Copyright © 2010 Pearson Education, Inc. Complex Numbers Perform arithmetic operations on complex numbersPerform arithmetic operations on complex numbers.

Copyright © 2010 Pearson Education, Inc.

Complex NumbersComplex Numbers

♦ Perform arithmetic operations on complex Perform arithmetic operations on complex numbersnumbers

♦ Solve quadratic equations having complex Solve quadratic equations having complex solutionssolutions

3.33.3

Slide 3.3 - 2Copyright © 2010 Pearson Education, Inc.

Properties of the Imaginary Unit i

Defining the number i allows us to say that the solutions to the equationx2 + 1 = 0 are i and –i.

i 1, i2 1

Slide 3.3 - 3Copyright © 2010 Pearson Education, Inc.

Complex Numbers

A complex number can be written in

standard form as a + bi where a and b

are real numbers. The real part is a and

the imaginary part is b. Every real

number a is also a complex number

because it can be written as a + 0i.

Slide 3.3 - 4Copyright © 2010 Pearson Education, Inc.

Imaginary Numbers

A complex number a + bi with b ≠ 0 is an imaginary number. A complex number

a + bi with a = 0 and b ≠ 0 is sometimes

called a pure imaginary number.

Examples of pure imaginary numbers

include 3i and –i.

Slide 3.3 - 5Copyright © 2010 Pearson Education, Inc.

The Expression

If a > 0, then a i a.

a

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Example 2Simplify each expression.

Solution

(b) 2 8 (a) 3 3

(b) 2 8 i 2 i 8 i2 16 1 4 1

(a) 3 3 i 3 i 3 i2 3 2 1 3 3

Slide 3.3 - 7Copyright © 2010 Pearson Education, Inc.

Example 3Write each expression in standard form. Support your results using a calculator.

a) (3 + 4i) + (5 i) b) (7i) (6 5i)

c) (3 + 2i)2 d)

Solution

a) (3 + 4i) + (5 i) = 3 + 5 + 4i i = 2 + 3i

b) (7i) (6 5i) = 6 7i + 5i = 6 2i

17

4 i

Slide 3.3 - 8Copyright © 2010 Pearson Education, Inc.

Solution continued

c) (3 + 2i)2 = (3 + 2i)(3 + 2i)

= 9 – 6i – 6i + 4i2

= 9 12i + 4(1)

= 5 12i

d)

17

4 i

17

4 i

17

4 i

68 17i

16 i2

68 17i

174 i

Slide 3.3 - 9Copyright © 2010 Pearson Education, Inc.

Quadratic Equations with Complex Solutions

We can use the quadratic formula to solve quadratic equations if the discriminant is negative.

There are no real solutions, and the graph does not intersect the x-axis.

The solutions can be expressed as imaginary numbers.

Slide 3.3 - 10Copyright © 2010 Pearson Education, Inc.

Example 4aSolve the quadratic equation

Support your answer graphically.

Solution

Rewrite the equation:

a = 1/2, b = –5, c = 17

x b b2 4ac

2a

5 5 2 4 0.5 17

2 0.5

1

2x2 17 5x.

1

2x2 5x 17 0.

5 9

5 3i

Slide 3.3 - 11Copyright © 2010 Pearson Education, Inc.

Example 4aSolution continued

The graphs do not intersect, so no real solutions, but two complex solutions that are imaginary.

Slide 3.3 - 12Copyright © 2010 Pearson Education, Inc.

Example 4bSolve the quadratic equation x2 + 3x + 5 = 0.Support your answer graphically.

Solution

a = 1, b = 3, c = 5

x b b2 4ac

2a

3 3 2 4 15

2 1

3 11

2

3 i 11

2

3

2

i 11

2

Slide 3.3 - 13Copyright © 2010 Pearson Education, Inc.

Example 4bSolution continued

The graph does not intersect the x-axis, so no real solutions, but two complex solutions that are imaginary.

Slide 3.3 - 14Copyright © 2010 Pearson Education, Inc.

Example 4cSolve the quadratic equation –2x2 = 3.

Support your answer graphically.

Solution

Apply the square root property.

2x2 3

x2 3

2

x 3

2 x i

3

2

Slide 3.3 - 15Copyright © 2010 Pearson Education, Inc.

Example 4cSolution continued

The graphs do not intersect, so no real solutions, but two complex solutions that are imaginary.