Complex numbers V-01 1

Reference: Croft & Davision, Chapter 14, p.621http://www.math.utep.edu/sosmath

1. Introduction

Extended the set of real numbers to find solutions of greater range of equations.

Let j be a root of the equation

Then and 012 j

012 x

12 j

1 j

COMPLEX NUMBERS

Complex numbers V-01 2

The number , and are called imaginary number

Example 1.1

Write down (a) , (b) , (c)

Solution:

(a)

(b)

(c)

4 24 2

j214144

j6212412424

j 11 222

j2 j62 j

0,

Complex numbers V-01 3

Example 1.2

j

jjjj

1

)1)(1)(1(

1225

1)1(

)(4

428

jj

Simplify (a) , (b)

Solution:

5j 8j

(a)

(b)

j

j

11

11

14

55or

Complex numbers V-01 4

is a real part and (or ) is an imaginary part

Example 1.3

Solve 012 xx

jj

x

23

21

231

231

1211411 2

21

23

23

Solution

The solution are known as complex number.

aacbbx

cbxax

24

02

2

Complex numbers V-01 5

Complex Number

• 16th Century Italian Mathematician – Cardano • z = a + bj (Rectangular Form)

– a: real number; real part of the complex number– b: real number; imaginary part of the complex

number– bj: imaginary number

a

b

Re

Ima + bj

Complex numbers V-01 6

2. Algebra of Complex Numbers Addition and Subtraction of Complex Numbers

If and then jbaz 111 jbaz 222

jbbaazz 212121 )(

jbbaazz 212121 )(

Complex numbers V-01 7

Example 2.1 If and , find and

Solution:

jz 321 jz 652 21 zz 21 zz

jjzz

33]6)3[()]5(2[21

jjzz

97]6)3[()]5(2[21

Complex numbers V-01 8

If and then jbaz 111 jbaz 222

Multiplication of complex numbers

jbababbaajbbjabjbaaa

jbajbazz

12212121

221212121

221121

Complex numbers V-01 9

Example 2.2 Find if and .

Solution:

21zz jz 231 jz 542

jj

jjjj

jjzz

232231012

23)1(1012)1015812(

)54)(23(2

21

Complex numbers V-01 10

Example 2.3

Find if and .

Solution:

21zz jz 231 jz 232

1349

)1(49)4669(

)23)(23(2

21

jjj

jjzz

Complex numbers V-01 11

ConjugateIf then

the complex conjugate of z is = and .

bjaz

z bja 22 bazz

Example 2.4

If , find and .

Solution:jz 34 z zz

25916)1(916)3(4 22 jzzjz 34

a

b

Re

Imz = a + bj

= a - bjz

θθ

-b

Complex numbers V-01 12

i.e.

2

2

2

1

2

1

zz

zz

zz

Division of two complex numbers

22

22

21122121

22

22

22

11

22

11

2

1

bajbababbaa

jbajba

jbajbajbajba

zz

Complex numbers V-01 13

Example 2.5 Simplify

Solution: jj

9225

j

j

j

jjj

jj

jj

jj

8549

858

85498

8549)1(1810

921845410

9292

9225

9225

22

2

Complex numbers V-01 14

Simplify

Solution:

jj

2592

2941

2928

294128

294451810

25)2(2)9)(5()9)(2()2)(5(

2525

25922592

2

22

2

2

1

j

j

jjj

jjj

jj

jjjj

zz

Example 2.6

Complex numbers V-01 15

Simplify

Solution: cossin

1j

cossincossincossin

cossincossin

cossin1

cossin1

22

j

jjj

jj

Example 2.7

Complex numbers V-01 16

djcbja ca db

Example 2.7

Find the values of x and y if .

Solution:

jyxjyx 35

If then and

145482

35

yx

xyxyx

Equality of Complex Numbers

Complex numbers V-01 17

3. Argand Diagram, Modulus and Argument

The representation of complex numbers by points in a plane is called an Argand diagram.

Example 3.1

Represent , and on an Argand diagram.

j1 j1 j32

Complex numbers V-01 18

Remark: 2zzz

Solution:

ab

tan

22 of modulus then the, bj a z If bazrz

1 2 3

1

2

3

-1

-2

-3

-1-2-3Real axis

Imaginary axis

a

b

Re

Ima + bj

z

θ

The angle θ is called the argument

Complex numbers V-01 19

Example 3.2

56.3)23(1tan and 13)3(232 (d) 22 θj

53.13)

34(1tan and 525)4()3(43 (c) 22 θj

Find the modulus and argument of (a) , (b) , (c) and (d) .

j32 j43j43 j32

133232 (a) 22 j 3.5623tan 1

54343 (b) 22 j 9.1263

4tan 1

Wrong !

Solution:

87.12653.13180 θ

Complex numbers V-01 20

Important Note !!!

• Argument of a complex number The argument of a complex number is the angle between the positive x-axis and the line representing the complex number on an Argand diagram. It is denoted arg (z).

Complex numbers V-01 21

4. Polar Form, Product and Quotient in Polar Form sin and cos zbza

r

jzjr

bjaz

sincos)sin(cos

which is the polar form expression

a

b

Re

Im

a + bj

zθ

Complex numbers V-01 22

Express the complex number in polar form.

Solution:

j43Example 4.1

13.53)3

4(tan3

4tan 1

87.1265)87.126sin87.126(cos5

525)4()3( 22

rjz

z

87.12613.53180

4

-3

θ

Wrong!

α

Complex numbers V-01 23

Express in true polar form.

Solution:

Example 4.2

)30360sin()30360cos(5 jz

30sin30cos5 jz

)330sin330(cos5 j

3300

300

A

C-300

Complex numbers V-01 24

Let and . 1111 sincos jrz 2222 sincos jrz

21212

1

2

1

21212121

sincos

sincos

jrr

zz

jrrzz

)sincossin(cossinsincoscos

1221

21212121

j

rrzz

Complex numbers V-01 25

Example 4.3 If and

find and .

30sin30cos21 jz 60sin60cos32 jz

21zz2

1

zz

Solution:

j

j

jzz

jjjzz

33.058.0

))5.0(866.0(32

6030sin6030cos32

6)0(66030sin6030cos32

2

1

21

Complex numbers V-01 26

Example 4.4 Express the conjugate of in true polar form.

Solution:

60sin60cos3 jz

)60360sin()60360cos(3 jz

)300sin300(cos3 j

60sin60cos3 jz

Complex numbers V-01 27

5. Exponential Form

To derive the exponential form we shall need to refer to the power series expansions of cos x, sin x …

.....!4!3!2

1

.....!5!3

sin

.....!4!2

1cos

432

53

42

xxxxe

xxxx

xxx

x

Complex numbers V-01 28

.....)!3

(!4!2

1

.....!4!3!2

1

.....!4!3!2

1

342

432

443322

jje

jje

jjjje

j

j

j

sincos

sincos

,

je

je

Therefore

j

j

Euler’s Formula !!!

Complex numbers V-01 29

Define and sincos je j sincos je j

=

=

sincos jrz jre

sincos jrz jre

which is the exponential form expression and is in radian.

Complex numbers V-01 30

Express the complex number and its conjugate in exponential form.

Solution:

j45

4.6)4()5( 22 z

c

ab

ab

5.234.141

66.38

)5

4(tantan

tan

0

0

11

Example 5.1

a

b

Re

Ima + bj

z

θ

which is the exponential form expression and is in radian.

jez 5.24.6 jez 5.24.6

Complex numbers V-01 31

Find (a) and (b) .

Solution:

2

jj ee jee jj

2

and sincos je j sincos je j

which is the exponential form expression and is in radian.

cos2

cos22

jj ee

sin2sin2

2

jj

jee jj

Example 5.2

Complex numbers V-01 32

Example 6.1 Use De Moivre’s theorem to write in an alternative form.

Solution

6. De Moivre’s Theorem

3sincos j

3sin3cossincos 3 jj

njnj n sincossincos

Complex numbers V-01 33

Example 6.2

If z = r(cos + j sin ), find z4 and use De Movire’s theorem to write your result in an alternative form.

Solution: z4 = r4 (cos + j sin )4

= r4 (cos 4 + j sin 4)

Complex numbers V-01 34

More concise form: If z = r then zn = rnn

For example, z4 = r44

Example 6.3(a) If z = 2/8 write down z4.(b) Express your answer in both polar and Cartesian form.

Solution (a) z4 = 24 (4)(/8) = 16 /2

(b) i) z4 = 16 (cos /2 + j sin /2)ii) a = 16 cos /2 = 0

b = 16 sin /2 = 16

Complex numbers V-01 35

Example 6.4If z = 3 (cos /12 + j sin /12) find z3 in Cartesian form

Solution

z3 = 33(cos(3)(/12) + j sin(3)(/12))= 27(cos /4 + j sin /4)

a = 27cos /4 = 19.09b = 27sin /4 = 19.09

Complex numbers V-01 36

Example 6.5

525)4()3( 22 z

13.53)34(tan

34tan 1

)13.53sin13.53(cos5 jz

(a) Express z = 3 + 4j in polar form.(b) Hence, find (3 + 4j)10, leaving your answer in polar

form.

Solution

)3.9sin3.9(cos5

))93.010(sin)93.010((cos510

1010

j

jz