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3.1 The Complex Numbers

Perform computations involving complex numbers.

Some functions have zeros that are not real numbers. In order to find the zeros of such functions, we must consider the complex-number system.

The Complex-Number System

We know that the square root of a negative number is not a real number. For example, $\sqrt{- 1}$ $\sqrt{- 1}$ is not a real number because there is no real number x such that $x^{2} = - 1$ $x^{2} = - 1$ . This means that certain equations, like $x^{2} = - 1$ $x^{2} = - 1$ or $x^{2} + 1 = 0$ $x^{2} + 1 = 0$ , do not have real-number solutions, and certain functions, like $f (x) = x^{2} + 1$ $f (x) = x^{2} + 1$ , do not have real-number zeros. Consider the graph of $f (x) = x^{2} + 1$ $f (x) = x^{2} + 1$ .

We see that the graph does not cross the x-axis and thus has no x-intercepts. This illustrates that the function $f (x) = x^{2} + 1$ $f (x) = x^{2} + 1$ has no real-number zeros. Thus there are no real-number solutions of the corresponding equation $x^{2} + 1 = 0$ $x^{2} + 1 = 0$ .

We can define a nonreal number that is a solution of the equation $x^{2} + 1 = 0$ $x^{2} + 1 = 0$ .

To express roots of negative numbers in terms of i, we can use the fact that

\sqrt{- p} = \sqrt{- 1 \cdot p} = \sqrt{- 1} \cdot \sqrt{p} = i \sqrt{p}

$\sqrt{- p} = \sqrt{- 1 \cdot p} = \sqrt{- 1} \cdot \sqrt{p} = i \sqrt{p}$

when p is a positive real number.

Example 1

Express each number in terms of i.

$\sqrt{- 7}$ $\sqrt{- 7}$
$\sqrt{- 16}$ $\sqrt{- 16}$
$- \sqrt{- 13}$ $- \sqrt{- 13}$
$- \sqrt{- 64}$ $- \sqrt{- 64}$
$\sqrt{- 48}$ $\sqrt{- 48}$

Solution

Now Try Exercises 1, 7, and 9.

The complex numbers are formed by adding real numbers and multiples of i.

Note that either a or b or both can be 0. When $b = 0$ $b = 0$ , $a + b i = a + 0 i = a$ $a + b i = a + 0 i = a$ , so every real number is a complex number. A complex number like $3 + 4 i$ $3 + 4 i$ or 17i, in which $b \neq 0$ $b \neq 0$ , is called an imaginary number. A complex number like 17i or −4i, in which $a = 0$ $a = 0$ and $b \neq 0$ $b \neq 0$ , is sometimes called a pure imaginary number. The relationships among various types of complex numbers are shown in the following figure.

Addition and Subtraction

The complex numbers obey the commutative, associative, and distributive laws. Thus we can add and subtract them as we do binomials. We collect the real parts and the imaginary parts of complex numbers just as we collect like terms in binomials.

Example 2

Add or subtract and simplify each of the following.

$(8 + 6 i) + (3 + 2 i)$ $(8 + 6 i) + (3 + 2 i)$
$(4 + 5 i) - (6 - 3 i)$ $(4 + 5 i) - (6 - 3 i)$

Solution

$\begin{matrix} (8 + 6 i) + (3 + 2 i) & = & (8 + 3) + (6 i + 2 i) \\ Collecting the real parts and the imaginary parts \\ = & 11 + (6 + 2) i = 11 + 8 i \end{matrix}$ $\begin{matrix} (8 + 6 i) + (3 + 2 i) & = & (8 + 3) + (6 i + 2 i) \\ Collecting the real parts and the imaginary parts \\ = & 11 + (6 + 2) i = 11 + 8 i \end{matrix}$
$\begin{array}{rcl} (4 + 5 i) - (6 - 3 i) & = & (4 - 6) + [5 i - (−3 i)] \\ Note that 6 and −3 i are both being subtracted . \\ = & −2 + 8 i \end{array}$ $\begin{array}{rcl} (4 + 5 i) - (6 - 3 i) & = & (4 - 6) + [5 i - (−3 i)] \\ Note that 6 and −3 i are both being subtracted . \\ = & −2 + 8 i \end{array}$

Now Try Exercises 11 and 21.

Multiplication

When $\sqrt{a}$ $\sqrt{a}$ and $\sqrt{b}$ $\sqrt{b}$ are real numbers, $\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$ $\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$ , but this is not true when $\sqrt{a}$ $\sqrt{a}$ and $\sqrt{b}$ $\sqrt{b}$ are not real numbers. Thus,

\begin{array}{rcll} \sqrt{- 2} \cdot \sqrt{- 5} & = & \sqrt{- 1} \cdot \sqrt{2} \cdot \sqrt{- 1} \cdot \sqrt{5} \\ = & i \sqrt{2} \cdot i \sqrt{5} \\ = & i^{2} \sqrt{10} = - 1 \sqrt{10} = - \sqrt{10} is correct! \end{array}

$\begin{array}{rcll} \sqrt{- 2} \cdot \sqrt{- 5} & = & \sqrt{- 1} \cdot \sqrt{2} \cdot \sqrt{- 1} \cdot \sqrt{5} \\ = & i \sqrt{2} \cdot i \sqrt{5} \\ = & i^{2} \sqrt{10} = - 1 \sqrt{10} = - \sqrt{10} is correct! \end{array}$

But

\begin{array}{l} \sqrt{- 2} \cdot \sqrt{- 5} = \sqrt{(- 2) (- 5)} = \sqrt{10} & is wrong! \end{array}

$\begin{array}{l} \sqrt{- 2} \cdot \sqrt{- 5} = \sqrt{(- 2) (- 5)} = \sqrt{10} & is wrong! \end{array}$

Keeping this and the fact that $i^{2} = - 1$ $i^{2} = - 1$ in mind, we multiply with imaginary numbers in much the same way that we do with real numbers.

Example 3

Multiply and simplify each of the following.

$\sqrt{- 16} \cdot \sqrt{- 25}$ $\sqrt{- 16} \cdot \sqrt{- 25}$
$(1 + 2 i) (1 + 3 i)$ $(1 + 2 i) (1 + 3 i)$
${(3 - 7 i)}^{2}$ ${(3 - 7 i)}^{2}$

Solution

$\begin{array}{rcl} \sqrt{- 16} \cdot \sqrt{- 25} & = & \sqrt{- 1} \cdot \sqrt{16} \cdot \sqrt{- 1} \cdot \sqrt{25} \\ = & i \cdot 4 \cdot i \cdot 5 \\ = & i^{2} \cdot 20 \\ = & - 1 \cdot 20 i^{2} = - 1 \\ = & - 20 \end{array}$ $\begin{array}{rcl} \sqrt{- 16} \cdot \sqrt{- 25} & = & \sqrt{- 1} \cdot \sqrt{16} \cdot \sqrt{- 1} \cdot \sqrt{25} \\ = & i \cdot 4 \cdot i \cdot 5 \\ = & i^{2} \cdot 20 \\ = & - 1 \cdot 20 i^{2} = - 1 \\ = & - 20 \end{array}$
$\begin{array}{rcll} (1 + 2 i) (1 + 3 i) & = & 1 + 3 i + 2 i + 6 i^{2} & Multiplying each term of one number by every term of the other (FOIL) \\ = & 1 + 3 i + 2 i - 6 & i^{2} = - 1 \\ = & - 5 + 5 i & Collecting like terms \end{array}$ $\begin{array}{rcll} (1 + 2 i) (1 + 3 i) & = & 1 + 3 i + 2 i + 6 i^{2} & Multiplying each term of one number by every term of the other (FOIL) \\ = & 1 + 3 i + 2 i - 6 & i^{2} = - 1 \\ = & - 5 + 5 i & Collecting like terms \end{array}$
$\begin{array}{rcll} {(3 - 7 i)}^{2} & = & 3^{2} - 2 \cdot 3 \cdot 7 i + {(7 i)}^{2} & Recall that {(A - B)}^{2} = A^{2} - 2 A B + B^{2} . \\ = & 9 - 42 i + 49 i^{2} \\ = & 9 - 42 i - 49 & i^{2} = - 1 \\ = & - 40 - 42 i \end{array}$ $\begin{array}{rcll} {(3 - 7 i)}^{2} & = & 3^{2} - 2 \cdot 3 \cdot 7 i + {(7 i)}^{2} & Recall that {(A - B)}^{2} = A^{2} - 2 A B + B^{2} . \\ = & 9 - 42 i + 49 i^{2} \\ = & 9 - 42 i - 49 & i^{2} = - 1 \\ = & - 40 - 42 i \end{array}$

Now Try Exercises 31, 39, and 55.

Recall that −1 raised to an even power is 1, and −1 raised to an odd power is −1. Simplifying powers of i can then be done by using the fact that $i^{2} = - 1$ $i^{2} = - 1$ and expressing the given power of i in terms of $i^{2}$ $i^{2}$ . Consider the following:

\begin{array}{rcl} i & = & \sqrt{- 1}, \\ i^{2} & = & - 1, \\ i^{3} & = & i^{2} \cdot i = (- 1) i = - i, \\ i^{4} & = & {(i^{2})}^{2} = {(- 1)}^{2} = 1, \\ i^{5} & = & i^{4} \cdot i = {(i^{2})}^{2} \cdot i = {(- 1)}^{2} \cdot i = 1 \cdot i = i, \\ i^{6} & = & {(i^{2})}^{3} = {(- 1)}^{3} = - 1, \\ i^{7} & = & i^{6} \cdot i = {(i^{2})}^{3} \cdot i = {(- 1)}^{3} \cdot i = - 1 \cdot i = - i, \\ i^{8} & = & {(i^{2})}^{4} = {(- 1)}^{4} = 1. \end{array}

$\begin{array}{rcl} i & = & \sqrt{- 1}, \\ i^{2} & = & - 1, \\ i^{3} & = & i^{2} \cdot i = (- 1) i = - i, \\ i^{4} & = & {(i^{2})}^{2} = {(- 1)}^{2} = 1, \\ i^{5} & = & i^{4} \cdot i = {(i^{2})}^{2} \cdot i = {(- 1)}^{2} \cdot i = 1 \cdot i = i, \\ i^{6} & = & {(i^{2})}^{3} = {(- 1)}^{3} = - 1, \\ i^{7} & = & i^{6} \cdot i = {(i^{2})}^{3} \cdot i = {(- 1)}^{3} \cdot i = - 1 \cdot i = - i, \\ i^{8} & = & {(i^{2})}^{4} = {(- 1)}^{4} = 1. \end{array}$

Note that the powers of i cycle through the values i, −1, −i, and 1.

Example 4

Simplify each of the following.

$i^{37}$ $i^{37}$
$i^{58}$ $i^{58}$
$i^{75}$ $i^{75}$
$i^{80}$ $i^{80}$

Solution

$i^{37} = i^{36} \cdot i = {(i^{2})}^{18} \cdot i = {(- 1)}^{18} \cdot i = 1 \cdot i = i$ $i^{37} = i^{36} \cdot i = {(i^{2})}^{18} \cdot i = {(- 1)}^{18} \cdot i = 1 \cdot i = i$
$i^{58} = {(i^{2})}^{29} = {(- 1)}^{29} = - 1$ $i^{58} = {(i^{2})}^{29} = {(- 1)}^{29} = - 1$
$i^{75} = i^{74} \cdot i = {(i^{2})}^{37} \cdot i = {(- 1)}^{37} \cdot i = - 1 \cdot i = - i$ $i^{75} = i^{74} \cdot i = {(i^{2})}^{37} \cdot i = {(- 1)}^{37} \cdot i = - 1 \cdot i = - i$
$i^{80} = {(i^{2})}^{40} = {(- 1)}^{40} = 1$ $i^{80} = {(i^{2})}^{40} = {(- 1)}^{40} = 1$

Now Try Exercises 79 and 83.

These powers of i can also be simplified in terms of $i^{4}$ $i^{4}$ rather than $i^{2}$ $i^{2}$ . Consider $i^{37}$ $i^{37}$ in Example 4(a), for instance. When we divide 37 by 4, we get 9 with a remainder of 1. Then $37 = 4 \cdot 9 + 1$ $37 = 4 \cdot 9 + 1$ , so

i^{37} = {(i^{4})}^{9} \cdot i = 1^{9} \cdot i = 1 \cdot i = i .

$i^{37} = {(i^{4})}^{9} \cdot i = 1^{9} \cdot i = 1 \cdot i = i .$

The other examples shown above can be done in a similar manner.

Conjugates and Division

Conjugates of complex numbers are defined as follows.

Each of the following pairs of numbers are complex conjugates:

- 3 + 7 i and - 3 - 7 i; 14 - 5 i and 14 + 5 i; and 8 i and - 8 i .

$- 3 + 7 i and - 3 - 7 i; 14 - 5 i and 14 + 5 i; and 8 i and - 8 i .$

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

Example 5

Multiply each of the following.

$(5 + 7 i) (5 - 7 i)$ $(5 + 7 i) (5 - 7 i)$
$(8 i) (- 8 i)$ $(8 i) (- 8 i)$

Solution

$\begin{array}{rcll} (5 + 7 i) (5 - 7 i) & = & 5^{2} - {(7 i)}^{2} & Using (A + B) (A - B) = A^{2} - B^{2} \\ = & 25 - 49 i^{2} \\ = & 25 - 49 (- 1) \\ = & 25 + 49 \\ = & 74 \end{array}$ $\begin{array}{rcll} (5 + 7 i) (5 - 7 i) & = & 5^{2} - {(7 i)}^{2} & Using (A + B) (A - B) = A^{2} - B^{2} \\ = & 25 - 49 i^{2} \\ = & 25 - 49 (- 1) \\ = & 25 + 49 \\ = & 74 \end{array}$
$\begin{array}{rcl} (8 i) (- 8 i) & = & - 64 i^{2} \\ = & - 64 (- 1) \\ = & 64 \end{array}$ $\begin{array}{rcl} (8 i) (- 8 i) & = & - 64 i^{2} \\ = & - 64 (- 1) \\ = & 64 \end{array}$

Now Try Exercise 49.

Conjugates are used when we divide complex numbers.

Example 6

Divide $2 - 5 i$ $2 - 5 i$ by $1 - 6 i$ $1 - 6 i$ .

Solution

We write fraction notation and then multiply by 1, using the conjugate of the denominator to form the symbol for 1.

\begin{array}{rcll} \frac{2 - 5 i}{1 - 6 i} & = & \frac{2 - 5 i}{1 - 6 i} \cdot \frac{1 + 6 i}{1 + 6 i} & \begin{array}{l} Note that 1 + 6 i is the conjugate of the divisor, 1 - 6 i . \end{array} \\ = & \frac{(2 - 5 i) (1 + 6 i)}{(1 - 6 i) (1 + 6 i)} \\ = & \frac{2 + 12 i - 5 i - 30 i^{2}}{1 - 36 i^{2}} \\ = & \frac{2 + 7 i + 30}{1 + 36} & i^{2} = - 1 \\ = & \frac{32 + 7 i}{37} \\ = & \frac{32}{37} + \frac{7}{37} i . & Writing the quotient in the form a + b i \end{array}

$\begin{array}{rcll} \frac{2 - 5 i}{1 - 6 i} & = & \frac{2 - 5 i}{1 - 6 i} \cdot \frac{1 + 6 i}{1 + 6 i} & \begin{array}{l} Note that 1 + 6 i is the conjugate of the divisor, 1 - 6 i . \end{array} \\ = & \frac{(2 - 5 i) (1 + 6 i)}{(1 - 6 i) (1 + 6 i)} \\ = & \frac{2 + 12 i - 5 i - 30 i^{2}}{1 - 36 i^{2}} \\ = & \frac{2 + 7 i + 30}{1 + 36} & i^{2} = - 1 \\ = & \frac{32 + 7 i}{37} \\ = & \frac{32}{37} + \frac{7}{37} i . & Writing the quotient in the form a + b i \end{array}$

Now Try Exercise 69.

3.1 Exercise Set

Express the number in terms of i.

1. $\sqrt{- 3}$ $\sqrt{- 3}$
2. $\sqrt{- 21}$ $\sqrt{- 21}$
3. $\sqrt{- 25}$ $\sqrt{- 25}$
4. $\sqrt{- 100}$ $\sqrt{- 100}$
5. $- \sqrt{- 33}$ $- \sqrt{- 33}$
6. $- \sqrt{- 59}$ $- \sqrt{- 59}$
7. $- \sqrt{- 81}$ $- \sqrt{- 81}$
8. $- \sqrt{- 9}$ $- \sqrt{- 9}$
9. $\sqrt{- 98}$ $\sqrt{- 98}$
10. $\sqrt{- 28}$ $\sqrt{- 28}$

Simplify. Write answers in the form $a + b i$ $a + b i$ , where a and b are real numbers.

11. $(- 5 + 3 i) + (7 + 8 i)$ $(- 5 + 3 i) + (7 + 8 i)$
12. $(- 6 - 5 i) + (9 + 2 i)$ $(- 6 - 5 i) + (9 + 2 i)$
13. $(4 - 9 i) + (1 - 3 i)$ $(4 - 9 i) + (1 - 3 i)$
14. $(7 - 2 i) + (4 - 5 i)$ $(7 - 2 i) + (4 - 5 i)$
15. $(12 + 3 i) + (- 8 + 5 i)$ $(12 + 3 i) + (- 8 + 5 i)$
16. $(- 11 + 4 i) + (6 + 8 i)$ $(- 11 + 4 i) + (6 + 8 i)$
17. $(- 1 - i) + (- 3 - i)$ $(- 1 - i) + (- 3 - i)$
18. $(- 5 - i) + (6 + 2 i)$ $(- 5 - i) + (6 + 2 i)$
19. $(3 + \sqrt{- 16}) + (2 + \sqrt{- 25})$ $(3 + \sqrt{- 16}) + (2 + \sqrt{- 25})$
20. $(7 - \sqrt{- 36}) + (2 + \sqrt{- 9})$ $(7 - \sqrt{- 36}) + (2 + \sqrt{- 9})$
21. $(10 + 7 i) - (5 + 3 i)$ $(10 + 7 i) - (5 + 3 i)$
22. $(- 3 - 4 i) - (8 - i)$ $(- 3 - 4 i) - (8 - i)$
23. $(13 + 9 i) - (8 + 2 i)$ $(13 + 9 i) - (8 + 2 i)$
24. $(- 7 + 12 i) - (3 - 6 i)$ $(- 7 + 12 i) - (3 - 6 i)$
25. $(6 - 4 i) - (- 5 + i)$ $(6 - 4 i) - (- 5 + i)$
26. $(8 - 3 i) - (9 - i)$ $(8 - 3 i) - (9 - i)$
27. $(- 5 + 2 i) - (- 4 - 3 i)$ $(- 5 + 2 i) - (- 4 - 3 i)$
28. $(- 6 + 7 i) - (- 5 - 2 i)$ $(- 6 + 7 i) - (- 5 - 2 i)$
29. $(4 - 9 i) - (2 + 3 i)$ $(4 - 9 i) - (2 + 3 i)$
30. $(10 - 4 i) - (8 + 2 i)$ $(10 - 4 i) - (8 + 2 i)$
31. $\sqrt{- 4} \cdot \sqrt{- 36}$ $\sqrt{- 4} \cdot \sqrt{- 36}$
32. $\sqrt{- 49} \cdot \sqrt{- 9}$ $\sqrt{- 49} \cdot \sqrt{- 9}$
33. $\sqrt{- 81} \cdot \sqrt{- 25}$ $\sqrt{- 81} \cdot \sqrt{- 25}$
34. $\sqrt{- 16} \cdot \sqrt{- 100}$ $\sqrt{- 16} \cdot \sqrt{- 100}$
35. $7 i (2 - 5 i)$ $7 i (2 - 5 i)$
36. $3 i (6 + 4 i)$ $3 i (6 + 4 i)$
37. $- 2 i (- 8 + 3 i)$ $- 2 i (- 8 + 3 i)$
38. $- 6 i (- 5 + i)$ $- 6 i (- 5 + i)$
39. $(1 + 3 i) (1 - 4 i)$ $(1 + 3 i) (1 - 4 i)$
40. $(1 - 2 i) (1 + 3 i)$ $(1 - 2 i) (1 + 3 i)$
41. $(2 + 3 i) (2 + 5 i)$ $(2 + 3 i) (2 + 5 i)$
42. $(3 - 5 i) (8 - 2 i)$ $(3 - 5 i) (8 - 2 i)$
43. $(- 4 + i) (3 - 2 i)$ $(- 4 + i) (3 - 2 i)$
44. $(5 - 2 i) (- 1 + i)$ $(5 - 2 i) (- 1 + i)$
45. $(8 - 3 i) (- 2 - 5 i)$ $(8 - 3 i) (- 2 - 5 i)$
46. $(7 - 4 i) (- 3 - 3 i)$ $(7 - 4 i) (- 3 - 3 i)$
47. $(3 + \sqrt{- 16}) (2 + \sqrt{- 25})$ $(3 + \sqrt{- 16}) (2 + \sqrt{- 25})$
48. $(7 - \sqrt{- 16}) (2 + \sqrt{- 9})$ $(7 - \sqrt{- 16}) (2 + \sqrt{- 9})$
49. $(5 - 4 i) (5 + 4 i)$ $(5 - 4 i) (5 + 4 i)$
50. $(5 + 9 i) (5 - 9 i)$ $(5 + 9 i) (5 - 9 i)$
51. $(3 + 2 i) (3 - 2 i)$ $(3 + 2 i) (3 - 2 i)$
52. $(8 + i) (8 - i)$ $(8 + i) (8 - i)$
53. $(7 - 5 i) (7 + 5 i)$ $(7 - 5 i) (7 + 5 i)$
54. $(6 - 8 i) (6 + 8 i)$ $(6 - 8 i) (6 + 8 i)$
55. ${(4 + 2 i)}^{2}$ ${(4 + 2 i)}^{2}$
56. ${(5 - 4 i)}^{2}$ ${(5 - 4 i)}^{2}$
57. ${(- 2 + 7 i)}^{2}$ ${(- 2 + 7 i)}^{2}$
58. ${(- 3 + 2 i)}^{2}$ ${(- 3 + 2 i)}^{2}$
59. ${(1 - 3 i)}^{2}$ ${(1 - 3 i)}^{2}$
60. ${(2 - 5 i)}^{2}$ ${(2 - 5 i)}^{2}$
61. ${(- 1 - i)}^{2}$ ${(- 1 - i)}^{2}$
62. ${(- 4 - 2 i)}^{2}$ ${(- 4 - 2 i)}^{2}$
63. ${(3 + 4 i)}^{2}$ ${(3 + 4 i)}^{2}$
64. ${(6 + 5 i)}^{2}$ ${(6 + 5 i)}^{2}$
65. $\frac{3}{5 - 11 i}$ $\frac{3}{5 - 11 i}$
66. $\frac{i}{2 + i}$ $\frac{i}{2 + i}$
67. $\frac{5}{2 + 3 i}$ $\frac{5}{2 + 3 i}$
68. $\frac{- 3}{4 - 5 i}$ $\frac{- 3}{4 - 5 i}$
69. $\frac{4 + i}{- 3 - 2 i}$ $\frac{4 + i}{- 3 - 2 i}$
70. $\frac{5 - i}{- 7 + 2 i}$ $\frac{5 - i}{- 7 + 2 i}$
71. $\frac{5 - 3 i}{4 + 3 i}$ $\frac{5 - 3 i}{4 + 3 i}$
72. $\frac{6 + 5 i}{3 - 4 i}$ $\frac{6 + 5 i}{3 - 4 i}$
73. $\frac{2 + \sqrt{3} i}{5 - 4 i}$ $\frac{2 + \sqrt{3} i}{5 - 4 i}$
74. $\frac{\sqrt{5} + 3 i}{1 - i}$ $\frac{\sqrt{5} + 3 i}{1 - i}$
75. $\frac{1 + i}{{(1 - i)}^{2}}$ $\frac{1 + i}{{(1 - i)}^{2}}$
76. $\frac{1 - i}{{(1 + i)}^{2}}$ $\frac{1 - i}{{(1 + i)}^{2}}$
77. $\frac{4 - 2 i}{1 + i} + \frac{2 - 5 i}{1 + i}$ $\frac{4 - 2 i}{1 + i} + \frac{2 - 5 i}{1 + i}$
78. $\frac{3 + 2 i}{1 - i} + \frac{6 + 2 i}{1 - i}$ $\frac{3 + 2 i}{1 - i} + \frac{6 + 2 i}{1 - i}$

Simplify.

79. $i^{11}$ $i^{11}$
80. $i^{7}$ $i^{7}$
81. $i^{35}$ $i^{35}$
82. $i^{24}$ $i^{24}$
83. $i^{64}$ $i^{64}$
84. $i^{42}$ $i^{42}$
85. ${(- i)}^{71}$ ${(- i)}^{71}$
86. ${(- i)}^{6}$ ${(- i)}^{6}$
87. ${(5 i)}^{4}$ ${(5 i)}^{4}$
88. ${(2 i)}^{5}$ ${(2 i)}^{5}$

Skill Maintenance

89. Write a slope–intercept equation for the line containing the point (3, −5) and perpendicular to the line $3 x - 6 y = 7$ $3 x - 6 y = 7$ . [1.4]

Given that $f (x) = x^{2} + 4$ $f (x) = x^{2} + 4$ and $g (x) = 3 x + 5$ $g (x) = 3 x + 5$ , find each of the following. [2.2]

90. The domain of $f - g$ $f - g$
91. The domain of $f / g$ $f / g$
92. $(f - g) (x)$ $(f - g) (x)$
93. $(f / g) (2)$ $(f / g) (2)$
94. For the function $f (x) = x^{2} - 3 x + 4$ $f (x) = x^{2} - 3 x + 4$ , construct and simplify the difference quotient

$\frac{f (x + h) - f (x)}{h}$ $\frac{f (x + h) - f (x)}{h}$
. [2.2]

Synthesis

Determine whether the statement is true or false.

95. The sum of two numbers that are conjugates of each other is always a real number.
96. The conjugate of a sum is the sum of the conjugates of the individual complex numbers.
97. The conjugate of a product is the product of the conjugates of the individual complex numbers.

Let $z = a + b i$ $z = a + b i$ and $\bar{z} = a - b i$ $\bar{z} = a - b i$ .

98. Find a general expression for $1 / z$ $1 / z$ .
99. Find a general expression for $z \bar{z}$ $z \bar{z}$ .
100. Solve $z + 6 \bar{z} = 7$ $z + 6 \bar{z} = 7$ for z.
101. Multiply and simplify:

$[x - (3 + 4 i)] [x - (3 - 4 i)] .$ $[x - (3 + 4 i)] [x - (3 - 4 i)] .$

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Table of Contents for 3.1 The Complex Numbers

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Table of Contents for
3.1 The Complex Numbers