Ch 1.3 Complex Numbers ppt

Download Report

Transcript Ch 1.3 Complex Numbers ppt

10 TH EDITION

COLLEGE ALGEBRA

LIAL HORNSBY SCHNEIDER

1.3 - 1

1.3

Complex Numbers

Basic Concepts of Complex Numbers Operations on Complex Numbers

Basic Concepts of Complex Numbers

There are no real numbers for the solution of the equation

x

2   1.

To extend the real number system to include such numbers as,  1, the number

i

is defined to have the following property;

i

2   1.

Basic Concepts of Complex Numbers

So…

i

  1 The number

i

is called the

imaginary unit

.

Numbers of the form

a

+

bi

, where

a

and

b

are real numbers are called

complex numbers

.

In this complex number,

a

is the

real part

and

b

is the

imaginary part

.

Complex numbers

a

+

bi

,

a

and

b

real Nonreal complex numbers

a

+

bi

,

b

≠ 0 Real numbers

a

+

bi

,

b

= 0 Irrational numbers Rational numbers Integers Non integers

Basic Concepts of Complex Numbers

Two complex numbers are equal provided that their real parts are equal and their imaginary parts are equal;

a

bi di

if and only if

a

c

and

b

d

Basic Concepts of Complex Numbers

For complex number

a

+

bi

, if

b

= 0, then

a

+

bi

=

a

So, the set of real numbers is a subset of complex numbers.

Basic Concepts of Complex Numbers

If

a

= 0 and

b

≠ 0, the complex number is

pure imaginary

.

A pure imaginary number or a number, like 7 + 2

i

with

a

≠ 0 and

b

≠ 0, is a

nonreal complex number

. The form

a

form

.

+

bi

(or

a

+

ib

) is called

standard

THE EXPRESSION 

a

If

a

 0, then

i a

.

Example 1

i a

Write as the product of a real number and

i

, using the definition of 

a

.

a.

 16

Solution

:  16 

i

16  4

i

Example 1

i a

Write as the product of a real number and

i

, using the definition of 

a

.

b.

 70

Solution

:  70 

i

70

Example 1

i a

Write as the product of a real number and

i

, using the definition of 

a

.

c.

 48

Solution

:  48 

i

48 

i

16 3  4

i

3 Product rule for radicals

Operations on Complex Numbers

Products or quotients with negative radicands are simplified by first rewriting 

a

as

i a

for a positive number.

Then the properties of real numbers are applied, together with the fact that

i

2   1.

Operations on Complex Numbers

Caution

When working with negative radicands, use the definition…

i a

before using any of the other rules for radicands.

Operations on Complex Numbers

Caution

In particular, the rule

c d

cd

is valid only when

c

and

d

are not both negative.

36  6, while  4  6

i

2   6 so  4  9 

a.

Example 2

FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS

Multiply or divide, as indicated. Simplify each answer.

 7  7

Solution

:  7

First write all square roots in terms of i.

 

i i

2  1   7 7

i

  7 2 7

i

2 = −1

b.

Example 2

FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS

Multiply or divide, as indicated. Simplify each answer.

 6  10

Solution

:  6  10 

i

6

i

10   1 2 15 

i

2 60   1 4 15   2 15

Example 2

FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS

Multiply or divide, as indicated. Simplify each answer.

c.

 20  2

Solution

:  20  2 

i i

20  2 20  2 10 Quotient rule for radicals

Example 2

FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS

Multiply or divide, as indicated. Simplify each answer.

d.

 48 24

Solution

:  48  24 

i i

48 2 4 

i

48 24 

i

2

Example 3 Write 4

SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND

128 in standard form

a

+

bi

.

Solution

: 4  12 8  4  64 2  4 8

i

2  64  8

i

Example 3 Write 4

SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND

128 in standard form

a

+

bi

.

Solution

:

Be sure to factor before simplifying

  4  4 8

i

2 2

i

2   64  8

i

2

i

4 2 Factor.

Lowest terms

Addition and Subtraction of Complex Numbers For complex numbers

a

+

bi

and

c

+

di

, (

a

bi

c

di

)  (

a

c b

 and (

a

bi

c

di

)  (

a

c b

Example 4

ADDING AND SUBTRACTING COMPLEX NUMBERS

Find each sum or difference.

a.

(3 

Solution

: ( 3  4

i

Add real parts.

Add imaginary parts.

 2  6

i

)   3  (  2 )    4  6 

i

Commutative, associative, distributive properties 1 2

i

Example 4

ADDING AND SUBTRACTING COMPLEX NUMBERS

Find each sum or difference.

b.

Solution

:   8

i

Example 4

ADDING AND SUBTRACTING COMPLEX NUMBERS

c.

Find each sum or difference.

Solution

:   

i

  

i

Example 4

ADDING AND SUBTRACTING COMPLEX NUMBERS

d.

Find each sum or difference.

Solution

:      2

i

Multiplication of Complex Numbers

The product of two complex numbers is found by multiplying as if the numbers were binomials and using the fact that

i 2

= –1, as follows.

(

a

 

di

) 

ac

adi

bic

bidi

FOIL 

ac

adi

bci

bd i

2 Distributive property;

i

2 = – 1 

ac

 (

ad

 

bd

(  1 )  (

ac

bd

ad

Multiplication of Complex Numbers For complex numbers

a

+

bi

and

c

+

di

, (

a

 

di

)  (

ac

bd ad

a.

Example 5

MULTIPLYING COMPLEX NUMBERS

Find each product.

Solution

:   2(3)  8

i

 9

i

 12

i

2

i

 18 

i

12 (  1 ) FOIL

i

2 = −1

Example 5

MULTIPLYING COMPLEX NUMBERS

Find each product.

b.

(4 

i

2

Solution

: (4 

i

2  4 2   16  24

i

 9

i

i

2 Square of a binomial

i

2

Remember to add twice the product of the two terms.

 16  24

i i

2 = −1 24

i

Example 5

MULTIPLYING COMPLEX NUMBERS

Find each product.

c.

(6  

Solution

: (6    6 2 

i

2 Product of the sum and difference of two terms  36   36  25  61, or 61 0

i i

2 = −1 Standard form

Simplifying Powers of

i

Powers of

i

can be simplified using the facts

i

2   1 and

i

4 

i

2  (  1 ) 2  1

Example 6

SIMPLIFYING POWERS OF i

Simplify each power of

i

.

a.

i

15

Solution

: Since

i

2 = –1 and

i

4 = 1, write the given power as a product involving

i

2 example,

i

3 

i

2

i

 (  1 )

i

or  

i

.

i

4 . For Alternatively, using

i

4

i

15 

i

12

i

3  and

i

3

i

  3

i

3 to rewrite

i

15  1 3 (

i i

gives

Example 6

SIMPLIFYING POWERS OF i

Simplify each power of

i

.

b.

i

 3

Solution

:

i

 3 

i

 4

i

 (

i

4 )  1

i

  1

i

i

Powers of

i i

1 

i i

2   1

i

3  

i i

4  1

i

5 

i i

6   1

i

7  

i i

8  1

i

9 

i i

10   1

i

11  

i i

12  1, and so on.

Ex 5c. showed that… (6    61 The numbers differ only in the sign of their imaginary parts and are called

complex conjugates

.

The product of a complex number and its conjugate is always a real number.

This product is the sum of squares of real and imaginary parts.

Property of Complex Conjugates For real numbers

a

and

b

, (

a

 

bi

) 

a

2 

b

2 .

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

a.

3 5   2

i i

Solution

: 3 5   2

i i

 (3 ( 5  

i i

(5 ) ( 5  

i i

) ) Multiply by the complex conjugate of the denominator in both the numerator and the denominator.

 15  3

i

25  10

i

i

2  2

i

2 Multiply.

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

a.

3 5   2

i i

Solution

:  15  3

i

 10

i

25 

i

2  2

i

2 Multiply.

 

i

26

i

2 = −1

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

a.

3 5   2

i i

Solution

:  

i

26  13 26  13

i

26

i

2 = −1

a

bi c c bi c

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

a.

3 5   2

i i

Solution

:  13 26  13

i

26

a

bi c c bi c

2 1

i

2 Lowest terms; standard form

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

b.

3

i

Solution

:

i

3  3 ( 

i

)

i

( 

i

)   3

i

i

2 –

i

is the conjugate of

i.

Example 7

DIVIDING COMPLEX NUMBERS

Write each quotient in standard form

a

+

bi

.

b.

3

i

Solution

:   3

i

i

2   3

i

1  

i

2 = −1(−1) = 1 

i

Standard form