Post on 31-Mar-2015
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
Slide 3.3 - 6Copyright © 2010 Pearson Education, Inc.
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.