using

inverses and reduction to triangular form, and be able to

compute inverses using Gaussian reduction and explain the

method in terms of elementary matrices;

•

be familiar with many topics in calculus like limits,

differentiations, and all methods of integrations;

• be familiar with first order differential equations (linear and

nonlinear) and their solution by many techniques;

• be familiar with many engineering applications of first order

differential equations like falling bodies, the time rate of change

in temperature of an object varies as the difference in

temperature between the object and surroundings, Chemical

Applications, time required for liquid tanks to get empty, Half

Life Of Nuclear Materials, and Electrical Circuits;

• be familiar with solution of higher order linear differential

equations with constant coefficients and Cuchy differential

equation and their solution by many techniques;

• be familiar with many engineering applications of higher order

differential equations like, Free Oscillation of suspended

bodies, Bending of Beams, and Electrical Circuits;

• be familiar with the benefits of Laplace and inverse Laplace

transforms, using Laplace transform for solving differentialequations, finding Laplace transform of any periodical and non


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periodical waveforms, using Laplace transform to solve many

application problems like Free Oscillation of suspended bodies,

Bending of Beams, and Electrical Circuits;

• be familiar with finding Fourier transform of any waveform by

advanced techniques like jump technique;

• be familiar with curve fitting by using least square technique and

using this technique to fiend Fourier transform for any waveform

numerically;

• be familiar with using power series for solving linear differential

equations of second order, and Bessel function;

•

be familiar with partial differentiation and solving partial

differential equations by many techniques as separation of

variables, Laplace transform, and Fourier transform;

• be familiar with solving differential equations which governs the

conduction of heat in solids;

•

be familiar with eigen values and eigen vectors and using them

for solving simultaneous linear differential equations;

• be familiar with special functions like Gamma and Beta functions;

• be familiar with many topics in numerical analysis like Numerical

solution of equations by many techniques like Simple Iteration,

Bisection, false position, Newton Raphson and secant method; and


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• be familiar with polynomial interpolation and numerical solution of

differential equations by many techniques like Euler’s and Runge-

Kutta’s method.


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Contents

Part 1

Chapter

No.

Title Page

No.

1 Mathematical Numbers 1

2 Matrices 42

3 Calculus 74

4 Ordinary Differential

Equations

101

5 Linear Differential Equations

Of Higher Order

151

6 Laplace Transforms 190

7 Fourier Series238

8 Least Square Technique 259


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References

Contents

Part 2

Chapter

No.

Title Page

No.

9 Power Series Solution Of

Differential Equations

285

10 Partial Differential Equations 346

11 Simultaneous Linear

Differential Equations

389

12 Special Functions 459

13 Numerical Analysis 478

Appendix 543

References 574


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Chapter 1

Mathematical Numbers

1.1 Natural Numbers

Natural numbers known as counting numbers are the numbers

beginning with 1, with each successive number greater than its

predecessor by 1. If the set of natural numbers is denoted by N, then

N = { 1, 2, 3, ......}

1.2 Whole Numbers

Whole numbers are the numbers beginning with 0, with each

successive number greater than its predecessor by 1. It combines the

set of natural numbers and the number 0. If the set of whole

numbers is denoted by W, then

W = { 0, 1, 2, 3, .......}

1.3 Integer Numbers

Integers are the numbers that are in either (1) the set of whole

numbers, or (2) the set of numbers that contain the negatives of the

natural numbers. If the set of integers is denoted by I, then

I = {......, -3, -2, -1, 0, 1, 2, 3, ......}

Positive integers are the numbers in I greater than 0. Negativenumbers are the numbers in I less than 0.

The number zero is neither positive nor negative, i.e., it is both

non-positive and non -negative.


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Mathematical Numbers2

Given the above definitions, the following statements about integers

can be made:

(1) N is the set of positive integers.

(2) W is the union of N and the number 0.

(3) The set of numbers that contain the negatives of the numbers in

N is the set of negative integers.

(4) I is the union of W and the set of negative integers.

1.4 Real Number Line

The set of real numbers can be pictorially represented by the real

number line. It is a straight line, whose "origin" is designated by the

number 0, and continues in both directions. All the positive integers

are ordered, in ascending order from left to right, to the right side of

0; all the negative integers are ordered, in descending order from

right to left, to the left side of 0. Notches are marked to denote the

position of these integers in the following figure (Fig.1).

Fig.1

Every point on the line corresponds to a real number, and every

real number can be paired with a point on this number line. If the

real number is an integer, its point on the number line coincides with

one of the notches for an integer; otherwise, its point lies between

two successive notches. All real numbers represented by points to


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Chapter One 3

the right of the number 0 are positive, while all real numbers

represented by points to the left of the number 0 are negative.

1.5 Absolute Values

The absolute value of a real number is the distance between its

corresponding point on the number line and the number 0. The

absolute value of the real number a is denoted by |a|.

From the diagram shown in Fig.2, it is clear that the absolute

value of non-negative numbers is the number itself, while the

absolute value of negative integers is the negative of the number.

Thus, the absolute value of a real number can be defined as follows:

For all real numbers a,

(1) If a > 0, then aa = .

(2) If a < 0, then aa −= .

Fig.2

Example 1:

| 2 | = 2| -4.5 | = 4.5

| 0 | = 0


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1.6 Complex Numbers

1.6.1 Introduction

The solution of a second order equation 02 =++ cbxax can be

obtained by the famous formula,a

acbb x

2

42 −±−=

For example, if 022 =−+ x , then we have:

( )2

31

2

91

2

811 ±−=

±−=

+±−= x

21 −=∴

or x As we see there is no problems with solving the above equation. But

if we solve the equation 0565 2 =+− x in the same way, we get:

( )10

646

10

100366 −±=

−±= x

And the next stage is now to determine the square root of (-64).

Is it (i) 8, (ii) -8, (iii) neither?It is, of course, neither, since + 8 and 8− are the square roots of

64 and not of (-64). In fact, ( )64− cannot be represented by an

ordinary number, for there is no real number whose square is a

negative quantity. However, 64*164 −=−

And therefore we can write:

( ) ( ) 1864*164*164 −=−=−=−


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Chapter One 5

Of course, we are still faced with ( )1− , which cannot be

evaluated as a real number, for the same reason as before, but, if we

can replace 1− with the letter j, then ( ) ( ) 88*164 j=−=− We now have a way of finishing off the quadratic equation we

started before as following:

0565 2 =+−

( )10

86

10

646

10

100366 j x

±=

−±=

−±=

8.06.0 +=∴ or 8.06.0 j x −=

1.6.2 Powers of j

( )

( ) ( ) 11

*1*

1

1

22

24

23

2

=−==

−=−==

−=

−=

j j

j j j j j

j

j

Note especially the last result: 14 = j . Every time a factor 4 j

occurs, it can be replaced by the factor 1, so that the power of j is

reduced to one of the four results above. In the same way we can

replace 12 −= j with –1.

The complex number 41 j x +=

, consists of two separate terms,1, and 4 j These terms cannot be combined any further, since the

second is an imaginary number (due to its having the factor j).


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In such an expression as 41 j+=

1 is called the real part of x

4 is called the imaginary part of x

The two together form what is called a complex number .

So, a Complex number = ( Real part ) + j( Imaginary part )

Complex numbers is very important especially in some

engineering application like electrical and mechanical engineering.

So we have to fully understand how to carry out the usual

arithmetical operations.

1.6.3 Addition and Subtraction of Complex Numbers.

Addition and Subtraction are quite easy as shown in the following

example:

Example 2 Find the results of the following arithmetical operations.

( ) ( )2673 j j −++ .

Solution :

( ) ( ) ( ) ( ) 59276326732673 j j j j j j +=−++=−++=−++So, in general, ( ) ( ) ( ) ( )d b jca jd c jba +++=+++

1.6.4 Multiplication of. Complex Numbers

The following example illustrate the multiplication process in

complex numbers.

Example 3 Find the results of the following arithmetical operations.

( )( )7532 J j ++


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Chapter One 7

Solution: These are multiplied together in just the same way as

you would determine the product ( )( )7532 j j ++ . Form the product

terms of

( )( )

2911

212910

21141510

7*37*25*35*27532

2

2

j

j

j j j

j j j j j

+−=

−+=

+++=

+++=++

If the expression contains more than two factors, we multiply the

factors together in stages:

( )( )( ) ( )

( )( )

( )( )

5136585122

58292222

212911

21212910

2121141510217532

2

2

j j

j j j

j j

j j

j j j j j j j

+=++−=

−++−=

−+−=

−−+=

−+++=−++

Example 4 Find the results of the following arithmetical

operations. ( )( )8585 j j −+

Solution:

( )( )

896425

644040258585 2

=+=

−−+=−+ j j j j j

In spite of what we said above, here we have a result containing no

imaginary term. The result is therefore entirely real. This is rather an

exceptional case. If we look at the two complex numbers we can


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find that they are identical except for the middle sign in the brackets

are different. These two complex numbers called conjugate complex

numbers and the product of two conjugate complex numbers is

always entirely real. In general we can say:

( )( ) 22 bababa −=−+ difference of two squares and there is no

any imaginary part.

1.6.5 Divison of. Complex Numbers

Division of a complex number by a real number is easy enough.

33.167.13

4

3

5

3

45 j j

j−=−=−

But how do we manage with dividing complex number with other

complex one? If we could, somehow, convert the denominator into a

real number, we could divide out as in the above example. So our

problem is really, how can we convert (4 + j3) into a completely real

denominator and this is explained in the previous item. We know

that we can convert (4 + j3) into a completely real number by

multiplying it by its conjugate (4 - j3). But if we multiply the

denominator by ( )34 j− , we must also multiply the numerator by

the same factor.

( )( )

( )( ) 25

3716

916

123728

3434

3447

34

47 j j

j j

j j

j

j −=

+

−−=

−+

−−=

+

−

48.164.025

37

25

16 j j −=−


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Chapter One 9

Then, to divide one complex number by another, therefore, we

multiply numerator and denominator by the conjugate of the

denominator. This will convert the denominator into a real number

and the final step can then be completed.

Example 5 Simplify the following expression:( )( )

43

2132

j

j j

+

−+

Solution:

( )( )

4.18.025

3520

169

43524

43

43*

43

8

43

8

43

62

43

2132

j j j

j

j

j

j

j

j

j

j

j

j j

−=−

=+

−−=

−

−

+

−=

+

−=

+

+−=

+

−+

Equal Complex Numbers

Now let us see what we can find out about two complex

numbers which we are told are equal.

Let the numbers , jba + and jd c + are equal

jd c jba +=+∴ Rearranging terms, we get ( )bd jca −=−∴

In this last statement the quantity on the left hand side is entirely

real, while that on the right hand side is entirely imaginary, i.e. a

real quantity equals an imaginary quantity. This seems contradictory

and in general it just cannot be true. But there is one special case for

which the statement can be true. That is when each side is zero.

( )bd jca −=−∴ can be true only if caieca ==− .,0

and if d biebd ==− .,0


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So we get this important result, If two complex numbers are equal

then,

(i) the two real parts are equal

(ii) the two imaginary parts are equal

For example, if 45 j jy +=+ , then we know 5= and 4= y .

1.6.6 Graphical Representation of a Complex Numbers

Although we cannot evaluate a complex number as a real number,

we can represent it diagrammatically, as we shall now see.

In the usual system of plotting numbers, the number 4 could be

represented by a line from the origin to the point 4 on the scale.

Likewise, a line to represent (-4) would be drawn from the origin to

the point (-4). These two lines are equal in length but are drawn in

opposite directions. Therefore, we put an arrow head on each to

distinguish between them as shown in Fig.3.

0 1 2 3 4

4

-1-2-3-4

-4

Fig.3

A line which represents a magnitude (by its length) and direction

(by the arrow head) is called a vector. We shall be using this word

quite a lot. Any vector therefore must include both magnitude (or

size) and direction. If we multiply (+4) by the factor (-1), we get(-4), i.e. the factor (-1) has the effect of turning the vector through

o180 as shown in Fig.4.


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Chapter One 11

0 1 2 3 4

4

o180

-1-2-3-4

-4

Fig.4

Multiplying by (-1) is equivalent to multiplying by2 j , i.e. by the

factor j twice. Therefore multiply in a single factor j will have half

the effect and rotate the vector through onlyo90 . So, the factor j

always turns a vector througho90 in the positive direction

measuring angles, i.e. anticlockwise. If we now multiply j 4 by a

further factor j, we get 42 j , i.e. (-4) and the following diagram

(Fig.5) agrees with this result. If we multiply (-4) by a further

factor j, sketch showing this new vector ( )4 j− is shown in Fig.6.

0-1-2-3-4

-4 4

o180

1 2 3 4

1

2

3

4

j40-1-2-3-4

-4 4

1 2 3 4

o

1801

2

3

4

j4

-1

-2

-3

-4

-j4

Fig.5 Fig.6

Let us denote the two reference lines by XX, and YY, as usual.

You will see that:

(i) The scale on the X-axis represents real numbers, XX1 is

therefore called the real axis.


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(ii) The scale on the Y-axis represents imaginary numbers, YY1 is

therefore called the imaginary axis.

If we now wish to represent 2+ 3 as the sum of two vectors, we

must draw them as a chain, the second vector starting where the first

one finishes as shown in Fig.7.

0 1 2 3 4 5

2 3

5

Fig.7

The two vectors, 2 and 3 are together equivalent to a single vector

drawn from the origin to the end of the final vector (giving naturally

that 2+3=5).

If we wish to represent the complex number (3 + j2), then we add

together the vectors which represent 3 and j2. Notice that the 3 is

now multiplied by a factor j which turns that vector through o90 .

The equivalent single vector to represent (2 + j3) is therefore the

vector from the beginning of the first vector (origin) to the end of

the last one. This graphical representation constitutes an Argand

diagram as shown in Fig.8.

0 1 2 3

1

2

3

2

j3

Fig.8


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Chapter One 13

Example 6 Draw an Argand diagram to represent the following

vectors: 231 j z += , 132 j z +−= , 423 j z −= , and 444 j z −−=

Solution: The Argand diagram of the above vectors are shown in

the following Fig.9.

0-1-2-3-4 1 2 3 4

1

2

3

-1

-2

-3

-4

-j4

3

j2 j1

j4

3

j4

z

1 z

2 z

3 z Fig.9.

1.6.7 Graphical Addition of Complex Numbers

Let us find the sum of 231 j z += and 422 j z −= by Argand

diagram. If we are adding vectors, they must be drawn as a chain.

We therefore draw at the end of 1 z , a vector representing 2 z in

magnitude and direction, and is parallel to it. In the same way, we

therefore draw at the end of 2 z , a vector representing 1 z in

magnitude and direction, and is parallel to it. Therefore we have a

parallelogram. Thus the sum of 1 z and 2 z is given by the vector

joining the starting point to the end of the last vector.


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The complex numbers 1 z and 2 z can thus be added together by

drawing the diagonal of the parallelogram formed by 1 z and

2 z .Thus, 25422321 j j j z z −=−++=+ which is clear that thisresults is the same as obtained from Fig10. So the sum of two

vectors on an Argand diagram is given by the diagonal of the

parallelogram of vectors.

0 1 2 3 4

1

2

-1

-2

-3

-4

-j4

3

j2

2 z

21 z z +5

1 z

Fig.10

Regarding to the subtraction it is quite similar to addition but the

only trick is simply this: ( )2121 z z z z −+=−

That is, we draw the vector representing 1 z and the negative

vector of 2 z and add them as before. The negative vector of 2 z issimply a vector with the same magnitude (or length) as 2 z but

pointing in the opposite direction.


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Chapter One 15

Example 7 If 231 j z += and 422 j z −= Find 21 z z −

Solution: It is clear from Argand diagram (Fig.11) that

6121

j z z +=− . We can now check for the above results:

61)42(23)42(2321 j j j j j Z Z +=+−++=−−+=−

0-1-2 1 2 3 4

1

2

3

4

-1

-2

-3

-4

-j4

3

j2

3

2 z

2 z −

u

5

6

21 z z −

1 z

Fig.1

1.6.8 Polar Form of a Complex Numbers

Complex numbers in the form jba + is called rectangular form.

Sometimes, it is convenient to express it in a different form. On an

Argand diagram shown in Fig.12, let OA be a vector jba + . Let r

= length of the vector and θ the angle made with OX. Since

jba z += , this can be written θ θ sincos jr r z += or

( )θ θ sincos jr z += This is called the polar form of the complex

number ba + , where: ( )

=+= −

aband bar 122 tan,, θ


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o

Ay

xa

b

r

θ

Fig.12

Example 8 Express 34 j z += in polar form.

Solution: First draw a sketch diagram (that always helps). We can

see that: 591634 22 =+=+=r and o87.364

3tan 1 =

= −θ

( )θ θ sincos jr jba z +=+= oo j z 87.36sin87.36cos5 +=∴

(i) r is called the modulus of the complex number z and is often

abbreviated to ( ) z mod or indicated by z

Thus if 43 j z += , ( ) 516943 22 =+=+=∴ z (ii) θ is called the argument of the complex number and can be

abbreviated to )(arg z . So, if 55 j z += then ( ) o z 45arg ==θ

Warning: In finding θ , there are of course two angles between o0

ando360 , the tangent of which has the value θ . We must be

careful to use the angle in the correct quadrant. Always draw a

sketch of the vector to ensure you have the right one. The follwing

table and Fig.13 show the correct angle range and quadrant.


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Chapter One 17

Value of a Value of b Angle range Quuadrant

+ve +ve oo 900


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But from Fig.14. This angle isoo 270180


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Chapter One 19

shorthand versionor θ ∠ to denote the polar form as shown in the

following examples:

Then, if 25 j z +−= ( ) 385.529425 ==+=r and fromabove 2.158=θ . Then, the full polar form is

oo j z 2.158sin2.158cos385.5 += and this can be shortened to

o z 2.158385.5 ∠= .

Example 11 express 34 − in shortened form.

Solution: ( ) 534 22 =+=r 75.0tan = E , o E 87.36=∴ o E 13.323360 =−=∴θ

ooo j z 8.323513.323sin13.323cos5 ∠=+=∴

Of course, given a complex number in polar form, you can convert

it into the basic rectangular form jba + simply by evaluating the

cosine and the sine and multiplying by the value of r .

Example 12 Find the rectangular form of the following: o z 355∠=

Solution:( )

868.3096.4

5736.08192.0535sin35cos5

j

j j z oo

+=

+=+=

1.6.9 Exponential Form of a complex numbers

There is still another way of expressing a complex number which

we must deal with. We shall Expalin it this way:

Many functions can be expressed as series. For example,


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......!4!3!2

1!

432

0

+++++== ∑∞

=

x x x x

m

xe

m

m x

(1)

.........!4!2

1)!2(

)1(cos42

0

2

+−+−=−= ∑∞

=

x xm x x

m

mm

(2)

.........!5!3)!12(

)1(sin

53

0

12

+−+−=+

−= ∑

∞

=

+ x x x

m

x x

m

mm

(3)

If we now take the series for x

e and write θ j in place of x, we get

the following series

( ) ( ) ( ) ( )!5!4!3!2

15432 θ θ θ θ

θ θ j j j j

je j +++++= (4)

( ) ( ) ( ) ( )+++++++=∴

!5!4!3!21

55443322 θ θ θ θ θ θ

j j j j je j

( ) ( ) ( ) ( )−−++−−+=∴

!5!4!3!21

5432θ θ θ θ

θ θ j j

je j

( ) ( ) ( ) ( )

−+−+

−+−=∴ ....

!5!3....

!4!21

5342 θ θ θ

θ θ θ j j je j (5)

It is clear from (2),(3) and (5) that the first bracket is in the form of

cosine and the second bracket in the form of sine.

θ θ θ sincos je j +=∴ (6)

Therefore, ( )θ θ sincos jr + can now be written asθ j

re . This is

called the exponential form of the complex number. It can be


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Chapter One 21

obtained from the polar form quite easily since the r value is the

same and the angle θ is the same in both.

The three ways of expressing a complex number are therefore

(i) jba z += (Rectangular form)

(ii) ( )θ θ sincos jr z += (Polar form)

(iii)θ jre z = (Exponential form)

And now a ward about negative angles. We know that:

θ θ θ sincos je j +=

if we replace θ by θ − in this result, we get the following:

( ) ( )θ θ

θ θ θ

sincos

sincos

j

je j

−=

−+−=−

So, θ θ θ sincos je j += And θ θ θ sincos je j −=−

There is one operation that we have been unable to carry out with

complex numbers before this. That is to find the logarithm of a

complex number. The exponential form now makes this possible,

since the exponential form consists only of products and powers.

For, if we haveθ jer z .= we can say: θ jr z += lnln

Example 13 Express4/1 π je − in the rectangular form:

Solution: Well now, we can write

( ) je

je

jeeee j j

−=

−=

−

== −−

122

1

2

1

4sin

4cos4/14/1

π π π π


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Since every complex number in polar form is of the same shape,

i.e. ( ) or jr θ θ θ ∠=+ sincos and differs from another complex

number simply by the values of r andθ

, we have a shorthandmethod of quoting the result in polar form.

Example 14 Express 34 j z −= in the polar form.

Solution: The vector has been drawn as shown in Fig. 16.

( ) 534 22 =+=r

Fig.16

From this 5=r , o E E 87.36,7.04

3tan −=∴−=−=

ooo 13.32387.36360 =−=∴θ

Then, the polar form isoo j z 13.323sin13.323cos5 += .

And the polar form iso je z 13.323.5= And the the shortened

form iso z 13.3235∠= . In this last example, we have

oo j z 13.323sin13.323cos5 +=


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Chapter One 23

Fig.17.

But the direction of the vector, measured from OX, could be given

as o87.36− , the minus sign showing that we are measuring the

angle in the opposite direction sense from the usual positive

direction. We could writeoo j z 87.36sin87.36cos5 −+−= .

But you already known as ( ) ( )θ θ coscos =− and

( ) ( )θ θ sinsin −=− .

oo

j z 87.36sin87.36cos5 −=∴

i.e. very much like the polar form but with a minus sign in the

middle. This comes about whenever we use negative angles.

In the same way we can say the following:

oooo j j z 110sin110cos5250sin250cos5 −+−=+=

It is sometimes convenient to use this form when the value of θ is

greater thano180 , i.e. in the 3rd and 4th quadrants. In the same

way we can write the following:


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ooooo j j z 1303130sin130cos3230sin230cos3 −∠=−=+=

( ) ( ) ooooo j j z 70470sin70cos4290sin290cos4 −∠=−=+= .

The polar form at first sight seems to be a complicated way ofrepresenting a complex number. However it is very useful as we

shall see.

Suppose we multiply together two complex numbers in this form:

Let ( )1111 sincos θ θ jr z += and ( )2222 sincos θ θ jr z +=

( ) ( )22211121 sincos*sincos* θ θ θ θ jr jr z z ++=∴

()21221

21212121

sinsinsincos

cossincoscos.*

θ θ θ θ

θ θ θ θ

j j

jr r z z

+++=∴

Rearranging the terms and remmembering 12 −= j we get:

( )

( )

++

−=

2121

21212121

sincoscossin

sinsincoscos.*

θ θ θ θ

θ θ θ θ

jr r z z

Now the brackets ( )2121 sinsincoscos θ θ θ θ − and

( )2121 sincoscossin θ θ θ θ + ought to ring the bell. What are they?

( ) ( )212121 cossinsincoscos θ θ θ θ θ θ +=−

( ) ( )212121 sinsincoscossin θ θ θ θ θ θ +=+

( ) ( )[ ]21212121 sincos.* θ θ θ θ +++=∴ jr r z z

Note: This is important result. Then we can say that to multiply

together two complex numbers in the polar form,

(i) Multiply the r 's together,

(ii) Add the angles, θ , together it is just easy as that.


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Chapter One 25

Example 15 Find the result of the following in the polar form:

403*302 ∠∠

Solution: It is easy to do that as we get:

( ) ( ) ooo 70640303*2403*302 ∠=+∠=∠∠ Now let us see if we can discover a similar set of rules for

division. We already know that to simplify 45 j+ we first

obtain a denominator that is entirely real by multiplying top and

bottom by the conjugate of the denominator i.e ( )45 j− Right.

Then let us do the same thing with 1 z and 2 z as following:

( )( )222

111

2

1

sincos

sincos

θ θ

θ θ

jr

jr

z

z

+

+=

( )( )

( )( )22

22

222

111

2

1

sincos

sincos*

sincos

sincos

θ θ

θ θ

θ θ

θ θ

j

j

jr

jr

z

z

−

−

+

+=∴

( )

( )22

22

21212121

2

1

2

1

sincos

sinsinsincoscossincoscos

θ θ

θ θ θ θ θ θ θ θ

+

+−+=∴

j j

r

r

z

z

( ) ( )1

sincoscossinsinsincoscos 21212121

2

1

2

1 θ θ θ θ θ θ θ θ −++=∴ j

r

r

z

z

( ) ( )( ) ( )212

12121

2

1

2

1 sincos θ θ θ θ θ θ −∠=−+−=∴r

r j

r

r

z

z

So, for division the rule is divide the r 's and subtract the angles θ 's.

Example 16 Simplify the following expressions: o

o

442

9510

∠

∠

,

Solution: ( ) oooo

o

51544955442

9510∠=−∠=

∠

∠


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1.6.10 DeMoivre’s Theorem

There is very important rule is called DeMoivre’s Theorem. It says

that to raise a complex number in polare form to any power n, we

raise the r to the power n and multiply the angle by n.

( )[ ] ( )θ θ θ θ n jnr jr nn sincossincos +=+∴

Example 17 Use DeMoivre’s Theorem to find the results of the

following expression in polar form: ( )[ ]3110sin110cos3 oo j+ Solution:

( )[ ] ( ) ( )( )oooo j j 110*3sin110*3cos3110sin110cos3 33 +=+

( )[ ] ( ) ( )( ) ooooo j j 33027330sin330cos27110sin110cos3 3 ∠=+=+∴This is where the polar form really comes into its own. For

DeMoivre's theorem also applies when we are raising the complex

number to a fractional power, i.e. when we are finding the roots of a

complex number as shown in the following example.

Example 18 Find the square root ofo z 449∠=

Solution:

We haveo

oo z 223

2

449449 ∠=

∠=∠=

Expansion of θ nsin and θ ncos

By DeMoiver ' s theorem, we know that:

( )n jn jn θ θ θ θ sincossincos +=+ where n is a positive integer.


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Chapter One 27

The method is simply to expand the right hand side as a binomial

series, after which we can equate real and imaginary parts. An

example will soon show you how it is done.

Example 19 To find expansions for θ 3cos and θ 3sin

Solution: We have ( ) ( )33sincos3sin3cos jsc j j +=+=+ θ θ θ θ

Where θ cos=c and θ sin= s just for simplicity.

Now expand this by the binomial series so that:

( ) ( ) ( ) ( )32233 33sincos3sin3cos js jsc jscc j j +++=+=+ θ θ θ θ

3223 333sin3cos jscs sc jc j −−+=+ θ θ

3223 333sin3cos s sc jcsc j −+−=+ θ θ

Now equating real parts and imaginary parts we get:

θ θ θ θ

θ θ θ θ

32

23

sinsincos33sin

sincos3cos3cos

−=

−=

If we wish, we can replace θ θ 22 cos1sin −= and

θ θ 22 sin1cos −= . So that we could write the results above as

following: θ θ θ cos3cos43cos 3 −= , θ θ θ 3sin4sin33sin −=∴ .

While these results are useful, it is really the method that counts. So

now do this one in just the same way as done before, obtain an

expansion for θ 4cos in terms of θ cos .

( ) ( )44sincos4sin4cos jsc j j +=+=+ θ θ θ θ


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( ) ( ) ( ) ( )432234 464 js jsc jsc jscc ++++=

432234 464 scs j sc sc jc +−−+=

334224 446 cs sc j s scc −++−=

Equating real parts:

( )( ) ( )22224

4224

116

64cos

cccc

s scc

−+−−=

+−=∴ θ

42424 2166 ccccc +−++−=

188 24 +−= cc

1cos8cos8 24 +−= θ θ

Similarly, ( ) ( )2233 4444sin sccscs sc −=−=θ

( ) scccs scs sc sccs

322

2222

4*414

14

==−=

+−−−=

θ θ θ sin*cos44sin 3=∴

Expansions for θ ncos and θ

nsin in terms of sines and cosines

of muiltiples of θ .

θ θ sincos j z += , θ θ sincos1 1 j z z

−==∴ −

θ cos21

=+∴ z

z and , θ sin21

j

z

z =−

Also by DeMoivre' s theorem θ θ n jn z n sincos +=


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Chapter One 29

And, θ θ n jn z n

sincos1

−=

θ n

z

z

n

n cos21

=+∴ And, θ n j

z

z n

n sin21

=−

Let us collect theses four results together: θ θ sincos j z +=

θ cos21

=+ z

z θ sin21

j z

z =−

θ n z

z n

n cos21

=+ θ n j z

z n

n sin21

=−

Example 20 Expand θ 3cos

Solution: From the previous results,

θ cos21

=+ z Q ( )33

cos21

θ =

+∴

z z

( )

3

3

32

233

1133

113

13cos2

z z z z

z z z

z z z

+++=

+

+

+=∴ θ

Now here is the trick: we rewrite this, collecting the terms up in

pairs from the two extreme ends, thus:

( )

++

+=

z z

z z

13

1cos2

3

33θ

But from the previous results

θ cos21

=+∴ z

z , and θ 3cos213

3 =+ z

z


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( ) θ θ θ cos2*33cos2cos2 3 +=∴

( )θ θ θ cos33cos41cos3 +=∴

Example 21 Expand θ 4sin

Solution:

θ sin21

j z

z =−Q , and, θ n j z

z n

n sin21

=−

( )44 1

sin2

−=∴

z z j θ

432

234 1141

61

4 z z

z z

z z

z z +

−

+

−=

61

41

2

2

4

4 +

+−

+=

z z

z z

Now, θ n z

z n

n cos21 =+Q

( )

62cos2*44cos2

61

41

sin22

2

4

44

+−=

+

+−

+=∴

θ θ

θ z

z z

z j

∴ 62cos2*44cos2sin16 4 +−= θ θ θ

( )32cos44cos81sin4 +−=∴ θ θ θ

θ θ θ cos63cos2cos8 3 +=∴


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Chapter One 31

1.6.11 The Roots of Unity

The problem here is to solve the equation 1=n z , where n is usually

a positive whole number.

Write both sides of the equation in polar form. Let z have polar form

( )θ θ sincos jr z += ( )θ θ n jnr z nn sincos +=∴

We know that ( )0sin0cos11 j+= . So our equation becomes:

( ) ( )0sin0cos*1sincos jn jnr z nn +=+=∴ θ θ

Now two complex numbers in standard polar form are equal if and

only if their modulus and arguments are equal. In the case of the

argument this statement has to be handled with care. It means are

equal if reduced to the proper range. So, for exampleo

10 and o370

count as equal from this point of view. So we can say that 1=nr

and that θ n and 0 are equal up to the addition of some multiple of

π 2 radians. π θ k nr

n

201 +== Where k is some wholenumber. Since r is real and positive, the only possibility for r is r = 1.

The other equation gives us:n

k π θ 20 +=

This, in principle, gives us infinitely many answers one for each

possible whole number k . But not all the answers are different.

Remember that changing the angle by π 2 does not change the

number z .

The distinct solutions, of which there are n, are given by 1=r and


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1,.........3,2,1,0,2 −== nk n

k π θ and we can write these

solutions as following: k k k j z θ θ sincos +=

Where 1,.........3,2,1,0,2 −== nk n

k π θ

That looks rather complicated. It becomes a lot simpler if you think

in terms of the Argand diagram. All the solutions have modulus 1

and so lie on the circle of radius 1 centered at the origin. The

solution with k = 1 is just z = 1. The other solutions are just 1−n

other points equally spaced round this circle, with angle n/2π

between one and the next. This is shown in Fig.18 for 17=n .

Fig.18 The n roots of 1.

Let's look at some specific examples. The cube roots of unity are the

solutions to 13

= z . There are three of them and they are:,3/2sin3/2cos,1 1 π π j z z o +==

3/4sin3/4cos2 π π j z +=


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Chapter One 33

Fig.19: The three cube roots of 1.

Note that 12 z z = ,212 z z = and 01 21 =++ z z . The roots are

shown in Fig.19.

Similarly the fourth roots of unity are the solutions of 14 = z and

these are: 1= z , j z = , 1−= z , and j z −=

A picture for n = 4 together with those for n = 5 and n= 6 is given

in Fig.20.

Fig.20 The nth roots of 1 for n = 4;5;6.

We can do other equations like this in much the same way.


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Example 22 Find the solutions of the equation j z =4 .

Solution:

Put ( )θ θ sincos jr z += . Then ( )θ θ 4sin4cos44

jr z += . We

know that: ( )2/sin2/cos1 π π j j += . So our equation becomes:

( ) ( )2/sin2/cos14sin4cos44 π π θ θ j jr z +=+=

Therefore; 1=r and π π

θ k 22

4 += or 28

π π θ

k +=

There are 4 distinct solutions, given by k = 0;1;2;3. They form a

square on the unit circle.

1.7 Polynomials

We have learned how to manipulate complex numbers, and

suggested that they will prove valuable in engineering calculations.

The original motivation for introducing them was to give the

equation 12 −= x two roots, namely j and j− , rather than it having

no roots. It turns out that this is all we have to do to ensure that

every polynomial has the right number of roots. We now discuss

this, and a number of other basic results about polynomials that are

quite useful to know.

A polynomial in x is a function of the form:

( ) on

nn

n a xa xa xa x p ++++= −

− 11

1 ....


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Chapter One 35

where the a's are (real or complex) numbers and 0≠na . For

example: ( ) ( ) 165,42 3483 −+−=+−= t t t t q x x x p

The highest power in the polynomial is called the degree of the

polynomial. The above examples have degrees 3 and 8.

A number a (real or complex) is said to be a root of the polynomial

( ) x p if ( ) 0=a p . Thus 1= x is a root of 0122 =+− x x

The first important result about polynomials is that a number a (real

or complex) is a root of the polynomial ( ) x p if and only if ( )a x −

is a factor of p( x), in the sense that we can write ( ) x p as:

( ) ( ) ( ) xqa x x p −= . Where ( ) xq is another polynomial. This result

is often called the remainder theorem . For example, 2= x is a root

of ( ) 2723 +−+= x x x x p and it turns out that

( ) ( ) 132 2 −+−= x x x x p

Note that necessarily the polynomial q has degree one less than the

degree of p. It may be the case that you can pull more than one

factor of a x − out of the polynomial. For example, 2 is a root of

( ) 12823 +−−= x x x x p and it turns out that

( ) ( )( )( )322 +−−= x x x x p

In such cases a is said to be a multiple root of ( ) x p . The multiplicity

of the root is the number of factors ( )a x − that you can take out. In

the above example, 2 is a root of multiplicity 2, or a double root . A


45/537


root is called a simple root if it produces only one factor. Multiple

roots are a considerable pain in the neck in many applications.

There is a simple test for multiplicity. Suppose a is a root of

( ) x p , so that ( ) 0=a p . If, in addition, ( ) 0=′ a p (derivative) then a

is a multiple root. To take the above example:

( ) 12823 +−−= x x x x pQ ( ) 823 2 −−=′∴ x x x p and ( ) 02 = p

and we have ( ) 02 =′ p , so we know that 2 is a multiple root.

2.9. Theorem (Fundamental Theorem of Algebra).

Let p be any polynomial of degree n. Then p can be factored into a

product of a constant and n factors of the form ( )a x − , where a

may be real or complex.

Also, the factorization is unique; you cannot find two essentially

different factorizations for the same polynomial. The factors need

not all be different because of multiple roots.

The fact that there cannot be more than n such factors is fairly

obvious, since we would have the wrong degree. What is not at all

obvious is that we have all the factors that we want. Note that this

result does not tell you how to find these factors; just that they must

be there!

The result is often stated loosely as: a polynomial of degree nmust have exactly n roots. You have to allow complex roots or the

theorem is not true. For example ( ) 12 += x x p has no real roots at


46/537

Chapter One 37

all. Its roots are j±= and it factorizes as ( ) ( )( ) j x j x x p +−= . In

fact, if 0≠ω then ( ) ω −= n z z p ( )1≥n always has exactly n

distinct roots because we know that it must have n roots in all and it

cannot have any multiple roots because ( ) 1−=′ nnz z p has only 0 as

a root and 0 is not a root of ( ) z p .

There is one other result about roots of polynomials that is worth

knowing. Suppose we have a polynomial with real , as opposed to

complex, coefficients. Suppose that the complex number z is a root

of the polynomial. Then the complex conjugate z is also a root. So

you get two roots for the price of one. You can see this in the

example of the previous paragraph. 12 + x has j as a root, so it

automatically must have j− as a root as well.

Example 23 Let ( ) 201694 234 +−+−= z z z z z p . Given that 2 +

j is a root, express ( ) z p as a product of real quadratic factors and

list all four roots, drawing attention to any conjugate pairs.

Solution:

Since p has real coefficients, and complex roots occur in pairs

consisting of a root and its complex conjugate. Given that j+2 is a

root, it follows that j−2 must also be a root, and so the quadratic:


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( )( ) ( )( ) 5422 2 +−=−−+− z z j z j z must be a factor. Dividing

the given polynomial by this factor gives

( ) 45420169422234

++−=+−+−= z z z z z z z z p

The roots of 42 + z are 2 j and its complex conjugate, j2− . Thus

the given polynomial, of degree four, has two pairs of complex

conjugate roots.

Example 24 Express 15 − z as a product of real linear and quadratic

factors.

Solution:

We rely on our knowledge of the nth roots of unity from the

previous section. Let

+

=

=

5

2sin

5

2cos

5

2exp

π π π α j j

Then the roots of 015 =− z are ,,,, 432 α α α α and, 1.

( ) ( )( ) 4322345 1111 α α α α −−−−−=++++−=− z z z z z z z z z z z

For convenience, write2α β = , and note that 3α β = while

4αα = . Our problem is to factorize 1234 ++++ z z z z as a

product of real quadratic factors. We know the roots are

β β α α and ,,, . Now construct the quadratic with roots α and α .

We have: ( )( ) ( ) ( ) 1222 +ℜ−=++−=−− α α α α α α α z z z z


48/537

Chapter One 39

where ( )α ℜ is the real part of α . Since ( )( ) β β −− z z behaves in

the same way, we have:

( ) ( ) ( ) 121211225

+ℜ−+ℜ−−=− β α z z z z

( )

+

−

+

−−=−∴ 1

5

4cos21

5

2cos211 225

π π z z z z

and this is a product of real linear and quadratic factors.

Problems

)

Express the complex number j z 31 −= exactly in

modulus - argument form. Hence find the modulus

and principal argument of4

z .

) Find all solutions w to the equation j273 −=ω and

mark them on an Argand diagram.

) Let 21 j z −= j+= 3ω be complex numbers.

Express each of the following complex numbers in

the rectangular form z z j j z

z −+++

31,2

, ω

ω

) Express the complex number 22 + exactly in

modulus - argument form. Hence find all solutions

w to the equation j22

3

+−=ω and mark them onan Argand diagram.


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) Let j z += 3 and j71 −=ω . Express z +ω

ω in a

rectangular form. Find also z z

ω

ω ,,

) Express the complex number j22 +− in polar

form. Hence solve the equation j z 223 +−=

expressing the solutions in polar form and marking

them in the Argand Diagram.

)

Let ( ) 882852345

+−−+−= z z z z z z p

Show that ( ) 02 = p . Show also that 222 +− z z is a factor of ( ) z p .

Hence write p as a product of linear factors.

) Show that ( ) j z +− 1 is a factor of the real

polynomial ( ) 862 23 +−+= z z z z p

Hence write p as a product of linear factors.

) Let ( ) 362753 234 −−+−= z z z z z p Show that

( ) 03 = j p . Hence write p as a product of linear

factors.

) Express in polar 35 j z −−=

) Express in rectangularo1562∠ and o375 −∠

) Ifoo j z 125sin125cos121 += and


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Chapter One 41

oo j z 72sin72cos52 += Then, find (i) 21 * z z and2

1

z

z giving the

results in polar form

) If oo j z 125sin125cos121 += , find3 z and

3

1

z

) If jy x z += , find the equations of the two loci

defined by: (i) 34 =− z and (ii) ( )6

2arg π

=+ z

)

If jy x z += , find the value of x and y when :

j j

z

j

z

−=+

− 3

43

1

3

) Express 32 j+ and 21 j− in polar form and

apply DeMoiver’s theorem to evaluate( )

21

324

j

j

−

+.

Express the result in rectangular and exponential

form.

) Find the fifth roots of 33 +− in polar and

exponential form.

) Express 125 j+ in polar form and hence

evaluate the principle value of ( )3

125 j+ givingthe results in rectangular form.


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) Obtain the expansion of θ 7sin in terms of

θ sin .


52/537

Chapter 2

Matrices

2.1 Introduction

A matrix is, by definition, a rectangular array of numeric or

algebraic quantities, which are subject to mathematical operations.

So a real matrix is an arrangement of real numbers into rows and

columns. Matrices can be defined in terms of their dimensions

(number of rows and columns). Let us take a look at a matrix with 4

rows and 3 columns (we denote it as a 4x3 matrix and call it A):

=

0

0

1

1

5

12

8

6

9

2

5

7

A

The dimensions of this matrix are 4 by 3. The dimensions of amatrix tell you the size of the matrix because they tell you the

number of rows and columns in the matrix. By convention, we list

the number of rows before the number of columns.

Definition 1 The dimensions of a matrix are the number of rows

and columns (listed in that order) of the matrix.

Each element of the matrix is named according to its position.

Typically, capital letters represent matrices and small letters with

subscripts represent elements in the matrix. Since vectors can be


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Chapter Two 43

considered to be matrices with only one row or one column, they

could be labeled with capital letters also. However, small letters

usually represents vectors. The element 6 is in the position a12 (read

a one two) because it is in row 1 and column 2. Also by convention,

we list the row number of the element before the column number.

An element in row i and column j would be denoted by ija . This

gives us a compact way to refer to specific elements of a matrix.

Can you represent the same information as before in a 3 by 4

matrix? Yes, you can. It would look like the matrix B which follows.

=

0

5

9

0

12

2

1

8

5

1

6

7

B

Matrix B is the transpose of A, and A is the transpose of B.

Transposing a matrix results in writing the columns as rows and the

rows as columns, but what really happens is that element ija is

placed in the position jib of the new matrix. Therefore, 12a moves

to the position 12b when we form the transpose of A. The transpose

of A is denoted byT

(read A transpose). Therefore, matrix B is

T .

Definition 2 By the transpose of the m by n matrix A, denoted by

T , we mean the n by m matrix, which has ija as its ( )

th ji,

element.


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Matrices44

Definition 3 We say that two m by n matrices, A and B are equal

if their corresponding elements are equal.

In other words, A = B if A and B have the same dimensions and

1111 ba = , 1212 ba = , etc. IsT = ? Usually not, but we have a

special word for a matrix which satisfiesT = .

Definition 4 A matrix is said to be symmetric ifT = .

Observe that the following matrix is symmetric:

=

3681

64058072

1529

A

Notice that jiij aa = for all i and j; as is true for all symmetric

matrices. Symmetric matrices are easy to spot because if you draw a

line down the main diagonal (from 9 to 3 in this matrix), then the

two halves are mirror images of each other. Symmetric matrices

have many special qualities that will be used when you study

matrices in more detail. The matrix A, given above, has another

special property; it is a square matrix because A has the same

number of rows as columns. Notice that A is a 4 by 4 square matrix.

We said that the main diagonal for A runs from 9 to 3. For any

square matrix, the main diagonal runs from the upper left corner to

the lower right corner.

Definition 5 We say that an m by n matrix is square if nm = .


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Chapter Two 45

2.2 Addition and Subtractions of Matrices

Definition 6 Matrices of the same dimensions are added by adding

corresponding elements.

For instance, ija corresponds to ijb because they both lie in the ith

row and jth column of their respective matrices. Therefore, we would

add, ijij ba + to obtain the ( ) ji,th element of + :

Example 1 Find the result of the following:

+

=+

0411

195

069

168

0

0

1

1

5

12

8

6

9

2

5

7

B A

Solution:

=

++

+

+

++

+

+

++

+

+

=+

09201217

11414

21215

0010

01

11

45912

68

66

11952

95

87

B A

Think about the similarities between addition and subtraction. How

do you think matrices are subtracted?

Definition 7 Matrices of the same dimensions are subtracted by

subtracting corresponding elements.

2.3 Multiplication of Matrices

Multiplying a matrix by a scalar value involves multiplying everyelement of the matrix by that value. Here we multiply our 4x3

matrix A by a scalar value k :


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Matrices46

=

=

0*

0*

1*

1*

5*

12*

8*

6*

9*

2*

5*

7*

0

0

1

1

5

12

8

6

9

2

5

7

**

k

k

k

k

k

k

k

k

k

k

k

k

k Ak

The multiplication operation on matrices differs significantly from

its real counterpart. One major difference is that multiplication can

be performed on matrices with different dimensions. The first

restriction is that the first matrix has to have the same amount of

columns as the second has rows. The reason for this will become

clear shortly. Another thing to note is that matrix multiplication is

not commutative i.e, (CD) does not equal ( DC ).

The procedure for matrix multiplication is rather simple. First, we

determine the dimensions of the resultant matrix. All we require is

that there are as many columns in the first matrix as there are rows

in the second. A simple way of determining is to look at the nearest

and farthest dimensions of two matrix symbols written next to eachother, for instance: C [2x3] D[3x2]. The nearest dimensions are both

equal to 3, and so we know that the operation is possible. The

farthest dimensions will give us the dimensions of the product

matrix, so our result will be a 2x2 matrix. The general rule says that

in order to perform the multiplication AB, where A is a mxn matrix

and B a k xl matrix, we must have n=k . The result will be a mxl

matrix.


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Chapter Two 47

Performing the operation product involves multiplying the cells

of a particular rows in the first matrix by the cells of a particular

column in the second matrix, adding the products, and storing the

result in the cell of the resultant matrix whose coordinates

correspond to the row of the first matrix and the column of the

second matrix. For instance, in AB = C , if we want to find the value

of c12, we must multiply the cells of row 1 in the first matrix by the

cells of column 2 in the second matrix and sum the results.

There are several interesting things to notice about matrix

multiplication. We multiplied a 1 by 3 matrix by a 3 by 4 matrix and

got a 1 by 4 matrix. The following picture expresses the

requirements on the dimensions:

Let's also look closely at how we multiply the matrices because we

will multiply matrices with larger dimensions later. This is a hands

on activity. Take your left pointer finger and place it at the

beginning of the first row of the first matrix (the only row we have

in this case). Take your right pointer finger and place it on the first

number of the first column of the second matrix. Multiply the two

numbers to which you are pointing. Each time you move, your left


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Matrices48

hand will go across the row, and your right hand will go down the

column. When you reach the end of the row and column, add the

numbers you have obtained from the multiplications. This number

goes in the first row and first column of your product matrix. This is

the same as taking the inner product of the first row of first matrix

and the first column of the second matrix. Now you can move to the

first row, second column doing the same thing. This number will go

in the first row, second column of your product matrix. In short,

position ij of your product matrix consists of the inner product of the

ith row of your first matrix and the jth column of the second matrix.

This is a lot easier to do than it is to describe! Your left hand will

move across and your right hand will move down. Do this for every

row and column combination to get your product matrix. This

picture depicts the motions necessary to find a product: Inner

product of row i with column j equals position ij

Definition 8 An identity matrix is a square matrix with ones along

the main diagonal and zeros elsewhere.

Example 2

If [ ]341=S And

=

23

1112

0

4

30

4

12

0122

R Find S *


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Matrices50

Example 4 Multiply the following matrices

=

=

14503

167

121548

125063

1204

9013708

51017

23

1112

0430

412

0122

* F R

2.4 Equations

Solving equations is an important part of mathematics. If we are

working with more than one unknown at a time, we need to solvesystems of equations. You may already know how to solve a system

of linear equations, but matrices provide a more compact way to

arrive at the solution. Matrices are also easier to manipulate on a

computer or calculator. Both of these facts will become more

important when you work with larger systems.

Example 5

Solve the following system of equations:

6624

9335

21

21

−=−−

=+

x x

x x

Solution: Let's look at a system of linear equations:

6624

9335

21

21

−=−−

=+

x x

x x

Can be written in matrix form as B X = where


61/537

Chapter Two 51

−−=

24

35 A ;

=

2

1

x

x X , and

−=

66

93 B

When you learned to solve systems of linear equations, you

learned that

(a) You arrive at the same solution no matter which equation you

write first,

(b) The solution doesn't change if you multiply an equation by a

scalar other than zero, and,

(c) You can replace an equation with the sum of that equation and

another equation without changing the solution.

These may not be exactly the words you used when you were

solving a system of linear equations, but you did all these things.

Experiment with the system above to convince yourself that these

statements are true. We can also solve this system entirely in matrix

form. We use the same rules, and we call them Elementary Row

Operations ( EROs). The EROs tell us that we can

(a) Interchange any two rows;

(b) Multiply any row by a non-zero scalar; and

(c) Replace any row by the sum of that row and any other row.

Proper use of EROs will leave us with a system that has the same

solution as our original system, but is much easier to solve. If you

were presented the system

b xa x == 21 ,


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Matrices52

You would be able to solve it instantly because you only have to

read the solution. If this system were written using matrix notation,

it would look like this:

=

b

a

x

x

2

1

10

01 The matrix

10

01is the 2 by 2 identity

matrix. Because you can just read of the solution when a system is

in this form, our first goal is to transform our system into this form.

Let's solve the system above using matrices. We can represent

this entire system with a 2 by 3 matrix, which looks like this:

−−− 66

93

24

35. This is called an augmented matrix because we

combined 2 matrices (a matrix and a vector for this system). In this

case, we combined the 2 by 2 coefficient matrix which is made of

the coefficients for our unknowns and the 2 by 1 matrix from the

right-hand side of the equations into one 2 by 3 matrix. In other

words, we put A to the left of the bar and put b to the right of the

bar. The application of an ERO to the augmented matrix does not

change the solution set of the linear system that the augmented

matrix represents because whatever you do to the left side of an

equation, you also do to the right side. Therefore, we will arrive at

the same solution whether we use augmented matrices or not, and

augmented matrices are more compact to write. Using matrix

notation, our goal is to transform our system into one that looks like

the following form:


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Chapter Two 53

b

a

10

01

In other words, we want the identity matrix to the left of the bar andthe solution to the right of the bar.

Remark 1 The bar is not a formal part of the matrix, so it is not

necessary. It is placed there so that we can refer to the different

parts of the augmented matrix and easily move back and forth

between the augmented matrix and the linear system that it

represents. In this book,1

r represents row 1 and so on.

−−− 66

93

24

35Original augment matrix

−−− 66

6.18

24

6.0151 ÷r

4.8

6.18

4.00

6.01214 r r +

4.021

6.18

10

6.012 ÷

r

12*6.021

6

10

01r r +−

When we convert this from augmented matrix notation back to the

algebraic notation for a system of equations, it looks like this:


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Chapter Two 55

0≠ A because aa /11 =− where 1−a is called the multiplicative

inverse or the reciprocal. There is something analogous to this with

matrices. It is also called the inverse. With scalars, 111

==

−−

aaaa .

Definition 9 The matrix1−

(called A inverse) is the inverse of a

square matrix A if I A == −− 11 where I is the identity matrix.

Once we find A 1; Ax = b can be solved by matrix multiplication

rather than Gauss Jordan elimination. We follow the algebraic steps

below to find an expression for x:

b Ax = b A Ax A 11 −−=∴ b A x I 1*

−=∴

This means that if we find1−; we only need to multiply to solve

systems with the same matrix A for different b vectors. Please

remember that11 −− ≠ Abb A , so you must multiply in the correct

order.

Remark 2 In computational mathematics, the inverse is very seldom found because other methods exist that serve the same

purpose and require fewer steps. However, the inverse will serve

our needs at this level and is important in the theory of matrices.

Example 7 Using the Gauss Jordan elimination method, let's find

1− where A

631

324

420

Solution:


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Matrices56

100

010

001

631

324

420

Original augmented matrix.

Switch 1r and 3r because we cannot have a zero on the main

diagonal, and we would prefer 1 rather 4.

−−−

001

410

100

420

21100

631

214 r r +−

−

001

4.01.00

100

420

1.210

631

( )10/2 −r

−−

− 8.02.014.01.00

100

2.0001.210

631

322 r r +−

−−

−

415

4.01.00

100

100

1.210

631

( )2.0/3 −r


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Chapter 3

Calculus

3.1 Limits

The concept of limits is essential to calculus. A good understanding

of limits will help explain many theories in calculus. So, it is

recommended to start studying calculus from limits.

Consider a function f defined for values of x, as x gets close to a

number a, not necessarily true for a= . If the value of ( ) x f

approaches a number b as x approaches a, then the limit of ( ) x f as

x approaches a is equal to b, denoted as :

b x f a x

=→

)(lim (1)

Example 1 Find the limit of 25)( += x x f as x approaches 3.

Solution: It is clear that as x approaches 3, 5 x approaches 15, and25 + x approaches 17. Thus; 1725lim

3=+

→ x

x

Example 2 Find the limits of102

1)(

−= x f as x approaches 5.

Solution: It is clear as x approach 5, 102 − x approaches zero the

102

1

− x

approaches

0

1 which is undefined. Thus;

)(102

1lim

5undefiend

x∞=

−→


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69/537

Calculus76

(or decreases ) indefinitely, then the limit of ( ) x f as x increases (or

decreases ) indefinitely is equal to b, denoted as :

b x f x

=+∞→

)(lim or b x f x

=−∞→

)(lim

A function ( ) x f is continuous at x = a if f is defined at x = a and

either; f is not defined anywhere near a, or f is defined arbitrarily

near a= and, )()(lim x f x f a x

=→

Conversely, A function ( ) x f is discontinuous at a x = if ( ) x f is

defined at a x = and ( ) x f is not continuous at a x = .

3.2 Derivatives

Suppose )( x f y = is shown in Fig.1, the slope of the curve is the

slope of the secant line between point A and another point P on the

graph is shown in the following equation:

( ) ( )( )( )

( ) ( )( )h

x f h x f

xh x

x f h x f m AP

−+=−+

−+=

Notice that h can change and with it the location of point P,

therefore h is the limiting factor of the slope of the curve. As h gets

close to point A, the slope of the curve becomes the tangent of the

graph at point A.

The tangent line of f at point A is:( ) ( )( )

h

x f h x f

h

−+

→0

lim

So, the Differentiation of function f at x is:

( ) ( )( )h

x f h x f

h

−+

→0lim


70/537

Chapter Three 77

If this limit exists, then it is called the derivative of function f

at x, which is denoted bydx

dyor x f )(′ .

So,( ) ( )( )h

x f h x f

dx

dyor x f

h

−+=′

→0lim)(

Fig.1 The Approximate slope of the curve at point A.

Fig.2 The slope of the curve at point A.

So, general rules of differentiation are shown in the appendix of this

book before going in the following example you have to take a look

to the rules of differentiation in the appendix.

Example 3 Find from the first principles( )( ) xe

dx

d tan


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Calculus78

Solution: Put ( ) xu tan= ue y =∴ uedu

dy=∴ and x

dx

du 2sec=

But from chain rule, dx

dy

du

dy

dx

dy

.=

,

( )( ) ( ) xeedx

d x x 2tantan sec*=∴

Example 4 Find

− 3 2 xdx

d

Solution:

−

−

−=−=

− 3

11

3

2

3 2

3

2

3

2 x x x

dx

d

Example 5 Find

−

−

1

3

2 xdx

d

Solution:

( )

−−=

−

− − 21

2

21*3

1

3 x

dx

d

xdx

d ( ) x x 2*1*2

323

2 −−=

( ) ( )222

2221

1*3

11

3

1

3

−

−=

−−=

−

−∴

x

x x

x x

x

xdx

d

Example 6 Find ( )4

75 + xdx

d

Solution: ( ) ( )334 75205*)75(475 +=+=+ x x xdx

d


72/537

Chapter Three 79

Example 7 Find ( )( )65sin + xdx

d

Solution: ( )( ) ( )65cos565sin +=+ x xdx

d

Example 8 Find ( )( )2cos xdx

d

Solution: ( )( ) ( ) ( )222 sin*22*sincos x x x x xdx

d −=−=

Example 9 Find ( )( ) xdx

d cos43ln −

Solution: ( )( ) ( ) x

x x

x x

dx

d

cos43

sin4sin4*

cos43

1cos43ln

−=

−=−

Example 10 Find ( )( )12log10 − xdx

d

Solution: ( )( )( ) ( ) ( ) ( )10ln12

22*

10ln12

112log10

−=

−=−

x x x

dx

d

Example 11 Find ( )( )12ln*5 − xedx

d x

Solution: Assume ( )12ln*5 −= xe y x , ( )12ln5 −== xvand eu x

uv y =∴ ,dx

duv

dx

dvu

dx

dy+=

( ) ( ) x x e x

xe

dx

dy 55 5*12ln2*12

1−+

−=∴

( ) ( )12ln5

12

2 55

−+−

=∴ xe x

e

dx

dy x x


73/537

Calculus80

Example 12 Find ( )( ) x xdx

d sinln3 5

Solution: Assume ( ) x x y sinln3 5= , ( ) xvand xu sinln3 5 ==

uv y =∴ ,dx

duv

dx

dvu

dx

dy+=

( )( ) ( ) 455 5*3*sinlncos*sin

13sinln3 x x x x

x x xdx

d +=∴

( )( ) ( ) x x x x x xdx

d sinln*15cot3sinln3 455 +=∴

Example 13 Find ( )

x

e

dx

d x

3ln

2

Solution: Assume ( ) xe

y x

3ln

2

= , ( ) xvand eu x 3ln2 ==

2v

dx

dvu

dx

duv

dx

dy −

=

( )

( )

( )( )2

222

3ln

3

3*2*3ln

3ln x

xee x

x

e

dx

d x x

x −=

∴

( ) ( )

( )( )

( ) ( )

( )( )2

2

2

2

3ln

)13ln2ln2(

3ln

)1

3ln2ln2(

x x

x x xe

x

x xe x

x

−+=

−+=

Example 14 Find( )

x

x x

dx

d

3cosh

)2sinh(2


74/537

Chapter Three 81

Solution: Assume( ) x x x

y3cosh

)2sinh(2= , 2u = , ( ) xv 2sinh= ,

and ( ) xw 3cosh=

Where

−+=

dx

dw

wdx

dv

vdx

du

uw

uv

w

uv

dx

d 111.

( ) ( ) ( ) ( )

( ) ( )

−

+=

∴

x x

x x

x x x

x x

x

x x

dx

d

3sinh3*3cosh

1

2cosh2*2sinh

12*

1*

3cosh

)2sinh(

3cosh

)2sinh(2

22

( ) ( )( )

( )( )( )

−+=

∴

x x

x x

x x x x

x x x

dxd

3cosh3sinh3

2sinh2cosh22*

3cosh)2sinh(

3cosh)2sinh(

22

( ) ( )( )

( )( )( )

−+=

∴

x

x

x

x

x x

x x

x

x x

dx

d

3cosh

3sinh3

2sinh

2cosh22*

3cosh

)2sinh(

3cosh

)2sinh( 22

( ) ( )( )

( )

( )( ) x x x x

x

x x

x

x x

x

x x

dx

d

3cosh3tanh)2sinh(3

3cosh

2cosh2

3cosh

)2sinh(2

3cosh

)2sinh(

2

22

−

+=

∴

Example 15 Find ( ) x x xdx

d 4cos2sin5

Solution:

Assume x x xuvw y 4cos2sin5== , where 5 xu = , xv 2sin= , and

xw 4cos=

Take the logarithm for both sides we get:


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Calculus82

( ) ( ) ( ) x x x x x x y 4cosln2sinlnln4cos2sinlnln 55 ++==∴

By differentiating both sides of the above equation we get:

( ) ( ) ( ) x

x x

x x xdx

dy y

4sin44cos

12cos22sin

1511 45 −++=∴

x x xdx

dy

y4tan42cot2

51−+=∴

−+=∴ x x

x x x x

dx

dy4tan42cot2

5*4cos2sin5

Example 16 Find

( )

32tan1 x

dx

d +

Solution: ( ) ( ) ( ) xdx

d x x

dx

d 2tan1*2tan132tan1

23 ++=+

( ) ( ) ( ) ( )

++=+∴ x

dx

d x x x

dx

d 22sec0*2tan132tan1 2

23

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