θ = r = log ≈ 0.6732, and so on. Observe that as θ increases its value, so does r. Therefore the matching graph should be C. θ 0 π 2π 3π r 1 0 -1 0 45. We can make a of values of points satisfying the equation to see which of the given graphs contains those points. So the matching graph should be D. θ 0 π 2π r 1 1 + -1 1 + 1 1 - 3 1 - 1 47. We can make a table of values of points satisfying the equation to see which of the given graphs contains those points. So the matching graph should be F. 49. r = 3cos(θ) r2 = 3rcos(θ) x2 + y2 = 3x in Cartesian coordinates. Using the method of completing square: , or . This is an equation of circle centered at with radius . Therefore the graph looks like: θ 0 π 2π r 0 3 0 -3 0 3 0 -3 0 51. We’ll start with a table of values, and plot them on the graph. So a graph of the equation is a 4-leaf rose.
Also, when r = 0, 1 + 3sin(θ) = 0, so: sin(θ) = θ ≈ - 0.34 + 2kπ θ ≈ π – (- 0.34) + 2kπ = π + 0.34 + 2kπ , k is an integer. So a graph of the equation is a limaçon symmetric about the y- axis. 61. We’ll start with a table of values. θ - π 0 π r - 2π - π 0 π 2π Observe that as θ increases its absolute so does r. Therefore a graph of the equation should contain two spiral curves that are symmetric about the y- axis. θ 0 π 2π r 4 3 + 5 1 3 - 2 3 - 1 undef. 5 3 + 4 63. We’ll start with a table of values. Also, when r = 0, 3 + sec(θ) = 0, so: sec(θ) = - 3 θ ≈ 1.9106 + 2kπ θ ≈ 2π - 1.9106 + 2kπ ≈ 4.3726 + 2kπ, where k is an integer. So a of the equation is a conchoid symmetric about the x-axis.
cos(θ) = = sin(θ) = = Therefore (x, y) is located in the quadrant II, and θ = . So . To put our answer in form: a = rcos(θ) = cos() ≈ 0.78824 b = rsin(θ) = sin() ≈ 1.903 Thus 0.78824 + 1.903i. 63. = . Let’s convert to polar form: 5 + 3i = x + yi. Then x = 5 and y = 3. Then r2 = x2 + y2 = 52 + 32 = 34 so r = (since r ≥ 0). Also: x = rcos(θ) and y = rsin(θ) 5 = cos(θ) 3 = sin(θ) cos(θ) = sin(θ) = Therefore (x, y) is located in the quadrant I, and θ ≈ 0.54042. So . To put our answer in form: a = rcos(θ) = cos(0.18014) ≈ 1.771 b = rsin(θ) = sin(0.18014) ≈ 0.3225 Thus 1.771 + 0.3225i. 65. If z5 = 2 then z = . In the complex plane, 2 would sit on the horizontal axis at an angle of 0, giving the polar form . Then = e0 = cos (0) + i sin(0) = ≈ 1.149. Since the angles 2π, 4π, 6π, 8π, and 10π are coterminal with the angle of 0, can be represented by turns as , , , , and to get all solutions. = ( = = cos() + isin() ≈ 0.355 + 1.092i = ( = = cos() + isin() ≈ -0.929 + 0.675i = ( = = cos() + isin() ≈ -0.929 - 0.675i = ( = = cos() + isin() ≈ 0.355 - 1.092i
= ( = = = = cos(2π) + isin(2π) ≈ 0.355 + 1.092i Observe that for the angles 2kπ, where k is an integer and k ≥ 5, the values of are repeated as the same as its values when k = 0, 1, 2, 3, and 4. In conclusion, all complex solutions of z5 = 2 are 1.149, 0.355 + 1.092i, -0.929 + 0.675i, -0.929 - 0.675i, and 0.355 - 1.092i. 67. If z6 = 1 then z = . In the complex plane, 1 would sit on the horizontal axis at an angle of 0, giving the polar form . Then = e0 = cos (0) + isin(0) = 1. Since the angles 2π, 4π, 6π, 8π, 10π, and 12π are coterminal with the angle of 0, can be represented by turns as , , , , , and . = = cos() + isin() = + i = = cos() + isin() = + i = = cos(π) + isin(π) = -1 = = cos() + isin() = - i = = cos() + isin() = - i = = cos(2π) + isin(2π) = 1 Observe that for the angles 2kπ, where k is an integer and k ≥ 6, the values of are repeated as the same as its values when k = 0, 1, 2, 3, 4, and 5. In conclusion, all complex solutions of z6 = 1 are 1, + i, + i, -1, - i, and - i. 8.4 Solutions to Exercises 1. Initial point (4,0); terminal point (0,2). The vector component form is .