Complex Numbers, Polar Equations, and Parametric Equations€¦ · 118 Chapter 8 Complex Numbers,...

114 Copyright © 2013 Pearson Education, Inc.

Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations Section 8.1 Complex Numbers

1. True 2. True

3. True 4. True

5. False Every real number is a complex number.

6. True

7. −4 is real and complex.

8. 0 is real and complex.

9. 13i is complex, pure imaginary and nonreal complex.

10. −7i is complex, pure imaginary and nonreal complex.

11. 5 + i is complex and nonreal complex.

12. −6 − 2i is complex and nonreal complex.

13. π is real and complex.

14. 24 is real and complex.

15. 25 5i- = is complex, pure imaginary and nonreal complex.

16. 36 6i- = is complex, pure imaginary and nonreal complex.

17. 5i

18. 6i

19. 10i

20. 15i

21. 12 2i

22. 10 5i

23. 3 2i-

24. 4 5i-

25. { }4i

26. { }6i

27. { }2 3i

28. { }4 3i

29. 2 2

3 3i

ì üï ïï ï- í ýï ïï ïî þ

30. 3 7

4 4i


31. { }3 5i

32. { }2 7i-

33. 1 6

2 2i

ì üï ïï ïí ýï ïï ïî þ

34. 1 6

3 3i


35. 1 3

2 2i


36. { }1 i

37. 13-

38. 17-

39. 2 6-

40 5 3-

41. 3

42. 10

43. 3i

44. 2i

45. 1

2

46. 1

3

47. 2-

48. 3-

Section 8.1 Complex Numbers 115

Copyright © 2013 Pearson Education, Inc.

49. 3 6i- -

50. 3 2i- -

51. 2 2 2i+

52. 10 2i+

53. 1 2

8 8i- +

54. 1 2

2 2i- +

55. 12 i-

56. 12 4i+

57. 2

58. 1

59. 0

60. 0

61. 13 4 2i- +

62. 3 7 2i- +

63. 8 i-

64. 2 16i- +

65. 14 2i- +

66. 17 i+

67. 5 12i-

68. 3 4i+

69. 10

70. 26

71. 13

72. 52

73. 7

74. 18

75. 25i

76. 53i

77. 12 9i+

78. 120 35i- -

79. 20 15i+

80. 20 60i-

81. 2 2i-

82. 4 i-

83. 3 4

5 5i-

84. 7 24

25 25i-

85. 1 2i- -

86. 2 i- +

87. 5 or 0 5i i+

88. 6 or 0 6i i+

89. 8 or 0 8i i+

90. 12 or 0 12i i+

91. 2 2

or 03 3

i i- -

Note: In the above solution, we multiplied the numerator and denominator by the complex conjugate of 3 ,i namely 3 .i- Since there is a reduction in the end, the same results can be achieved by multiplying the numerator and denominator by .i-

92. 5 5

or 09 9

i i- -

93. i

94. i

95. 1-

96. 1-

97. i-

98. i-

99. 1

100. 1

101. i-

102. 1-

103. i-

104. 1

105. Answers will vary. This method works

because 2 1.i =-

106. Answers will vary.

116 Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations


107. We need to show that 2

2 2.

2 2i i

æ ö÷ç ÷ç + =÷ç ÷÷çè ø

( )

2

2 2

2 2

2 2

2 2

2 2 2 22

2 2 2 2

2 2 2 1 12

4 4 4 2 21 1 1 1

12 2 2 2

i

i i

i i i i

i i i

æ ö÷ç ÷ç + ÷ç ÷÷çè ø

æ ö æ ö÷ ÷ç ç÷ ÷ç ç= + ⋅ ⋅ +÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

= + ⋅ + = + +

= + + - = + - =

108. We need to show that 3

3 1.

2 2i i

æ ö÷ç ÷ç + =÷ç ÷÷çè ø

3

2

2 2

2

3 1

2 2

3 1 3 1

2 2 2 2

3 1 3 3 1 12

2 2 2 2 2 2

3 1 3 3 1

2 2 4 2 4

3 1 3 3

2 2 4 2

i

i i

i i i

i i i

i i

æ ö÷ç ÷ç + ÷ç ÷÷çè ø

æ öæ ö÷ ÷ç ç÷ ÷ç ç= + +÷ ÷ç ç÷ ÷÷ ÷ç çè øè øé ùæ ö æ ö æ öê ú÷ ÷ç ç ÷ç÷ ÷ç ç= + + ⋅ ⋅ +ê ú÷ç÷ ÷ç ç ÷ç÷ ÷÷ ÷ è øç çê úè ø è øê úë û

æ ö é ù÷ç ê ú÷ç= + + +÷ç ê ú÷÷çè ø ê úë ûæ ö÷ç ÷ç= + + +÷ç ÷÷çè ø

( )11

4

é ùê ú-ê úê úë û

3 1 3 3 1

2 2 4 2 4

3 1 2 3

2 2 4 2

i i

i i

æ ö é ù÷ç ê ú÷ç= + + -÷ç ê ú÷÷çè ø ê úë ûæ öæ ö÷ ÷ç ç÷ ÷ç ç= + +÷ ÷ç ç÷ ÷÷ ÷ç çè øè ø

( )

2

2

3 1 1 3

2 2 2 2

3 1 3 3 1 1 1 3

2 2 2 2 2 2 2 2

3 3 1 3

4 4 4 4

3 4 3 3 31

4 4 4 4 4

i i

i i i

i i i

i i i

æ öæ ö÷ ÷ç ç÷ ÷ç ç= + +÷ ÷ç ç÷ ÷÷ ÷ç çè øè ø

= ⋅ + ⋅ + ⋅ +

= + + +

æ ö÷ç ÷ç= + + - = + + - =÷ç ÷÷çè ø

109. −2 + i is a solution of the equation

110. −3 + 4i is a solution of the equation

111. 30 60i+

112. 50 98i+

113. 233 119

37 37

i+

114. 215 95

26 26i+

115. Z = 110 + 32.

116. 16.22»

Section 8.2 Trigonometric (Polar) Form of Complex Numbers

1. length (or magnitude)

2. The argument of a complex number is the angle formed by the vector and the positive x-axis.

3.

4.

5.

6.

Section 8.2 Trigonometric (Polar) Form of Complex Numbers 117


7.

8.

9.

10.

11. 1 − 4i

12. −4 − i

13. 3 i-

14. 2 2i- +

15. 3i-

16. 3

17. 3 3i- +

18. 6 2i-

19. 6 8i- -

20. 9 2i-

21. 7 9i+

22. 2 4i+

23. 7 7

6 6i+

24. 8 13

35 28i-

25. 2 2i+

26. 2 2 3i+

27. 10i

28. 8i-

29. 2 2 3i- -

30. 3 i-

31. 3 3 3

2 2i- +

32. 3 i+

33. 5 5 3

2 2i-

34. 3 2 3 2i- +

35. 1 i- -



36. 6 6

2 2i-

37. 2 3 2i-

38. 2 6

2 2i-

39. ( )6 cos 240 sin 240i+

40. ( )2 cos60 sin 60i+

41. ( )2 cos330 sin 330i+

42. ( )8 cos 30 sin 30 .i+

43. ( )5 2 cos 225 sin 225 .i+

44. ( )2 2 cos 135 sin 135i+

45. ( )2 2 cos 45 sin 45i+

46. ( )4 2 cos 45 sin 45i+

47. ( )5 cos90 sin 90i+

48. ( )2 cos 270 sin 270i+

49.

( )4 cos180 sin180i+

50. ( )7 cos0 sin 0i+

51. ( )13 cos56.31 sin 56.31i+

52. 0.8192 0.5736i+

53. 1.0261 2.8191i- -

54. 17(cos 165.96 sin 165.96 )i+

55. 12(cos 90° + i sin 90°)

56. 3 0 or 3i- + -

57. ( )34 cos 59.04 sin 59.04i+

58. 0.3502 0.9367i- +

59. Since the modulus represents the magnitude of the vector in the complex plane, z = 1 would represent a circle of radius one centered at the origin.

60. When graphing x + yi in the plane as (x, y), if x and y are equal, we are graphing points in the form (x, x). These points make up the line y = x.

61. Since the real part of z = x + yi is 1, the graph of 1 + yi would be the vertical line x = 1.

62. When graphing x + yi in the plane as (x, y), since the imaginary part is 1, the points are of the form (x, 1). These points constitute the horizontal line y = 1.

63. z is in the Julia set 64. (a) Let 1z a bi= + and its complex

conjugate be 2 .z a bi= -

2 21z a b= + and

( )22 2 22 1 .z a b a b z= + - = + =

(b) ( )2 2 21 1 1 2z a b abi- = - - +

( )2 2 22 1 1 2z a b abi- = - - -

(c)−(d)The results are again complex conjugates of each other. At each iteration, the resulting values from

1 2and z z will always be complex

conjugates. Graphically, these represent points that are symmetric with respect to the x-axis, namely points such as (a, b) and (a, −b)

(e) Answers will vary.

65. Let ( )cos sinz r iθ θ= + . This represents a

vector with magnitude r and angle θ . The conjugate of z is the vector with magnitude r pointing in the θ- direction. Thus,

( ) ( )cos 360 sin 360r iθ θé ù- + -ë û satisfies

these conditions as shown below.

( ) ( )

( )( )

( )( )

( )( ) ( )

cos 360 sin 360

cos360 cos sin 360 sin

sin 360 cos cos360 sin

1 cos 0 sin

0 cos 1 sin

cos sin

cos sin

r i

ri

ri

r i

r i

θ θ

θ θθ θ

θ θθ θ

θ θ

θ θ

é ù- + -ë ûé ù + = ê úê ú+ - ê úë ûé ù⋅ + ⋅= ê úê ú+ ⋅ - ⋅ê úë û

= -

é ù= - + -ë û

This represents the conjugate of z, which is a reflection over the x-axis.

Section 8.3 The Product and Quotient Theorems 119


66. Let ( )cos sinz r iθ θ= + . This represents a

vector with magnitude r and angle θ . Then −z should be the vector with magnitude r that points in the opposite direction. The angle that makes this vector point in the opposite direction is θ π+ . Thus

( ) ( )cos sinr iθ π θ πé ù+ + +ë û satisfies these

conditions as shown below.

( ) ( )

( )( )

( )( )( )( )

( )( )

cos sin

cos cos sin sin

sin cos cos sin

cos 1 sin 0

sin 1 cos 0

cos sin

cos sin

r i

ri

ri

r i

r i z

θ π θ π

θ π θ πθ π θ π

θ θ

θ θ

θ θθ θ

é ù+ + +ë ûé ù-= ê úê ú+ +ê úë ûé ù- - ⋅= ê úê úê ú+ - + ⋅ë û

= - -

=- + =-

This represents −z, which is a rotation about the origin.

67. The answer is B 68. The answer is C

69. The answer is A

70. Let z a bi= + . Thus 2 2z a b= + . Now,

let

( ) 2iz i a bi ai bi ai b b ai= + = + = - =- + .

Thus,

( )2 2 2 2 2 2iz b a b a a b z= - + = + = + =

The graphs of z and iz are perpendicular.

Section 8.3 The Product and Quotient Theorems

1. multiply; add

2. divide; subtract

3. 3 3 3i- +

4. 10 3 10i- +

5. 12 3 12i+

6. 20 20 3i+

7. 4

8. 16 0 or –16i- +

9. 0 3 or – 3i i-

10. 6i

11. 15 2 15 2

2 2i- +

12. 21 3 21

2 2i- -

13. 3 i+

14. 6 6 3i- +

15. 2 0 or 2i- + -

16. 2i

17. 1 3

6 6i- -

18. 1 3i- -

19. 2 3 2i-

20. 3 1

2 2i- -

21. 1 1

2 2i- -

22. 1 1

4 4i+

23. 3 i+ 24. 3i



25. 0.6537 7.4715i+

26. 8.3380 3.8881i+

27. 30.8580 18.5414i+

28. 13.5747 24.4894i+

29. 1.9563 0.4158i+

30. 2.0732 2.1684i+

31. 3.7588 1.3681i- -

32. 511.3793 295.2450i- -

33. 2

34. 2 cis 135 ;w = 2 cis 225 .z =

35. 2 cis 0

36. 2; It is the same.

37. i-

38. cis ( 90 )-

39. ;i-

It is the same.


41. The two results will have the same magnitude because in both cases you are finding the product of 2 and 5. We must now determine if the arguments are the same. In the first product, the argument of the product will be 45°+ 90° = 135°. In the second product, the argument of the product will be −315°+ (−270°) = −585°. Now, −585° is

coterminal with −585°+ 2·360° = 135°. Thus, the two products are the same since they have the same magnitude and argument.

42. ( )cos sinz r iθ θ= +

Since ( )1 1 0 1 cos0 sin 0i i= + = + ,

( )( )

( ) ( )

( ) ( )

[ ]

1 cos0 sin 01

cos sin

1cos 0 sin 0

1cos sin

1cos sin

iz r i

ir

ir

ir

θ θ

θ θ

θ θ

θ θ

+ =

+

é ù= - + -ë û

é ù= - + -ë û

= -

43. 1.18 0.14i-

44. 1.7 2.8i+

45. 27.43 11.50i+ 46. 22.75»

Section 8.4 DeMoivre’s Theorem; Powers and Roots of Complex Numbers

1. 0 27 or 27i i+

2. 16 0 or –16i- +

3. 1 0 or 1i+ ⋅

4. 8 0 or 8i+

5. 27 27 3

2 2i-

6. 27 27 3

2 2i- +

7. 16 3 16i- +

8. 128 128 3i- +

9. 0 4096 or 4096i i+ 10. 1 0 or 1i+ 11. 128 128i+ 12. 8 8i- -

13. (a) ( )cos0 sin 0 1 cos0 sin 0i i+ = +

We have 1r = and 0 .θ = Since

( ) ( )3 cos3 sin 3 1 cos 0 sin 0 ,r i iα α+ = +

then we have 3 1 1r r= = and 0 360

3 0 3603

0 120 120 , any integer.

kk

k k k

α α + ⋅= + ⋅ =

= + ⋅ = ⋅

If 0, then 0 .k α= =

If 1, then 120 .k α= =

If 2, then 240 .k α= =

So, the cube roots are cos 0 sin 0 , cos120 sin120 ,i i+ +

and cos 240 sin 240 .i+

(b)

Section 8.4 DeMoivre’s Theorem; Powers and Roots of Complex Numbers 121


14. (a) cos30 sin 30 , cos150 sin150 ,

and cos 270 sin 270 .

i ii

+ + +

(b)

15. (a) 2 cis 20 , 2 cis140 , and 2 cis 260 .

(b)

16. (a) 3 cis100 ,3 cis 220 , and 3 cis 340 .

(b)

17. (a) ( )( )( )

2 cos90 sin 90 ;

2 cos 210 sin 210 ;

2 cos330 sin 330

i

i

i

+

+

+

(b)

18. (a) ( )( )( )

3 cos30 sin 30 ;

3 cos150 sin150 ;

3 cos 270 sin 270

i

i

i

+

+

+

(b)

19. (a) ( )( )( )

4 cos 60 sin 60 ;

4 cos180 sin180 ;

4 cos300 sin 300

i

i

i

+

+

+

(b)

20. (a) ( )3 cos0 sin 0 ;i+

( )( )

3 cos120 sin120 ;

3 cos 240 sin 240

i

i

+

+

(b)

21. (a) ( )

( )

( )

3

3

3

2 cos 20 sin 20 ;

2 cos140 sin140 ;

2 cos 260 sin 260

i

i

i

+

+

+

(b)



22. (a) ( )

( )

( )

3

3

3

4 cos100 sin100 ;

4 cos 220 sin 220 ;

4 cos340 sin 340

i

i

i

+

+

+

(b)

23. (a) ( )

( )

( )

3

3

3

4 cos50 sin 50 ;

4 cos170 sin170 ;

4 cos 290 sin 290

i

i

i

+

+

+

(b)

24. (a) ( )

( )

( )

3

3

3

2 cos110 sin110 ;

2 cos 230 sin 230 ;

2 cos350 sin 350

i

i

i

+

+

+

(b)

25. cos 0 sin 0 , and cos180 sin180i i+ + (or 1

and –1)

26. cos 0 sin 0 , cos90 sin 90 ,i i+ +

cos180 sin180 , and cos 270 sin 270i i+ +

(or 1, i, –1 and –i)

27. cos 0 sin 0 , cos 60 sin 60 ,i i+ +

cos120 sin120 , cos180 sin180 ,i i+ +

cos 240 sin 240 ,and cos300 sin 300 .i i+ +

1 3 1 3or 1, , , 1,

2 2 2 2

1 3 1 3, and

2 2 2 2

i i

i i

æçç + - + -ççèö÷÷- - - ÷÷÷ø

28. cos 45 sin 45 andi+ cos 225 sin 225i+

2 2 2 2or and

2 2 2 2i i

æ ö÷ç ÷ç + - - ÷ç ÷÷çè ø

29. cos30 sin 30 ,i+ cos150 sin150 , andi+

cos 270 sin 270 .i+

Section 8.4 DeMoivre’s Theorem; Powers and Roots of Complex Numbers 123


30. cos 22.5 sin 22.5 ,i+

cos112.5 sin112.5 ,cos 202.5 sin 202.5 ,i i+ + and cos 292.5 sin 292.5 .i+

31. {

}cos 0 sin 0 , cos120 sin120 ,

cos 240 sin 240

i i

i

+ +

+

or 1 3 1 3

1, ,2 2 2 2

i iì üï ïï ï- + - -í ýï ïï ïî þ

32. {cos60 sin 60 , cos180 sin180 ,i i+ +

}cos300 sin300i+ or

1 3 1 3, 1,

2 2 2 2i i

ì üï ïï ï+ - -í ýï ïï ïî þ

33. {cos90 sin 90 , cos 210 sin 210 ,i i+ +

}cos330 sin330i+ or

3 1 3 1, ,

2 2 2 2i i i

ì üï ïï ï- - -í ýï ïï ïî þ

34. {

}

cos 67.5 sin 67.5 ,

cos157.5 sin157.5 ,

cos 247.5 sin 247.5 ,

cos337.5 sin 337.5

iiii

+

+ +

+

35. ( ) ( ){2 cos 0 sin 0 , 2 cos120 sin120 ,i i+ +

( )}2 cos 240 sin 240i+ or

{ }2, 1 3 , 1 3i i- + - -

36. ( ) ( ){3 cos 60 sin 60 , 3 cos180 sin180 ,i i+ +

( )}3 cos300 sin 300i+ or

3 3 3 3 3 3, 3,

2 2 2 2i i

ì üï ïï ï+ - -í ýï ïï ïî þ

37. {cos 45 sin 45 , cos135 sin135 ,i i+ +

}cos 225 sin 225 , cos315 sin315i i+ + or

2 2 2 2 2 2, , ,

2 2 2 2 2 2

2 2

2 2

i i i

i

ìïï + - + - -íïïîüïï- ýïïþ

38. ( ) ( ){( )

2 cos 45 sin 45 ,2 cos135 sin135 ,

2 cos 225 sin 225 ,

i i

i

+ +

+

( )}2 cos315 sin 315i+ or

{ }2 2, 2 2, 2 2, 2 2i i i i+ - + - - -

39. {

}

cos 22.5 sin 22.5 , cos112.5 sin112.5 ,

cos 202.5 sin 202.5 ,

cos 292.5 sin 292.5

i iii

+ +

+

+

40. ( ){

}

cos18 sin18 , cos90 sin 90 or 0 ,

cos162 sin162 , cos 234 sin 234 ,

cos306 sin 306

i i

i ii

+ +

+ +

+

41. ( ) ( ){( )}

2 cos 20 sin 20 , 2 cos140 sin140 ,

2 cos 260 sin 260

i i

i

+ +

+

42. ( ) ( ){( )( )}

2 cos15 sin15 , 2 cos105 sin105 ,

2 cos195 sin195 ,

2 cos 285 sin 285

i i

i

i

+ +

+

+

43. 1 3 1 3

1, , .2 2 2 2

x i i= - + - - We see that the

solutions are the same as Exercise 31.

44. 3 3 3 3 3 3

3, , .2 2 2 2

x i i=- + - We see that the

solutions are the same as Exercise 36.

45. De Moivre’s theorem states that

( ) ( )2 2cos sin 1 cos 2 sin 2

cos 2 sin 2

i ii

θ θ θ θθ θ

+ = +

= +

46. ( )2cos sin cos 2 sin 2i i + = +

47. 2 2cos sin cos 2 - =

48. 2sin cos sin 2θ θ θ=



49. (a) Yes

(b) No

(c) Yes 50. (a) blue

(b) red

(c) yellow

51. Using the trace function, we find that the other four fifth roots of 1 are: 0.30901699 + 0.95105652i, –0.809017 + 0.58778525i, –0.809017 – 0.5877853i, 0.30901699 – 0.9510565i.

52. Using the trace function, we find that three of the tenth roots of 1 are: 1, 0.80901699 + 0.58778525i, 0.30901699 + 0.95105652i.

53. 4;2 2 3i- -

54. { }0.3436 1.4553 , 0.3436 1.4553i i+ - -

55. { }1.8174 0.5503 ,1.8174 0.5503i i- + -

56. {

}1.3606 1.2637 , 1.7747 0.5464 ,

0.4141 1.8102

i i

i

+ - +

-

57.

{

}

0.8771 0.9492 , 0.6317 1.1275 ,

1.2675 0.2524 , 0.1516 1.2835 ,

1.1738 0.54083

i ii i

i

+ - +

- - - -

-

58. The number 1 has 64 complex 64th roots. Two of them are real, 1 and –1, and 62 of them are not real.

59. false

60. false

61. If z is an nth root of 1, then 1.nz = Since

1 1 11 ,

1

n

n zz

æ ö÷ç= = = ÷ç ÷çè ø then

1

z is also an nth root

of 1.

62. – 64. Answers will vary.

Chapter 8 Quiz (Sections 8.1−8.4)

1. (a) 6 2-

(b) 1

3i

2. (a) 1 6i- +

(b) 7 4i+

(c) 17 17i- -

(d) 7 23

17 17i- -

3. (a) 2 2i- -

(b) i , or 0 + i

4. 1 47

6 6i


5. (a) ( )4 4 cos 270 sin 270i i- = +

(b) ( )1 3 2 cos300 sin 300i i- = +

(c) ( )3 10 cos 198.4 sin198.4 .i i- - = +

6. (a) 2 2 3i+

(b) 3.2139 3.8302i- +

(c) 0 7i-

(d) 2 or 2 0i+

7. (a) ( )36 cos130 sin130i+

(b) 2 3 2i+

(c) 27 3 27

2 2i- +

(d) 864 864 3i- -

8. ( ) ( ){( )

2 cos 45 sin 45 ,2 cos135 sin135 ,

2 cos 225 sin 225 ,

i i

i

+ +

+

( )}2 cos315 sin 315i+ or

{ }2 2, 2 2, 2 2, 2 2i i i i+ - + - - -

Section 8.5 Polar Equations and Graphs 125


Section 8.5 Polar Equations and Graphs

1. (a) II (since 0 and 90 180 )r θ> < <

(b) I (since r > 0 and 0° < 90θ < )

(c) IV (since r > 0 and –90° < θ < 0°)

(d) III (since r > 0 and 180° < θ < 270°)

2. (a) positive x-axis

(b) negative x-axis

(c) negative y-axis

(d) positive y-axis (since 450° – 360° = 90°)

For Exercises 3(b)−14(b), answers may vary.

3. (a)

(b) (1, 405°) and (–1, 225°)

(c) 2 2

,2 2

æ ö÷ç ÷ç ÷ç ÷÷çè ø

4. (a)

(b) (3, 480°) and (–3, 300°)

(c) 3 3 3

,2 2

æ ö÷ç ÷ç- ÷ç ÷÷çè ø

5. (a)

(b) (–2, 495°) and (2, 315°)

(c) ( )2, 2-

6. (a)

(b) (−4, 390°) and (4, 210°)

(c) ( )2 3, 2- -

7. (a)

(b) (5, 300°) and (–5, 120°)

(c) 5 5 3

,2 2

æ ö÷ç ÷ç - ÷ç ÷÷çè ø

8. (a)

(b) (2, 315°) and (–2, 135°)

(c) ( )2, 2-

9. (a)

(b) (–3, 150°) and (3, –30°)

(c) 3 3 3

,2 2




10. (a)

(b) (–1, 240°) and (1, 60°)

(c) 1 3

,2 2

æ ö÷ç ÷ç ÷ç ÷÷çè ø

11. (a)

(b) 11 2

3, and 3,3 3

æ ö æ ö÷ ÷ç ç-÷ ÷ç ç÷ ÷ç çè ø è ø

(c) 3 3 3

,2 2


12. (a)

(b) 4, and 4,2 2

æ ö æ ö÷ ÷ç ç- -÷ ÷ç ç÷ ÷ç çè ø è ø

(c) (0, −4)

13. (a)

(b) 7 4

2, and 2,3 3

æ ö æ ö÷ ÷ç ç- ÷ ÷ç ç÷ ÷ç çè ø è ø

(c) ( )1, 3- -

14. (a)

(b) 11

5,6

πæ ö÷ç ÷ç ÷çè ø and

175,

6

æ ö÷ç- ÷ç ÷çè ø

(c) 5 3 5

,2 2


For Exercises 15(b)–26(b), answers may vary.

15. (a)

(b) ( )2, 315 ; a second possibility is

( )2,135 .-

16. (a)


( )2, 225 .-

17. (a)


( )3, 270 .-



18. (a)

(b) ( )3, 270 ; second possibility is

( )3, 90 .-

19. (a)


( )2, 225 .-

20. (a)


( )2, 315 .-

21. (a)


( )3, 240 .-

22. (a)


( )1, 30 .-

23. (a)


( )3, 180 .-

24. (a)

(b) (2, 180°); a second possibility is (−2, 0°).

25. (a)




26. (a)


27. 4

cos sinr

=

-

28. 7

cos sinr

-

=+

29. r = 4 or r = −4

30. r = 3 or r = −3

31. 5

2cos sinr

=

+

32. 6

3cos 2sinr

=

-

33. sinr kθ = 34. sin

krθ

=

35. sin

krθ

= cscr k θ=

36. y = 3



37. cosr kθ = 38. cos

krθ

=

39. seccos

kr r k θθ

= =

40. x = 3

41. C

42. D

43. A

44. B

45. 2 2cosr θ= + (cardioid)

46. 8 6cos (limaçon)r θ= +

47. 3 cos r θ= + (limaçon)

48. 2 cos r θ= - (limaçon)

49. 4cos 2 r θ= (four-leaved rose)

50. 3cos5 r θ= (five-leaved rose)

51. 2 4cos 2 2 cos 2 (lemniscate)r rθ θ= =

52. 2 4sin 2 2 sin 2 (lemniscate)r rθ θ= =



53. 4 4cosr θ= - (cardioid)

54. 6 3cos (limaçon)r θ= -

55. 2sin tanr θ θ= (cissoid)

56. cos 2

(cissoid with a loop)cos

r θθ

=

57. r = 2 sin θ

Multiply both sides by r to obtain 2 2 sin .r r θ= Since 2 2 2 and sin ,r x y y r θ= + = 2 2 2x y y+ = .

Complete the square on y to obtain

( )22 2 22 1 1 1 1.x y y x y+ - + = + - =

The graph is a circle with center at (0, 1) and radius 1.

58.

59.

60.



61.

62.

63.

64.

65.

66.

67.

68.

69. In rectangular coordinates, the line passes through ( )1,0 and ( )0,2 . So

2 0 22

0 1 1m -= = =-

- - and

( ) ( )0 2 1 2 2y x y x- =- - =- +

2 2.x y+ = Converting to polar form

cos sin

cra bθ θ

=+

, we have:

2.

2cos sinr

θ θ=

+




71. (a) ( ),r θ-

(b) ( ),r π θ- or ( ),r θ- -

(c) ( ),r π θ+ or ( ),r θ-

72. (a) θ- (b) π θ-

(c) –r; –θ (d) –r

(e) π θ+ (f) the polar axis

(g) the line 2

πθ =

73.

74.

75.

76.

77. r = 4 sinθ , r = 1 + 2 sinθ , 0 ≤ θ < 2π 4sin 1 2sin 2sin 1

1 5sin or

2 6 6

θ θ θπ πθ θ

= + =

= =

The points of intersection are

4sin , 2, 6 6 6

π π πæ ö æ ö÷ ÷ç ç=÷ ÷ç ç÷ ÷ç çè ø è ø and

5 5 54sin , 2, .

6 6 6

π π πæ ö æ ö÷ ÷ç ç=÷ ÷ç ç÷ ÷ç çè ø è ø 78. (3, 60°); (3, 300°)

79. 4 2 4 2 5

, ; , 2 4 2 4

æ ö æ ö+ -÷ ÷ç ç÷ ÷ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

80. 0, ;2

æ ö÷ç ÷ç ÷çè ø 1, ;

4

æ ö÷ç ÷ç ÷çè ø

31,

4

æ ö÷ç- ÷ç ÷çè ø

81. (a) Plot the following polar equations on the same polar axis in radian mode:

Mercury: 20.39(1 0.206 )

;1 0.206cos

rθ

-=

+

Venus: 20.78(1 0.007 )

;1 0.007 cos

rθ

-=

+

Earth: 21(1 0.017 )

1 0.017 cosr

θ-

=+

;

Mars: 21.52(1 0.093 )

.1 0.093cos

rθ

-=

+



(b) Plot the following polar equations on the same polar axis:

Earth: 21(1 0.017 )

1 0.017 cosr

θ-

=+

;

Jupiter: 25.2(1 0.048 )

1 0.048cosr

θ-

=+

;

Uranus: 219.2(1 0.047 )

1 0.047 cosr

θ-

=+

;

Pluto: 239.4(1 0.249 )

1 0.249cosr

θ-

=+

.

(c) We must determine if the orbit of Pluto is always outside the orbits of the other planets. Since Neptune is closest to Pluto, plot the orbits of Neptune and Pluto on the same polar axes.

Neptune: 230.1(1 0.009 )

;1 0.009cos

rθ

-=

+

Pluto: 239.4(1 0.249 )

1 0.249cosr

θ-

=+

The graph shows that their orbits are very

close near the polar axis. Use ZOOM or change your window to see that the orbit of Pluto does indeed pass inside the orbit of Neptune. Therefore, there are times when Neptune, not Pluto, is the farthest planet from the sun. (However, Pluto’s average distance from the sun is considerably greater than Neptune’s average distance.)

82. (a) In degree mode, graph 2 40,000cos 2 .r θ=

Inside the “figure eight” the radio signal

can be received. This region is generally in an east-west direction from the two radio towers with a maximum distance of 200 mi.



(b) In degree mode, graph 2 22,500sin 2 .r θ=

Inside the “figure eight” the radio signal

can be received. This region is generally in a northeast-southwest direction from the two radio towers with a maximum distance of 150 mi.

Section 8.6 Parametric Equations, Graphs, and Applications

1. C

2. D

3. A

4. B

5. (a)

(b) 2 2( 2) or 4 4.y x y x x= - = - + Since t is in [–1, 1], x is in [–1 + 2, 1 + 2] or [1, 3].

6. (a)

(b) 12

xy = + Since t is in [–2, 3], x is in [2(–

2), 2(3)] or [–4, 6].

7. (a)

(b) 2 ,x t= we have 23 4.y x= - Since t is

in [0, 4], x is in [ 0, 4] or [0, 2].

8. (a)

(b) ( )22 2 4x t y y= = = or 4y x= . Since t

is in [0, 4], x is in 2 2[0 , 4 ] , or [0, 16].

9. (a)

(b) 2y x= - for x is in (–∞, ∞)

10. (a)

(b) ( )211 2

4y x= + + for x is in (–∞, ∞)

Section 8.6 Parametric Equations, Graphs, and Applications 135


11. (a)

(b) 2 2 4.x y+ = for x is in [–2, 2]

12. (a)

(b) 2 2 2 2

1 15 35 3

x y x yæ ö æ ö÷ ÷ç ç÷ ÷+ = + =ç ç÷ ÷ç ç÷ ÷ç çè ø è ø

13. (a)

(b) 2

2 19

xy = + for x is in (–∞, ∞)

14. (a)

(b) 21y x= +

15. (a)

(b) 1

,yx

= where x is in (0, 1]

16. (a)

(b) 1

,yx

= where x is in (0, ∞).

17. (a)

(b) 2 2y x= + , x is in (–∞, ∞).

18. (a)

(b) 4 1y x= - ; x is in ) [ )0, or 0, .é ¥ ¥êë



19. (a)

(b) ( ) ( )2 22 1 1.x y- + - =

20. (a)

(b) ( ) ( )2 21 2

14 9

x y- -+ =

Also, –1 sin 1, 2 2sin 2

1 1 2sin 3.

t tt

£ £ - £ £

- £ + £

21. (a)

(b) 1

;yx

= x is in ( , 0) (0, ).-¥ È ¥

22. (a)

(b) 2

;yx

= x is in (– , 0) (0, )¥ È ¥

23. (a)

(b) 6;y x= - x is in (–∞, ∞)

24. (a)

(b) 6;y x= - x is in [ )2,¥

25.

2 2 9x y+ =

26.

2 2 4x y+ =

Section 8.6 Parametric Equations, Graphs, and Applications 137


27.

2 2

19 4

x y+ =

28.

2 2

116 9

x y+ =

In Exercises 29 – 32, answers may vary.

29. ( )23 1y x= + -

( )2, 3 1x t y t= = + - for t in ( ),-¥ ¥ ;

23, 1x t y t= - = - for t in ( ),-¥ ¥

30. ( )24 2y x= + +

( )2, 4 2x t y t= = + + for t in ( ),-¥ ¥ ;

24, 2x t y t= - = + for t in ( ),-¥ ¥

31. ( )22 2 3 1 2y x x x= - + = - +

( )2, 1 2x t y t= = - + for t in ( ),-¥ ¥ ;

21, 2x t y t= + = + for t in ( ),-¥ ¥

32. ( )22 4 6 2 2y x x x= - + = - +

( )2, 2 2x t y t= = - + for t in ( ),-¥ ¥ ;

22, 2x t y t= + = + for t in ( ),-¥ ¥

33. [ ]2 2sin , 2 2cos , for in 0,4x t t y t t π= - = -

34. [ ]sin , 1 cos , for in 0,4x t t y t t π= - = -

Exercises 35−38 are graphed in parametric mode in the following window.

35. 2cos , 3sin 2x t y t= =

36. 3cos 2 , 3sin 3x t y t= =

37. 3sin 4 , 3cos3x t y t= =

38. 4sin 4 , 3sin 5x t y t= =



For Exercises 39–42, recall that the motion of a projectile (neglecting air resistance) can be modeled

by: 20 0( cos ) , ( sin ) 16x v t y v t tθ θ= = - for t in

[0, k].

39. (a) 24x t= 216 24 3y t t=- +

(b) 2

336

xy x=- +

(c) 2.6 sec;62 ft

40. (a) 75x t= 216 75 3y t t=- +

(b) 2163

5625y x x=- +

(c) 8.1 sec; 609 ft

41. (a) ( )88cos 20x t=

2(88sin 20 ) 16 2y t t= - +

(b) ( )2

2tan 20 2

484cos 20

xy x= - +

(c) 1.9 sec; 161 feet 42. (a) ( )136cos 29 tx =

( )22.5 16 136sin 29y t t= - +

(b) ( )2

2tan 29 2.5

1156cos 29

xy x= - +

(c) 4.2 sec; 495 feet

43. (a) 213 8;

256y x x=- + +

This is a parabolic path.

(b) approximately 7 sec and 448 feet

44. 1456 feet

45. (a) 32x t= 232 3 16 3y t t= - +

(b) approximately 112.6 feet.

(c) The maximum height of 51 ft is reached

at 3 1.73» sec. Since 32 ,x t= the ball

has traveled horizontally 32 3 55.4» ft.

(d) Yes

46. (a)

(b) 20.0 »

(c) ( )

( )2

88 cos 20.0 ,

16 88 sin 20.0

x t

y t t

=

=- +

47. (a)

(b) 50.0 »

(c) ( )88 cos50.0 ,x t=

( )216 88 sin 50.0y t t=- +

For Exercises 48−51, many answers are possible.

48. ( )1 1y m t x y= - + or ( )21 1y m t x y= - + .

49. 2( )y a t h k= - + or 2y at k= +

50. secx a θ= . We therefore have tan ,b y =

2

2 2 22

by t aa

51. To find a parametric representation, let x = a sin t. We therefore have cos ;y b t=

22 2

2; 1

tx t y ba

Chapter 8 Review Exercises 139


52. To show that r aθ= is given parametrically by cos , sin ,x a y aθ θ θ θ= =

( )for in , ,θ -¥¥ we must show that the

parametric equations yield ,r aθ= where 2 2 2.r x y= +

( ) ( )

( )

2 2 2

2 22

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

cos sin

cos sin

cos sin

cos sin

or just

r x y

r a a

r a a

r a a

r a r a

r a r a

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ

= +

= +

= +

= +

= + =

= =

This implies that the parametric equations satisfy .r aθ=

53. To show that r aθ = is given parametrically

by cos sin

, ,a ax yθ θ

θ θ= =

( ) ( )for in ,0 0, ,θ -¥ È ¥ we must show that

the parametric equations yield ,r aθ = where 2 2 2.r x y= +

( )

2 2 2

2 22

2 2 2 22

2 2

2 22 2 2

2 2

2 22 2 2 2

2 2

cos sin

cos sin

cos sin

cos sin

or just

r x y

a ar

a ar

a ar

a ar r

a ar r

θ θθ θ

θ θθ θ

θ θθ θ

θ θθ θ

θ θ

= +

æ ö æ ö÷ ÷ç ç= + ÷ ÷ç ç÷ ÷ç çè ø è ø

= +

= +

= + =

= =

This implies that the parametric equations satisfy .r aθ =

54. The second set of equations x = cos t, y = –sin t, t in [0, 2π] trace the circle out clockwise. A table of values confirms this.

55. If ( )x f t= is replaced by ( )x c f t= + , the

graph will be translated c units horizontally.

56. If ( )y g t= is replaced by ( ),y d g t= + the

graph is translated vertically d units.

Chapter 8 Review Exercises

1. 3i

2. 2 3i

3. { }9i

4. 3 23

4 4i


5. 2 3i- -

6. 8 15i-

7. 5 4i+

8. 5 6i- -

9. 29 37i+

10. 22 3i+

11. 32 24i- +

12. 27 36i-

13. 2 2i- -

14. 2 11i+

15. 2 5i-

16. 3 7

2 2i- -

17. 3 11

26 26i- +

18. 2 3i-

19. i

20. i-

21. 0 – 30 or – 30i i

22. 3 3 3i- -

23. 1 3

8 8i- +

24. 2 0 or 2i- + -

25. 8i

26. 128 128i- +

27. 1 3

2 2i- -

28. The vector representing a real number will lie on the x-axis in the complex plane.



29.

30.

31.

32. The resultant of 7 + 3i and −2 + i is ( ) ( )7 3 2 5 4i i i+ + - + = + .

33. ( )2 2 cos135 sin135i+

34. 0 3 or 3i i+

35. 2 2i- -

36. ( )8 cos120 sin120i+

37. ( )2 cos315 sin 315i+

38. 2 2 3i- -

39. 4(cos 270° + i sin 270°).

40. 4.4995 5.3623i-

41. Since the imaginary part of z is the negative of the real part of z, we are saying y = –x. This is a line.

42. Since the modulus of z is 2, the graph would be a circle, centered at the origin, with radius 2.

43. ( )6 2 cos105 sin105 ,i+

( )6 2 cos 225 sin 225 ;i+

( )6 2 cos345 sin 345i+

44. ( )10 8 cos 27 sin 27 ;i+

( )10 8 cos99 sin 99 ;i+

( )10 8 cos171 sin171 ;i+

( )10 8 cos 243 sin 243 ;i+

( )10 8 cos315 sin 315i+

45. None

46. one

47.

( ) ( ){( )( )}

2 cos 45 sin 45 ,2 cos135 sin135 ,

2 cos 225 sin 225 ,

2 cos315 sin 315

i i

i

i

+ +

+

+

48. ( ) ( ){( )}

5 cos 60 sin 60 ,5 cos180 sin180 ,

5 cos300 sin 300

i i

i

+ +

+

49. {cos 135º + i sin 135º, cos 315º + i sin 315º}

50. 5 2 5 2

, 2 2


51. (2, 120°)

52. Since r is constant, the graph will be a circle.

53. r = 4 cos θ is a circle.

Chapter 8 Review Exercises 141


54. r = –1 + cos θ is a cardioid.

55. r = 2 sin 4θ is an eight-leaved rose.

56. line;

57. 2 236 or 6 9 0

2y x y x

æ ö÷ç=- - + - =÷ç ÷çè ø

58. 2 2

2 21 1 1 or 0

2 2 2x y x y x yæ ö æ ö÷ ÷ç ç- + - = + - - =÷ ÷ç ç÷ ÷ç çè ø è ø

59. 2 2 4x y+ =

60. sin cos or tan 1 = =

61. tan

tan sec or cos

r r

= =

62. 5 or 5r r= =-

63. B

64. A

65. C

66. B

67. 2

or 2seccos

r r

= =

68. 2

or 2cscsin

r r

= =

69. 4

cos 2sinr

=

+

70. 2 or 2r r=- =-

71.

72. Convert the polar coordinates to rectangular coordinates, apply the distance formula: ( )1 1,r θ in rectangular coordinates is

( )1 1 1 1cos , sin .r rθ θ

( )2 2,r θ in rectangular coordinates is

( )2 2 2 2cos , sin .r rθ θ

( )2 21 2 1 2 1 22 cosd r r r r = + - -

73. 2 1.y x= + , x is in [0, )¥

74. 3 5x y- = x is in [–13, 17]

75. 2

3 125

xy = + , x is in (–∞, ∞)

76. 1

4y

x=

-;x is in [5, ∞)

77. ( )2 211 or 2 1 0

2y x y x=- - + - = ,

x is in [–1, 1]

78. 3 5cosx t= + , 4 5siny t= + ,

where t in [0, 2π]

79. (a) ( )118cos 27x t= and

( ) 2118sin 27 16 3.2y t t= - +

(b) ( )22

43.2 tan 27

3481cos 27y x x= - +

(c) t = 3.4 sec, the baseball traveled 358 ft 80. 1 + i is not in the Mandelbrot set



Chapter 8 Test

1. (a) 4 3-

(b) 1

2i

(c) 1

3

2. (a) 7 3i-

(b) 3 5i- -

(c) 14 18i-

(d) 3 11

13 13i-

3. (a) i-

(b) 2i

4. 1 31

4 4i


5. (a) ( )3 cos90 sin 90i+

(b) ( )5 cos 63.43 sin 63.43i+

(c) 2(cos 240 sin 240 )i+

6. (a) 3 3 3

2 2i+

(b) 4 cis 40° = 3.06 + 2.57i

(c) 0 3 3i i+ =

7. (a) ( )16 cos50 sin 50i+

(b) 2 3 2i+

(c) 4 3 4i+

8. ( )2 cos67.5 sin 67.5 ;+

( )( )( )

2 cos157.5 sin157.5 ;

2 cos 247.5 sin 247.5 ;

2 cos337.5 sin 337.5

i

i

i

+

+

+

9. Answers may vary. (a) (5, 90°);a second possibility is (5, –270°)

(b) ( )2 2, 225 , a second possibility is

( )2 2, 135-

10. (a) 3 2 3 2

, 2 2

æ ö- ÷ç ÷ç ÷ç ÷÷çè ø

(b) ( )0, 4-

11. 1 cosr θ= - is a cardioids

12. 3cos3r θ= is a three-leaved rose

13. (a) 2 4x y- =-

(b) 2 2 36.x y+ =

Chapter 8 Test 143


14.

15. x = 2cos 2t, y = 2sin 2t for t in [ ]0,2π

16. 2 1 1 2z i- =- -

Since

( ) ( )2 21 2 1 4 5 2,r = - + - = + = > z is

not in the Julia set.

Complex Numbers, Polar Equations, and Parametric Equations€¦ · 118 Chapter 8 Complex Numbers,...

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Transcript of Complex Numbers, Polar Equations, and Parametric Equations€¦ · 118 Chapter 8 Complex Numbers,...