Chapter 8. Complex Numbers, Polar Equations, and Parametric Equations. Section 8.1: Complex Numbers. 26. { ± 6i}

Transcription

1 Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations 6. { ± 6i} Section 8.1: Complex Numbers 1. true. true. true 4. true 5. false (Every real number is a complex number. 6. true 7. 4 is real and complex is real and complex. 9. 1i is complex, pure imaginary and nonreal complex i is complex, pure imaginary and nonreal complex i is complex and nonreal complex i is complex and nonreal complex. 1. π is real and complex is real and complex = 5i is complex, pure imaginary and nonreal complex = 6i is complex, pure imaginary and nonreal complex i 18. 6i 7. { ± i } 8. { ± 4i } i ± i ± 1. { ± i 5}. { ± i 7} i ± 1 6 i ± 1 i ± 6. { 1± i} i i i. 10i 5. i 4. 4i 5 5. { ± 4i} i 44. i i 111

2 11 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations i i i i i i i i i 6. + i 6. 8 i i i i i i i 76. 5i i i i i 81. i 8. i i 86. i i i Answers will vary. 94. Answers will vary. 95. i i i i i i i or 0+ 5i i or 0+ 6i i or 0+ 8i i or 0 + 1i i or 0 i Note: In the above solution, we multiplied the numerator and denominator by the complex conjugate of, i namely. i Since there is a reduction in the end, the same results can be achieved by multiplying the numerator and denominator by i. 5 5 i or 0 i 9 9

3 Section 8.: Trigonometric (Polar Form of Complex Numbers We need to show that + i i + = i. = + i+ i 1 1 = + i+ i = + i+ i = + i+ ( 1 = + i = i 108. We need to show that 1 + i 1 i + = i. 1 1 = + i + i = + i + i+ i 1 1 = + i i i = + i i ( = + i + i = + i i = + i + i = + i+ i+ i 1 = + i+ i+ i = + i+ ( 1 = + i+ i = i Z = I = + i Z = i Section 8.: Trigonometric (Polar Form of Complex Numbers 1. length (or magnitude. The argument of a complex number is the angle formed by the vector and the positive x-axis i i i i

4 114 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations i 1. 4 i 1. i i i i 0. 9 i i. 6 8i i i i 6. + i 7. 10i 8. 8i 9. i 0. i i 1. + i. + i. 5 5 i 15. i 4. + i 5. 1 i i 7. i 8. 6 i 6 cos 40 + i sin ( 40. ( cos60 + i sin 60

5 Section 8.: Trigonometric (Polar Form of Complex Numbers ( cos0 + i sin ( cos 0 + i sin ( cos 5 + i sin ( cos 15 + i sin ( cos 45 + i sin ( cos 45 + i sin cos90 ( + i sin ( cos 70 + i sin ( cos180 + i sin cos0 ( + i sin ( cos i sin i i (cos i sin (cos 90 + i sin i or ( cos i sin i 59. Since the modulus represents the magnitude of the vector in the complex plane, r = 1 would represent a circle of radius one centered at the origin. 60. When graphing x+ yi in the plane as ( xy,, if x and y are equal, we are graphing points in the form ( xx,. 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 = When graphing x+ yi in the plane as ( xy,, since the imaginary part is 1, the points are of the form (x, 1. These points constitute the horizontal line y = yes 64. (a Let z 1 = a + bi and its complex conjugate be z = a bi. 1 z = a + b and ( = + = + = 1. z a b a b z ( ( z1 1= a b 1 + abi z 1= a b 1 abi (c (d The results are again complex conjugates of each other. At each iteration, the resulting values from z1 and z will always be complex conjugates. Graphically, these represent points that are symmetric with respect to the x-axis,, a, b. namely points such as ( ab and ( (e Answers will vary. 65. Let z r( cosθ isinθ = +. 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, ( θ i ( θ r cos 60 + sin 60 satisfies these conditions as shown below. r cos ( 60 θ + isin ( 60 θ = r ( cos60 cosθ + sin 60 sinθ i ( sin 60 cosθ cos 60 sinθ + = r ( 1cos θ + 0sin θ i ( 0 cosθ 1 sinθ + = r( cosθ isinθ = r cos ( θ + isin ( θ This represents the conjugate of z, which is a reflection over the x-axis. 66. Let z r( cosθ isinθ = +. 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 r cos ( θ + π + isin ( θ + π satisfies these conditions as shown below.

6 116 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations ( θ π isin ( θ π ( cosθ cosπ sinθ sin π + i ( sinθ cosπ + cosθ sin π ( cosθ ( 1 sinθ 0 + i ( sinθ ( 1 + cosθ 0 ( cosθ isinθ r( cosθ isinθ z r cos = r = r = r = + = This represents z, which is a rotation about the origin i or i i = 0i i 1. i 1. i i i or 16. i i i 67. The answer is B. 68. The answer is C. 69. The answer is A. 70. Let z = a+ bi. Thus z = a + b. Now, let iz = i ( a + bi = ai + bi = ai b = b + ai. Thus, iz = ( b + a = b + a = a + b = z The graphs of z and iz are perpendicular. Section 8.: The Product and Quotient Theorems 1. multiply;add. divide; subtract. + i i i i i or 4i i or i i 1 1 i i i 4. i i i i i i i i i. 4. w = cis 15. z = cis cis 0 6. ; It is the same. 7. i

7 Section 8.4: DeMoivre s Theorem; Powers and Roots of Complex Numbers cis ( i ; It is the same. 40. Answers will vary. 41. The two results will have the same magnitude because in both cases you are finding the product of and 5. We must now determine if the arguments are the same. In the first product, the argument of the product will be = 15. In the second product, the argument of the product will be 15 + ( 70 = 585. Now, 585 is coterminal with = 15. Thus, the two products are the same since they have the same magnitude and argument. 4. z = r( cosθ + isinθ Since 1 1 0i 1( cos0 isin0 1 1cos0 ( + i sin0 = z r( cosθ + isinθ = + = +, 1 = cos ( 0 i sin ( 0 r θ + θ 1 = cos ( θ i sin ( θ r + 1 = [ cos θ i sin θ] r i i i Section 8.4: DeMoivre s Theorem; Powers and Roots of Complex Numbers i or 7i i or i or i or i i or 4096i i or i i 1. (a cos 0 + isin 0 = 1( cos 0 + isin 0 We have r = 1 and θ = 0. Since r ( cos α + i sin α = 1 ( cos 0 + i sin 0, then we have r = 1 r = 1 and k α = k α = = k = 10 k, k any integer. If k = 0, then α = 0. If k = 1, then α = 10. If k =, then α = 40. So, the cube roots are cos 0 + isin 0, cos10 + isin10, and cos 40 + i sin (a cos0 + isin 0, cos150 + isin150, and cos 70 + i sin (a cis 0, cis140, and cis i i i 16. (a cis100,cis0, and cis40.

8 118 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 17. (a cos90 ( + isin90, ( + i ( + i cos 10 sin 10, and cos0 sin (a ( cos 0 + isin 0, ( + i ( + i cos140 sin140, and cos 60 sin (a cos0 ( + i sin0, ( + i ( + i cos150 sin150, and cos 70 sin 70.. (a 4 ( cos100 + i sin100, ( + i 4 ( cos40 + i sin cos 0 sin 0, and 19. (a 4cos60 ( + i sin60, ( + i ( + i 4 cos180 sin180, and 4 cos00 sin 00.. (a 4( cos50 + i sin50, ( + i ( + i 4 cos170 sin170, and 4 cos 90 sin (a cos0 ( + i sin0, ( + i ( + i cos10 sin10, and cos 40 sin (a ( cos110 + i sin110, ( + i ( cos50 + i sin 50. cos 0 sin 0, and

9 Section 8.4: DeMoivre s Theorem; Powers and Roots of Complex Numbers cos 0 + isin 0, and cos180 + isin180. (or 1 and 1 Exercise 8 Exercise 9 9. cos 0 + i sin 0, cos150 + i sin150, and cos 70 + i sin cos.5 + i sin.5, cos isin11.5,cos isin 0.5, and cos i sin cos 0 + isin 0, cos90 + isin 90, cos180 + isin180, and cos 70 + isin 70 (or 1, i, 1 and i 7. cos 0 + isin 0, cos 60 + isin 60, cos10 + isin10, cos180 + isin180, cos 40 + isin 40, and cos 00 + isin or 1, + i, + i, 1, 1 1 i, and i 8. cos 45 + i sin 45 and cos 5 + i sin 5 or + i and i 1. { cos 0 + isin 0, cos10 + isin10, cos 40 + i sin or 1, + i, i. {cos 60 + isin 60, cos180 + isin180, cos00 + i sin 00 or i, 1, i. {cos 90 + isin 90, cos 10 + isin 10, cos0 + i sin 0 or 1 1 0, i, i } } 4. { cos isin 67.5, cos isin157.5, cos i sin 47.5, cos i sin ( + i ( + i } { cos 0 sin 0, cos10 sin10, ( i } cos 40 + sin 40 or {, 1+ i, 1 i} }

10 10 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 6. ( + i ( + i { cos 60 sin 60, cos180 sin180, ( i } cos00 + sin 00 or + i,, i 7. {cos 45 + isin 45, cos15 + isin15, cos 5 + isin 5, cos15 + isin 15 or + i, + i, i, i { 8. ( + i ( + i ( cos 5 + i sin 5, cos 45 sin 45, cos15 sin15, ( i } cos15 + sin 15 or { + i, + i, i, i } 9. { cos.5 + isin.5,cos isin11.5, cos i sin 0.5, cos i sin 9.5 { i + i ( cos18 sin18, cos90 sin 90 or 0, cos16 + isin16, cos 4 + isin 4, cos06 + i sin 06 { 41. ( + i ( + i } cos 0 sin 0, cos140 sin140, cos 60 sin 60 { ( + i } 4. ( + i ( + i ( cos195 + isin195, 4. cos15 sin15, cos105 sin105, } ( + i } cos 85 sin x = 1, + i, i. We see that the solutions are the same as Exercise x =, + i, i. We see that the solutions are the same as Exercise De Moivre s theorem states that ( cosθ + isinθ = 1 ( cos θ + isin θ = cos θ + i sin θ 46. ( cosθ + i sinθ ( cos θ sin θ i ( sinθ cosθ = + = cos θ + isin θ } 47. cos θ sin θ = cos θ. 48. sinθ cosθ = sin θ. 49. (a Yes No (c Yes 50. (a blue red (c yellow 51. Using the trace function, we find that the other four fifth roots of 1 are: i, i, i, i. 5. Using the trace function, we find that three of the tenth roots of 1 are: 1, i, i. 5. 4; i 54. { i, i} 55. { i, i} 56. { i, i, i } 57. { i, i, i, i, i } 58. The number 1 has 64 complex 64th roots. Two of them are real, 1 and 1, and 6 of them are not real. 59. false 60. false n 61. If z is an nth root of 1, then z = 1. Since n = =, 1 z n = then 1 is also an nth root z z of Answers will vary.

11 Section 8.5: Polar Equations and Graphs 11 Chapter 8 Quiz (Sections (a 6 1 i. (a 1+ 6i 7+ 4i (c 17 17i 7 (d i (a i i, or 0 + i i ± 5. (a 4i = 4( cos 70 + isin 70 1 i = ( cos00 + isin 00 (c i = ( + i 10 cos sin (a + i i (c 0 7i 7. (a 6( cos10 + i sin10 Section 8.5: Polar Equations and Graphs 1. (a II (since r > 0 and 90 < θ < 180 I (since r > 0 and 0 < θ < 90 (c IV (since r > 0 and 90 < θ < 0 (d III (since r > 0 and 180 < θ < 70. (a positive x-axis negative x-axis (c negative y-axis (d positive y-axis (since = 90 For Exercises 1, answers may vary.. (a Two other pairs of polar coordinates for (1, 45 are (1, 405 and ( 1, 5. (c Since x = rcosθ x = 1 cos 45 = and y = rsinθ y = 1 sin 45 =, the point is,. 4. (a + i (c { i 8. ( + i ( + i ( cos 5 + i sin 5, cos 45 sin 45, cos15 sin15, ( i } cos15 + sin 15 or { + i, + i, i, i } (, 10, (, 480, (, 00. (c,.

12 1 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 5. (a 9. (a 6. (a (, 15, (, 495, (, 15. (c (,. (, 10 ; (, 150 and (, 0. (c,. 10. (a 7. (a ( 4, 0 ; ( 4,90 and ( 4, 10. (c (,. ( 1, 10 ; ( 1, 40 and (1, (c,. 11. (a (5, 60 ; (5, 00 and ( 5, 10. (c 5 5,. 5π, ; 11π π, and,. 8. (a (c,. 1. (a (, 45 ; (, 15 and (, 15. (c (,. π π π 4, ; 4, and 4,. (c ( 0, 4.

13 Section 8.5: Polar Equations and Graphs 1 For Exercises 1, answers may vary. 1. (a 17. (a 14. (a (, 15 ; a second possibility is (, (a (, 45 or (, 5. (, 15 or (, (a (, 45 ; a second possibility is (, (a 0. (a (, 60 or (, (a (, 90 or (, (a ( 1, 10 ; ( 1, 0. (, 70 or (, 90. (, 0 ; (,180.

14 14 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations. (a 6. r = or r =. (,180 ; (, r =. cosθ + sinθ 4. 7 r =. cosθ + sinθ 8. 6 r =. cosθ sinθ 5. r = 4 or r = rsinθ = k k r = r = kcscθ sinθ. y = r = k sinθ

15 Section 8.5: Polar Equations and Graphs 15. rcosθ = k 4. k 5. r = r = ksecθ cosθ 6. x = r = k cosθ 44. r = cos θ (limaçon 45. r = 4cos θ (four-leaved rose 7. r = represents the set of all points units from the pole. The correct choice is C. 8. D. 9. A 40. B. 41. r = + cosθ (cardioid 46. r = cos5 θ (five-leaved rose 4. r = 8 + 6cos θ (limaçon 47. r = 4 cos θ r = ± cos θ (lemniscate 4. r = + cos θ (limaçon 48. r = 4 sin θ r = ± sin θ (lemniscate

16 16 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 49. r = 4 4cosθ (cardioid 50. r = 6 cos θ (limaçon 5. r = sin θ Multiply both sides by r to obtain r = rsin θ. Since r = x + y and y = rsin θ, x + y = y. Complete the square on y to obtain x + y y+ 1= 1 x + ( y 1 = 1. The graph is a circle with center at ( 0,1 and radius r = sinθ tanθ (cissoid r is undefined at θ = 90 and θ = Notice that for [180, 60, the graph retraces the path traced for [0, 180. cos θ r = (cissoid with a loop cosθ 55. Notice that for [180, 60, the graph retraces the path traced for [0, 180.

17 Section 8.5: Polar Equations and Graphs

18 18 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 65. In rectangular coordinates, the line passes through ( 1, 0 and ( 0,. So 0 m = = = and y 0 = x 1 y = x+ ( ( x+ y =. Converting to polar form c r =, we have: acosθ + bsinθ r =. cosθ + sinθ 66. Answers will vary. 67. (a ( r, θ ( r, π θ or ( r, θ (c ( r, π + θ or ( r, θ 68. (a θ π θ (c r; θ (d r (e π + θ (f the polar axis π (g the line θ = r = 4 sinθ, r = 1 + sinθ, 0 θ < π 4sinθ = 1+ sinθ sinθ = 1 1 π 5π sin θ = θ = or 6 6 The points of intersection are 4sin π, π, π = and sin π, π, π = (,60,(, π 4 5π, and,. 4 4 π 76. 0,, π 1,, 4 and π 1,. 4

19 Section 8.5: Polar Equations and Graphs (a Plot the following polar equations on the same polar axis in radian mode:.9(1.06 Mercury: r = ; cosθ.78(1.007 Venus: r = ; cosθ 1(1.017 Earth: r = ; cosθ 1.5(1.09 Mars: r = cosθ (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. 0.1(1.009 Neptune: r = ; cosθ 9.4(1.49 Pluto: r = cosθ Plot the following polar equations on the same polar axis: 1(1.017 Earth: r = ; cosθ 5.(1.048 Jupiter: r = ; cosθ 19.(1.047 Uranus: r = ; cosθ 9.4(1.49 Pluto: r = cosθ 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.

20 10 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 78. (a In degree mode, graph r = 40,000cos θ. 6. (a 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 00 mi. In degree mode, graph r =,500sin θ. x y = + 1, x is in [(, (] or [ 4, 6]. 7. (a 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. 8. (a y = x 4. Since t is in [0, 4], x is in [ 0, 4] or [0, ]. Section 8.6: Parametric Equations, Graphs, and Applications 1. C.. D.. A. 4. B. 5. (a 9. (a 4 4 x = t = y = y or y = x. Since t is ( in [0, 4], x is in [0, 4 ], or [0, 16]. y = x, for x is in (,. y = ( x or y = x 4x+ 4.

21 Section 8.6: Parametric Equations, Graphs, and Applications (a y = 1+ x 9, for x is in (,. 14. (a y = 1 x , for x is in (,. ( 11. (a y = 1+ x 15. (a x + y = 4, for x is in [, ]. 1. (a 16. (a 1 y =, where x is in (0, 1]. x 1. (a x y x y + = 1 + = Since cot t =, tan t (0,. 1 y =, where x is in x 17. (a y = x +., x is in (,.

22 1 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 18. (a. (a 19. (a 4 y = x 1 for, x is in 0, or [ 0,.. (a = y = x, x is in (, 0 (0,. 0. (a ( x ( y + 1 = (a y = x 6, x is in (,. ( x 1 ( y + = Also, 1 sin t 1, sin t 1 1+ sint. 5. y x, x is in [ = 6,. 1. (a x + y =9 1 y = x, x is in (, 0 (0,.

23 Section 8.6: Parametric Equations, Graphs, and Applications x = t sin t, y = cos t, for t in [ 0,4π ] 7. x + y = 4 4. x = t sin t, y = 1 cos t, for t in [ 0,4π ] 8. x y + = Exercises 5 8 are graphed in parametric mode in the following window. x y + = In Exercises 9, answers may vary. y = x ( =, = ( + 1 for t in (, =, = 1 for t in (, x t y t x t y t y = x ( ; =, = ( for t in (, = 4, = + for t in (, x t y t x t y t ; 5. x = cos t, y = sint 6. x = cos t, y = sint 1. ( y = x x+ = x 1 + =, = ( 1 + for t in (, = + 1, = + for t in (, x t y t x t y t. ( y = x 4x+ 6= x + ; =, = ( + for t in (, = +, = + for t in (, x t y t x t y t ;

24 14 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 7. x = sin4 t, y = cost 8. x = 4sin4 t, y = sin5t feet 45. (a x = t y = t 16t + approximately 11.6 feet (c The maximum height of 51 ft is reached at 1.7 sec. Since x = t, the ball has traveled horizontally 55.4 ft. (d Yes 46. (a For Exercises 9 4, recall that the motion of a projectile (neglecting air resistance can be modeled by: x = ( v0cos θ t, y = ( v0sin θ t 16t for t in [0, k]. 9. (a x = 4t y = 16t + 4 t x y = + 6 (c.6 ; 6 ft 40. (a x = 75t x y = 16t + 75 t 16 y = x + x 565 (c 8.1 seconds; 609 ft 41. (a x = ( 88cos 0 y = (88sin 0 t 16t + ( t x y = tan 0 x + 484cos 0 (c The softball traveled 1.9 sec and 161 feet. 4. (a x = ( 16cos 9 t = ( 16sin 9 y t t x y = tan 9 x cos 9 ( (c 4. sec and 495 feet. 4. (a 1 y = x + 56 x + 8. This is a parabolic path. approximately 7 sec and 448 feet. 47. (a θ 0.0 (c x = 88( cos 0.0 t, = + ( y 16t 88 sin 0.0 t. θ 50.0 (c x = 88( cos50.0 t, = + ( y 16t 88 sin 50.0 t. For Exercises 48 51, many answers are possible. y = m t x1 + y 1 or = ( ( 49. y = a( t h + k or ( y = a t + h h + k = at + k. y m t x y.

25 Chapter 8: Review Exercises x = asecθ. We therefore have b btan θ = y, y = ( t a a 51. To find a parametric representation, let x = a sin t. y = bcos t,, t x = t y = b 1 a 5. To show that r = aθ is given parametrically by x = aθ cos θ, y = aθ sin θ, for θ in (,, we must show that the parametric equations yield r = aθ, where r = x + y. r = x + y r = ( aθ cosθ + ( aθ sinθ r = a θ cos θ + a θ sin θ r = a θ cos θ + a θ sin θ r = a θ ( cos θ + sin θ r = a θ r =± aθ or just r = aθ This implies that the parametric equations satisfy r = aθ. 5. To show that rθ = a is given parametrically acosθ asinθ by x =, y =, θ θ for θ in (,0 ( 0,, we must show that the parametric equations yield rθ = a, where r = x + y. r = x + y acosθ asinθ r = + θ θ a cos θ a sin θ r = + θ θ a a r = cos θ + sin θ θ θ a a r = ( cos θ + sin θ r = θ θ a a r =± or just r = θ θ This implies that the parametric equations satisfy rθ = a. 54. The second set of equations x = cos t, y = sin t, t in [0, π] 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. i. i. { ± 9i} i ± 5. i i i i i i i i 1. i i 15. 5i 16. 5i 5i 1 i i 5i+ 5i = = 1+ i 1+ i 1 i 1 i 7i 7 = = i i i 19. i 0. i i or 0i. i 1. + i i or

26 16 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 5. 8i i 7. 1 i 8. The vector representing a real number will lie on the x-axis in the complex plane The resultant of 7 + i and + i is 7+ i + + i = 5+ 4i. ( ( 9. ( + i 4 cos 70 sin Since the modulus of z is, the graph would be a circle, centered at the origin, with radius. 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 ( cos 7 + i sin 7, 10 8 ( cos99 + i sin99, 10 8 ( cos171 + i sin171, 10 8 ( cos 4 i sin 4, 10 8( cos15 i sin15. + and ( + i sin105 ( i ( + i 44. one 45. none 6 cos5 + sin5, and 6 cos45 sin 45. { 46. ( + i ( + i 5 cos60 sin 60,5 cos180 sin180, 5cos00 { ( + i sin00 } 47. ( + i ( + i ( cos 5 + i sin 5, cos 45 sin 45, cos15 sin15, ( + i } cos15 sin {cos 15º + i sin 15º, cos 15º + i sin 15º} 49. (, ,.. ( cos15 + i sin i or i 51. The angle must be quandrantal. 5. Since r is constant, the graph will be a circle. 5. r = 4 cos θ is a circle. 5. i 6. ( + i 8 cos10 sin ( cos15 + i sin i

27 Chapter 8: Review Exercises 17 Graph is retraced in the interval (180, r = 1 + cos θ is a cardioid. 64. C. 65. B r = or r cscθ sinθ = r = or r secθ cosθ = 68. r = or r = r = cosθ + sinθ 55. r = sin 4θ is an eight-leaved rose. 56. line, The graph continues to form eight petals for the interval [0, Convert the polar coordinates to rectangular coordinates, apply the distance formula: r, θ in rectangular coordinates is 71. ( 1 1 ( r1cos θ1, r1sin θ 1. (, ( r θ r θ r θ in rectangular coordinates is cos, sin. ( 1 1 cos θ1 θ d = r + r rr 7. x y = 5, x is in [ 1, 17]. 7. y = x +1, x is in [0,. 57. y = 6 or + 6 9= 0 x y x y = x 4,, x is in [5,. x y = 1+, x is in (, or 0 x + y = x + y x y = x + y = sinθ = cos θ or tanθ = 1 tanθ 61. r = tanθsec θ or r = cosθ 6. B. 6. A. 1 y = x 1 or y + x 1= 0, x is in [ 1, 1]. 76. ( 77. x = + 5cost, y = 4 + 5sin t, where t in [0, π]

28 18 Chapter 8: Complex Numbers, Polar Equations, and Parametric Equations 78. (a Let z 1 = a + bi and its complex conjugate be z = a bi.. (a 7 i 1 z = a + b and ( = + = + = 1. z a b a b z Let z 1 = a + bi and z = a bi. z1 + z1 = ( a+ bi + ( a+ bi = ( a + abi + b i + ( a + bi = a + abi + b ( 1 + a + bi = a b + abi + a + bi = a b + a + ab+ b i = c+ di ( ( ( ( z + z = a bi + a bi = ( a abi + b i + ( a bi = a abi + b ( 1 + a bi = a b abi + a bi = a b + a ab+ b i = c di ( ( (c (d The results are again complex conjugates of each other. At each iteration, the resulting values from z1 and z will always be complex conjugates. Graphically, these represent points that are symmetric with respect to the x-axis, namely points such as ( ab, a, b. and ( 79. (a x = ( 118cos 7 ( θ ( t and y = vsin t 16t + h y = 118sin 7 t 16t +. 4 y =. x + ( tan 7 x 481cos 7 (c.4 sec, the baseball traveled 58 ft. Chapter 8 Test 1. (a 4 1 i (c 1 5i (c wz = 14 18i w 11 (d = i z 1 1. (a i i i ± 5. (a i = ( cos90 + isin90. ( + i 5 cos6.4 sin 6.4. (c (cos 40 + i sin (a + i 4 cis 40 = i (c 0+ i = i 7. (a 16( cos50 + i sin 50 + i (c 4 + 4i 8. ( cos sin 67.5, ( cos i sin157.5, ( + i ( cos7.5 + i sin 7.5. cos 47.5 sin 47.5, and 9. Answers may vary. (a 5, 90, (5, (a (, 5. Alternatively, if θ = 5 60 = 15, a second possibility is (, 15.,. ( 0, 4.

29 Chapter 8: Quantitative Reasoning cardioid. 15. x = cos t, y = sin t for t in [ ] 0, π 1. three-leaved rose. 16. z 1= 1 i r = 1 + = 1+ 4 = 5 >, z Since ( ( is not in the Julia set. Chapter 8: Quantitative Reasoning Graph is retraced in the interval (180, (a x y = 4 Since k y =, k =.5, and n= 1.15, y = n x in...4 in in x x + y =