§ 7.7 complex numbers. blitzer, intermediate algebra, 5e – slide #3 section 7.7 complex numbers...

§ 7.7

Complex Numbers

Complex Numbers

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.7

Complex Numbers

The Imaginary Unit iThe imaginary unit i is defined as

The Square Root of a Negative NumberIf b is a positive real number, then

.1 where,1 2 ii

.or 1)1( ibbibbb


Complex Numbers

EXAMPLEEXAMPLE

Write as a multiple of i: .520(b)300(a)

SOLUTIONSOLUTION

13001300300(a)

1520520(b)

31013100 i

ii 520or 520

1520


Complex Numbers

Complex Numbers & Imaginary NumbersThe set of all numbers in the form

with real numbers a and b, and i, the imaginary unit, is called the set of complex numbers. The real number a is called the real part, and the real number b is called the imaginary part of the complex number If , then the complex number is called an imaginary number.

bia

.bia 0b


Complex Numbers

Adding & Subtracting Complex Numbers1)

In words, this says that you add complex numbers by adding their real parts, adding their imaginary parts, and expressing the sum as a complex number.

2)

In words, this says that you subtract complex numbers by subtracting their real parts, subtracting their imaginary parts, and expressing the difference as a complex number.

idbcadicbia

idbcadicbia


Complex Numbers

EXAMPLEEXAMPLE

Perform the indicated operations, writing the result in the form a + bi: (a) (-9 + 2i) – (-17 – 6i) (b) (-2 + 6i) + (4 - i).

SOLUTIONSOLUTION

(a) (-9 + 2i) – (-17 – 6i)

= -9 + 2i + 17 + 6i

= -9 + 17 + 2i + 6i

= (-9 + 17) + (2 + 6)i

= 8 + 8i

Remove the parentheses. Change signs of the real and imaginary parts being subtracted.

Group real and imaginary terms.

Add real parts and imaginary parts.

Simplify.


Complex Numbers

(b) (-2 + 6i) + (4 - i)

= -2 + 6i + 4 - i

= -2 + 4 + 6i - i

= (-2 + 4) + (6 - 1)i

= 2 + 5i

Remove the parentheses.

Group real and imaginary terms.

Add real parts and imaginary parts.

Simplify.

CONTINUECONTINUEDD


Complex Numbers

EXAMPLEEXAMPLE

Find the products: (a) -6i(3 – 5i) (b) (-4 + 2i)(-4 - 2i).

SOLUTIONSOLUTION

(a) -6i(3 – 5i)Distribute -6i through the parentheses.

Multiply.

iii 5636

23018 ii Replace with -1. 13018 i 2i

Simplify and write in a + bi form.i1830


Complex Numbers

CONTINUECONTINUEDD

148816 ii 12 i

ii 88416 Group real and imaginary terms.

20 Combine real and imaginary terms.

(b) (-4 + 2i)(-4 – 2i)248816 iii Use the FOIL method.


Complex Numbers

Multiplying Complex NumbersBecause the product rule for radicals only applies to real numbers, multiplying radicands is incorrect. When performing operations with square roots of negative numbers, begin by expressing all square roots in terms of i. Then perform the indicated operation.


Complex Numbers

EXAMPLEEXAMPLE

Multiply:

SOLUTIONSOLUTION

Express square roots in terms of i.

.416

The square root of 64 is 8.

14116416

ii 416 264i

164

8

. and 64416 2iii

12 i


Complex Numbers

In the next chapter we will study equation whose solutions involve the square roots of negative numbers. Because the square of a real number is never negative, there are no real number solutions to those equations. However, there is an expanded system of numbers in which the square root of a negative number is defined. This set is called the set of complex numbers.

The imaginary number i is the basis of this new set.

So come… now go with us to never-never land , a place where you have not been before…


Complex Numbers



bia

.bia 0b


Complex Numbers

EXAMPLEEXAMPLE

Divide and simplify to the form a + bi:

SOLUTIONSOLUTION

Multiply by 1.

.24

36

i

i

i

i

i

i

i

i

24

24

24

36

24

36

The conjugate of the denominator is 4 – 2i. Multiplication of both the numerator and the denominator by 4 – 2i will eliminate i from the denominator.

Use FOIL in the numerator and 22 BABABA in thedenominator.

22

2

24

6121224

i

iii


Complex Numbers

Simplify.2

2

416

62424

i

ii

CONTINUECONTINUEDD

1416

162424

i 12 i

Perform the multiplications involving -1.416

62424

i

Combine like terms in the numerator and denominator.20

2418 i

Express answer in the form a + bi.i20

24

20

18

Simplify.i5

6

10

9


Complex Numbers

EXAMPLEEXAMPLE

Simplify:

SOLUTIONSOLUTION

.cba 1340046 iii

11a 2323246 ii

11b 2002002400 ii

iiiiiiii 11c 6621213


In Summary…

To add or subtract complex numbers, add or subtract their real parts and then add or subtract their imaginary parts. Adding complex numbers is easy.

To multiply complex numbers, use the rule for multiplying binomials. After youare done, remember that 12 i

and make the substitution. In fact, if you can only remember one thing from this section – remember this fact, that is, when your square i, you get -1.

To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator. This gives you a real number in the denominator, and you will know how to proceed from that point.


Complex Numbers



bia

.bia 0b


Complex Numbers

Simplifying Powers of i1) Express the given power of i in terms of

2) Replace with -1 and simplify. Use the fact that -1 to an even power is 1 and -1 to an odd power is -1.

.2i2i


Complex Numbers

EXAMPLEEXAMPLE

Divide and simplify to the form a + bi:

SOLUTIONSOLUTION

Multiply by 1.

.4

5

i

i

The conjugate of the denominator, 0 - 4i, is 0 + 4i. Multiplication of both the numerator and the denominator by 4i will eliminate i from the denominator.

Multiply. Use the distributive property in the numerator.2

2

16

420

i

ii

i

i

i

i

i

i

4

4

4

5

4

5


Complex Numbers

Perform the multiplications involving -1.

116

1420

i

CONTINUECONTINUEDD

12 i

16

420

i

i16

20

16

4

Express the division in the

form a + bi.

i4

5

4

1 Simplify real and imaginary

parts.

§ 7.7 complex numbers. blitzer, intermediate algebra, 5e – slide #3 section 7.7 complex numbers...

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terms of i

multiple of i

real parts

imaginary parts

set of complex numbers

real number b

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intermediate algebra