Chapter 2- Complex Number

7/30/2019 Chapter 2- Complex Number

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EBE 101

NAME OF SCHOOLFACULTY OF ENGINEERING AND INFORMATIONTECHNOLOGY

PREPARED BY:

SYED ZAHURUL ISLAM

CHAPTER2CHAPTER2

ENGINEERING MATHEMATICS I

COMPLEX FUNCTIONS & VECTORALGEBRA

COMPLEX FUNCTIONS & VECTORALGEBRA


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CHAPTER 2: Complex Function & Vector Algebra

TOPIC

Introduction

Shorthand notation

Multiplication in polar coordinatesDivision in polar coordinates

deMoivres theorem

Roots of a complex number

Trigonometric expansions

Loci problems

Programme 2: Complex numbers 2


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TOPIC

Introduction


TheThe polar form of a complex number is

readily obtained from the Argand diagram

of the number in Cartesian form.

Given:

then:

and

z a jb= +

2 22 2 2 sor a b r a b= + = +

1tan so tanb ba a

= =

= (a,b)

Im


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TOPIC

Example:

z = 4 j3 = 4 + j(-3)Here, a = 4

b = -3

r = {(16) + 9 = 25 = 5

tan = (-3/4)

= tan-1 (-3/4) = - 36.8698

180 36.8698 = 143.1302

- Sin 30 = sin (180 + 30) = Sin 210 = sin 330

1st Q2nd Q

3rd Q4th

Q

ALL +sin and Cosine

tan, cot cos and sec


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TOPIC

Example:

z = 4 j3 = 4 + j(-3)

Here, a = 4

b = -3

r = {(16) + 9 = 25 = 5

tan = (-3/4)

= tan-1 (-3/4) = - 36.8698

180 36.8698 = 143.1302

- Sin 30 = sin (180 + 30) = Sin 210 = sin 3304, -3

r = 5

x

Y, (Im)


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TOPIC

Example 1: Expressz = 4 j3 in polar form

ANS. 5 (cos 323o8 + j sin323o8)

Example 2: Expressz = - 5 i3 in polar form

z= r (cos + i sin)

z = 34 (cos 30.96 + i sin 30.96)

Example 3: Expressz = 1 + i3 in polar form

ANS. z = 2 (cos /3 + i sin /3)


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TOPIC

Example 3. If z1 = r1 (cosA + jsinA) and z2 = r2 (cosB + jsinB)

show that z1.z2 = r1r2 [cos (A B) + j sin(A+B)]


show that z1/z2 = (r1/r2) [cos (A B) + j sin(A - B)]

Example 5. if z1 = 8(cos 65 + jsin 65) and z2 = 4(cos 23 + jsin 23)

Then, find z1.z2Find z1/z2


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TOPIC


show that z1.z2 = r1r2 [cos (A B) + j sin(A+B)]

z1.z2 = r1.r2 [(cosA + jsinA) . (cosB + jsinB)]= r1.r2 [cosA cosB + cosA.jsinB + jsinA cosB + j2 sinA sinB]

= r1.r2 [cosA cosB + cosA.jsinB + jsinA cosB - sinA sinB]

= r1.r2 [cosA cosB - sinA sinB + j (sinA cosB + cosA sinB)]

= r1.r2 [cos (A - B) + j sin (A+B)]

j = -1


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TOPIC

Shorthand notation

Positive angles


The shorthand notation for a positive angle (anti-clockwise rotation) is

given as, for example:

With the modulus outside the bracket and the angle inside the bracket.

z r=


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TOPIC

Shorthand notation

Negative angles


The shorthand notation for a negative angle (clockwise rotation) is given as,

for example:

With the modulus outside the bracket and the angle inside the bracket.

z r=


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TOPIC

Multiplication in polar coordinates


When two complex numbers, written in polar form, are multiplied the

product is given as a complex number whose modulus is the product of the

two moduli and whose argument is the sum of the two arguments.

( ) ( )

[ ] [ ]( )

1 1 2 21 1 2 2

1 2 1 21 2 1 2

If cos sin and cos sin

then cos sin

z r j z r j

z z r r j

= + = +

= + + +


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TOPIC

Division in polar coordinates


When two complex numbers, written in polar form, are divided the quotient

is given as a complex number whose modulus is the quotient of the two

moduli and whose argument is the difference of the two arguments.

( ) ( )

[ ] [ ]( )

1 1 2 21 1 2 2

1 11 2 1 2

2 2

If cos sin and cos sin

then cos sin

z r j z r j

z rj

z r

= + = +

= +


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TOPIC

Introduction

Shorthand notation



deMoivres theorem



Loci problems



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TOPIC

deMoivres theorem


If a complex number is raised to the powern the result is a complex number

whose modulus is the original modulus raised to the powern and whose

argument is the original argument multiplied by n.

( )

( ) ( )

If cos sin

then cos sin = cos sinn

n n

z r j

z r j r n j n

= +

= + +


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TOPIC

Applications of De Moivres Theorem

Example: Find3

cos sin6 6

i

+

3

3

3 3cos sin cos sin6 6 6 6

cos sin2 2

0

cos sin6 6

i i

i

i

i i

+ +

= +

= +

+ =


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TOPIC

In general (a + bi)n is

Apply to

Try

( ) ( )( )cos sinn nz r n i n = +

( ))

4

3 cos330 sin 330i+ o o

12

2 2

2 2i

+


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TOPIC

In general (a + bi)n is

Try

( ) ( )( )cos sinn nz r n i n = +

12

2 2

2 2i

+

a b r= 1

= tan-1 (b/a) = -450

z= [cos(-45)+i sin (-45)]12

= cos(-45)12 + isin (-45)12

= -1+ i .0

= -1


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TOPIC

Find (1 + 3i)9

Ans. -512


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TOPIC

Example : Prove that cos4A = 8 cos4A 8 cos2 A + 1

cos4A + j sin 4A = (cosA + j sinA)4

= ((c + js) 2 )2= c4 + 4 c3 js + 6 c2 (js)2 + 4c (js)3 +(js)4

= c4 + 4 c3 js - 6 c2 s2 - 4cjs3 +s4 [j = -1; j2 = -1]

= (c4 - 6 c2 s2+s4 ) + j (4 c3s - 4cs3)

cos4A + j sin 4A = (c4 - 6 c2 s2+s4 ) + j (4 c3s - 4cs3)

Therefore, cos4A = c4 - 6 c2 s2+s4

cos4A = cos4 A - 6 cos2 A sin2 A+sin4 A= cos4 A 6 cos2 A (1- cos2 A) + (1 - cos2 A)*(1 - cos2 A)

= cos4 A 6 cos2 A + 6 cos4 A + 1- 2cos2 A + cos4 A

= 8 cos4 A 8 cos2 A + 1

j3 = j2.j =-j

sin2 A + cos2 A =1

sin2 A = 1- cos2 A

cos2 A = 1- sin2 A


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TOPIC

Introduction

Shorthand notation



deMoivres theorem



Loci problems



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TOPIC



There are n distinct values of the nth roots of a complex numberz. Each root

has the same modulus and is separated from its neighbouring root by

2radians

n


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TOPIC

Introduction

Shorthand notation



deMoivres theorem



Loci problems



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TOPIC



Since:

then by expanding the left-hand side by the binomial theorem we can find

expressions for:

( )cos sin cos sinn

j n j n + = +

cos and sin in terms of powers of cos and sinn n

CHAPTER 2 C l F i & V Al b


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TOPIC



Let:

so that:

1cos sin then cos sinz j j

z = + =

from which we can expand cos and sin in terms of powers of cos and sinn n

1 12cos 2sin

1 12cos 2sin

n n

n n

z z jz z

z n z j nz z

+ = =

+ = =

CHAPTER 2 C l F ti & V t Al b


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TOPIC

Learning outcomes

Use the shorthand form for a complex number in polar form

Write complex numbers in polar form using negative angles

Multiply and divide complex numbers in polar form

Use deMoivres theorem

Find the roots of a complex number

Demonstrate trigonometric identities of multiple angles using complex numbers

Solve loci problems using complex numbers




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TOPIC

Introduction: scalar and vector quantities

Vector representation

Components of a given vector

Vectors in space

Direction cosines

Scalar product of two vectors

Vector product of two vectors

Angle between two vectors

Direction ratios

Programme 6: Vectors



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TOPIC




Vectors in space

Direction cosines




Direction ratios




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TOPIC



(a) Ascalar quantity is defined completely by a single number with

appropriate units

(b) A vector quantity is defined completely when we know not only its

magnitude (with units) but also the direction in which it operates

Physical quantities can be divided into two main groups, scalar quantitiesand vector quantities.



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TOPIC




Vectors in space

Direction cosines




Direction ratios




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TOPIC



A vector quantity can be represented graphically by a line, drawn so that:

(a) The length of the line denotes the magnitude of the quantity(b) The direction of the line (indicated by an arrowhead) denotes the

direction in which the vector quantity acts.

The vector quantity AB is

referred to as ora____

B



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TOPIC


Two equal vectors

Types of vectors

Addition of vectors

The sum of a number of vectors




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p g

TOPIC


Two equal vectors


If two vectors, a and b, are said to be equal, they have the same magnitude

and the same direction



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p g

TOPIC



If two vectors, a and b, have the same magnitude but opposite direction then

a = b



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TOPIC


Types of vectors


(a) A position vector occurs when the point A is fixed

(b) A line vector is such that it can slide along its line of action

(c) A free vector is not restricted in any way. It is completely defined by itslength and direction and can be drawn as any one of a set of equal length

parallel lines

____B



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TOPIC


Addition of vectors


The sum of two vectors and is defined as the single vector____AB

____C

____C

____ ____ ____

or

B BC AC+ =

+a b = c



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TOPIC




Draw the vectors as a chain.

____ ____ ____ ____ ____

____

or

B BC CD DE AE

AE

+ + + =

+ + + =a b c d



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TOPIC




If the ends of the chain coincide the sum is 0.

+ + + =a b c d 0



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TOPIC



Just as can be replaced by so any single vector can

be replaced by any number of component vectors so long as the form a chain

beginning atPand ending at T.

____ ____ ____ ____

AB BC CD DE+ + +____

AE____

T

____

T = + + +a b c d



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TOPIC

Vectors in space


In three dimensions a vector can be defined in terms of its components inthe three spatial direction Ox, Oy and Ozas:

where k is a unit vector in the Ozdirection

a b c= + +r i j k

The magnitude of r can then be

found from Pythagoras theorem to

be:

2 2 2

r a b c= + +



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TOPIC

Vector Dot Product




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TOPIC



Ifa and b are two vectors, thescalar

productofa and b is defined to be the

scalar (number):

where a and b are the magnitudes of the

vectors and is the angle between them.

The scalar product (dot product) is denoted

by:

cosab

cosab =a.b



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TOPIC

.3+4=and2=betweenangletheFind jivjiu

vu

vu

=cos( ) ( ) 5383142 ==+= vu

( )u = + =2 1 52 2

v = + = + = =4 3 16 9 25 52 2

5

1

55

5cos =

==

vu

vu

v i j= +4 3

jiu = 2=

4.63

5

1cos

1



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TOPIC

Example

.6-3,-3,zand2-1,3,-w

:ifzandwctorsbetween veangletheFind

==rr

rr

Solution

2

03699419

1239

6,3,32,1,3

6,3,32,1,3cos

=

=++++

+=

=

=

vu

vurr

rr

True or False? Whenever two non-zero vectors are

perpendicular, their dot product is 0.

Think before you click.Congratulate yourself if you

chose True!



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TOPIC

True or False?

Whenever you find the angle between two non-zero vectors theformula

will generate angles in the interval

vu

vurr

rr

=cos

.2

0

This is False. For example, consider the vectors:

.2,1-vand1,4 ==

rr

u

vu

vucos rr

rr

=

Finish this on your own then click for the answer.

2 2 4

1

1

2

x

y


T F l ? Wh

fi d th l b t


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TOPIC

True or False? Wheneveryou find the angle between

two non-zero vectors the formula

will generate angles in the interval

vuvu rr

rr

=cos

.2

0

This is False. For example, consider the vectors:

.2,1-vand1,4 ==rr

u

o

rr

rr

5.167or9.2

...9762.085

9

517

18

1,21,4

1,21,4

vu

vucos

=

==

==

When the cosine is negative the angle between the two vectors

is obtuse.



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TOPIC

Find the angle between the vectors v = 3i + 2j and w = 6i + 4j

The vectors have the same direction.

We say they are parallel becauseremember vectors can be moved

around as long as you don't change

magnitude or direction.

wv

wv

=cos 5213

818+

=676

26

= =1

== 01cos 1What does it mean when theangle between the vectors is 0?

jiv 23 +=

jiw 46 +=


If the angle between 2 vectors is what would their dot

2i + 8j


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TOPIC

0= wv

Determine whether the vectors v = 4i -j andw = 2i + 8j are orthogonal.

( ) ( ) 08124 =+= wv

The vectors v and w are orthogonal.

If the angle between 2 vectors is , what would their dot

product be? 2

vu

vu =cos

Since cos is 0, the

dot product must be 0.

2

2

2

Vectors u and v in this case are called orthogonal.

(similar to perpendicular but refers to vectors).

compute their dot product

and see if it is 0

w = 2i + 8j

v = 4i -j



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TOPIC



Ifa and b are two parallel vectors, thescalar productofa and b is then:

Therefore, given:

then:

1 2 3 1 2 3

1 1 2 2 3 3

anda a a b b b

a b a b a b

= + + = + +

= + +

a i j k b i j k

a.b

cos0ab ab= =a.b


Example


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TOPIC

( ( 1542 += wv

= + = 8 5 3

Example

1

This is called the dot product. Notice the answer is just

a number NOT a vector.

find,4and52If wvjiwjiv +=+=


Properties of the Dot Product


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TOPIC

Properties of the Dot Product


P 6 V


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TOPIC

Vector Cross Product



P 6 V t


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TOPIC



The vector product(cross product) ofaand b, denoted by:

is a vectorwith magnitude:

and a direction such that a, b and

form a right-handed set.

sinab

a b

a b


P 6 V t


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TOPIC



If is a unit vector in the direction of:

then:

Notice that:

sinab =a b n

a b

= b a a b

n




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TOPIC



Since the coordinate vectors are mutually perpendicular:

and

= = =

i j k

j k i

k i j

= = =i i j j k k 0


Properties of Cross Product


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TOPIC




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TOPIC



So, given:

then:

That is:

1 2 3 1 2 3anda a a b b b= + + = + +a i j k b i j k

2 3 3 2 1 3 3 1 1 2 2 1( ) ( ) ( )a b a b a b a b a b a b = + a b i j k

1 2 3

1 2 3

a a ab b b

=

i j k

a b



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TOPIC



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TOPIC

+ k(v1.w2 v2.w1)i (v2.w3 v3.w2)j (v1.w3 v3.w1)

Multiply


Example


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TOPIC


Applications ofvxw - the perpendic

ular vector


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TOPIC

The resulting vector of the cross product ofvxw will always beperpendicular to v and w, as shown in the diagram below.

vxw

vw

Find the vector that is perpendicular to:

-3i-6j-12k

and


Applications ofvxw - the perpendic

ular vector


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TOPIC

The resulting vector of the cross product ofvxw will always beperpendicular to v and w, as shown in the diagram below.

vxw

vw

Find the vector that is perpendicular to:

3i+ 2j- k

and

3j- 5k.

-7i + 15j + 9k


Find the area of the triangle with ver

tices A(1,1,2) B(-1,3,2) and C(4,1,5).


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TOPICFind two sides of the triangle, the vectors

AB and AC.

and

Now find the cross product of the two

vectors:

The area is given as follows:

Area = (1/2) |ABAC|


Practice


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TOPIC

A = 5i +2j-3kand B = 15i + aj -9k

For which value ofa, A and B will be parallel?

Tips:

A and B will be parallel ifAB =0



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TOPIC

Assignment 03


Question 1


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TOPIC

If the two vectors are :A = it2 jt + (2t+1) k

B = 5it +jt - kt3

Find the followings

)().( BA

dt

dandBA

dt

d


Question 2


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TOPIC

Find the area of the triangle PQR

P(1,3,2)

Q(2,-1,1)

R(-1,2,3)




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TOPICLearning outcomes

Define a vector

Represent a vector by a directed straight line

Add vectors

Write a vector in terms of component vectors

Write a vector in terms of component unit vectors

Set up a system for representing vectorsObtain the direction cosines of a vector

Calculate the scalar product of two vectors

Calculate the vector product of two vectors

Determine the angle between two vectors

Evaluate the direction ratios of a vector

Documents

Chapter 2- Complex Number