The Complex Numbers 10.8. The Number i i is the unique number for which and so we have i 2 = –1....

The Complex Numbers10.8

The Number ii is the unique number for which and so we have i

2 = –1.1i

We can now express the square root of a negative number in terms of i.

Imaginary NumbersAn imaginary number is a number that can be written in the form a + bi, where a and b are real numbers and 0.b

Solution

Express in terms of i:a) 9

b) 48

b) 48 1 16 3

a) 9 1 9

1 9 3, or 3 .i i

1 16 3 4 3 4 3i i

Example

Complex NumbersA complex number is any number that can be written in the form a + bi, where a and b are real numbers. (Note that a and b both can be 0.)

The following are examples of imaginary numbers:

7 2

12

3

11

i

i

i

Here a = 7, b =2.

Here 2, 1 / 3. a b

Here 0, 11. a b

Addition and Subtraction

The complex numbers obey the commutative, associative, and distributive laws. Thus we can add and subtract them as we do binomials.

Add or subtract and simplify.

a) (3 2 ) (7 8 )

b) (10 2 ) (9 )

i i

i i

Example

MultiplicationTo multiply square roots of negative real numbers, we first express them in terms of i. For example, 6 5 1 6 1 5

6 5i i 2 30i

1 30 30.

Caution!With complex numbers, simply multiplying radicands is incorrect when both radicands are negative:

3 5 15.

Multiply and simplify. When possible, write answers in the form a + bi.

Solution

a) 2 10

b) 2 5 3

c) 2 4 3

i i

i i

a) 2 10 1 2 1 10

2 10i i 2 20 1 2 5 2 5i

Example

Solution continued

b) 2 5 3 2 5 2 3i i i i i

2c) 2 4 3 8 6 4 3i i i i i

210 6i i

10 6 6 10i i

8 2 3i

11 2i

Conjugate of a Complex NumberThe conjugate of a complex number a + bi is a – bi, and the conjugate of a – bi is a + bi.

Solution

Find the conjugate of each number.

a) 4 3

b) 6 9

c)

i

i

i

b) 6 9i

The conjugate is 4 – 3i.

The conjugate is –6 + 9i.

The conjugate is –i.

a) 4 3i

c) i

Example

Conjugates and Division

Conjugates are used when dividing complex numbers. The procedure is much like that used to rationalize denominators.

Solution

Divide and simplify to the form a + bi.3 2

a)

4b)

2 3

i

ii

i

3 2 3 2a)

i i

i i

i

i

2

23 2 3 2

1

i i i

i

3 2 2 3i i

Example

Solution continued

4 4b)

2 3 2 3

i i

i i

2 3

2 3

i

i

2

2(4 )(2 3 ) 8 12 2 3

(2 3 )(2 3 ) 4 9

i i i i i

i i i

8 14 3 5 14

4 9 13

i i

5 14

13 13i

Powers of i

Simplifying powers of i can be done by using the fact that i

2 = –1 and expressing the given power of i in terms of i

2. Consider the following:

i 23 = (i

2)11i1

= (–1)11i = –i

Solution

Simplify: 40

33

a) ;

b) .

i

i

2040 2a) i i 201 1

1633 2b) i i i 161 1i i i

Example