User manual | Chapter 9 Polar Coordinates and Complex Numbers

Chapter 9 Polar Coordinates and Complex Numbers
11.
Polar Coordinates
9-1
90˚
120˚
Check for Understanding
180˚
1. There are infinitely many ways to represent the
angle v. Also, r can be positive or negative.
2. Draw the angle v in standard position. Extend the
terminal side of the angle in the opposite
direction. Locate the point that is r units from
the pole along this extension.
3. Sample answer: 60° and 300°
Plot (4, 120) such that v is in standard position
and r is 4 units from the pole. Extend the
terminal side of the angle in the opposite
direction. Locate the point that is 4 units from the
pole along this extension.
r 4
v 120 180 or v 120 180
60
300
4. The points 3 units from the origin in the opposite
direction are on the circle where r 3.
5. All ordered pairs of the form (r, v) where r 0.
6.
90˚
120˚
150˚
330˚
210˚
240˚
8.
120˚
270˚
90˚
C
180˚
1 2 3 4
240˚
270˚
19
6
13
7
, 2,
13
25
15a.
→ 2,
6
Chapter 9
(1) → 2,
(3) → 2,
19
6
90˚
120˚
60˚
30˚
5 10 15 20
0˚
330˚
300˚
270˚
15b. 210 (30) 240
N
2
A
360 (r )
240
2
360 (20 )
838 ft2
11
6
5
3
Pages 558–560
7
6
, 2, ,
16.
120˚
Exercises
90˚
0˚
330˚
240˚
274
30˚
1 2 3 4
210˚
17.
60˚
E
180˚
(r, v (2k 1))
→ 2,
180˚
→2, 6 2(2)) → 2, 6
7
6
300˚
270˚
150˚
→2, 6 2(1) → 2, 6
6
0˚
330˚
150˚
(r, v 2k)
5
3
3
2
4.37
0
25
6
11
6
7
6
2.52 (3)2 2(2.5)(3
) cos 4 6
5
6.25
9 15
cos 12
5.25
15 cos
51
2
3
3
2
1 2 3 4
14. P1P2 1 2 3 4
4
3
0
60˚
1 2 3 4
240˚
7
6
3
30˚
240˚
6
D
2
6
4
3
210˚
0˚
90˚
120˚
210˚
5
3
5
6
300˚
10. Sample answer: 2,
2
270˚
180˚
11
6
3
2
2
3
330˚
210˚
2,
9.
30˚
150˚
B
300˚
150˚
1 2 3 4
4
3
60˚
13.
0
7
6
300˚
240˚
6
0˚
330˚
210˚
3
5
6
0˚
1 2 3 4
2
2
3
30˚
A
180˚
7.
60˚
1 2 3 4
2
3
5
6
30˚
150˚
Pages 557–558
12.
60˚
270˚
300˚
2
3
2
3
6
5
6
F
0
1 2 3 4
11
6
7
6
4
3
3
2
5
3
18.
90˚
120˚
19.
60˚
180˚
0˚
2 4 6 8
6
H
0
1 2 3 4
G
330˚
210˚
240˚
20.
270˚
2
2
3
7
6
300˚
21.
150˚
90˚
J
4
3
22.
3
2
2
2
3
23.
3
90˚
L
4
3
24.
90˚
120˚
240˚
5
3
3
2
25.
60˚
180˚
330˚
210˚
240˚
26.
2
2
3
27.
0
180˚
1 2 3 4
Q
7
6
4
3
3
2
30˚
1 2 3 4
270˚
3
7
3
2(1) → 2,
3
6
0
1 2 3 4
11
6
7
6
35.
6
2
5
6
4
3
3
120˚
5
3
3
2
90˚
60˚
30˚
150˚
180˚
0
1 2 3 4
1 2 3 4
4
3
300˚
36.
120˚
90˚
30˚
180˚
1 2 3 4
240˚
275
0˚
330˚
210˚
(r, v (2k 1)180°)
→ (2, 60° (1)180°) → (2, 240°)
→ (2, 60° (3)180°) → (2, 600°)
37.
60˚
150˚
240˚
5
3
3
2
270˚
300˚
0˚
330˚
210˚
11
6
7
6
→ 2, 3 2(0) → 2, 3
→ 2,
2
2
3
0˚
7
300˚
270˚
28. Sample answer: 2, 3, 2, 3, (2, 240°),
(2, 600°)
(r, v 2k)
0˚
330˚
2
3
330˚
240˚
5
3
1 2 3 4
5
6
60˚
210˚
11
6
30˚
240˚
34.
33.
60˚
210˚
5
3
150˚ R
90˚
180˚
1 2 3 4
90˚
10
150˚
3
3
2
120˚
11
6
120˚
6
5
6
32.
0
P
4
3
3
300˚
7
6
300˚
270˚
0˚
6
N
4
31. Sample answer: (4, 675°), (4, 1035°), (4, 135°),
(4, 495°)
(r, v 360k°)
→ (4, 315 360(1)°) → (4, 675°)
→ (4, 315 360(2)°) → (4, 1035°)
(r, v (2k 1)180°)
→ (4, 315 (1)180°) → (4, 135°)
→ (4, 315 (1)180°) → (4, 495°)
330˚
5
6
0˚
1 2 3 4
2
2
3
30˚
150˚
270˚
→ 1, 3 (1) → 1, 3
60˚
M
210˚
11
6
7
6
13
→ 1, 3 (3) → 1, 3
1 2 3 4
1 2 3 4
(r, v (2k 1))
30˚
180˚
7
→ 1, 3 2(2) → 1, 3
300˚
150˚
0
270˚
→ 1, 3 2(1) → 1, 3
330˚
120˚
6
5
6
(r, v 2k)
0˚
1 2 3 4
240˚
5
3
10
1, 3 60˚
210˚
11
6
7
13
4
, 1, ,
30. Sample answer: 1, 3, 1, 3 3
30˚
1 2 3 4
7
6
5
3
K
180˚
0
3
2
120˚
6
5
6
11
6
4
3
3
29. Sample answer: (1.5, 540°), (1.5, 900°), (1.5, 0°),
(1.5, 360°)
(r, v 360k°)
→ (1.5, 180° 360(1)°) → (1.5, 540°)
→ (1.5, 180° 360(2)°) → (1.5, 900°)
(r, v (2k 1)180°)
→ (1.5, 180° (1)180°) → (1.5, 0°)
→ (1.5, 180° (1)180°) → (1.5, 360°)
3
5
6
30˚
150˚
2
2
3
2
3
270˚
2
300˚
3
6
5
6
0
1 2 3 4
7
6
11
6
4
3
3
2
5
3
Chapter 9
38.
90˚
120˚
39.
60˚
30˚
150˚
180˚
1 2 3 4
0˚
150˚
240˚
40.
90˚
120˚
30˚
1 2 3 4
0˚
50a.
330˚
210˚
240˚
300˚
270˚
49a. When v 120°, r 17. The maximum speed at
120° is 17 knots.
49b. When v 150°, r 13. The maximum speed at
150° is 13 knots.
60˚
r0
180˚
330˚
210˚
90˚
120˚
270˚
1 2 3 4
240˚
270˚
0˚
50b.
300˚
3
2
3 3 or 120°
N
N
2
2
A
360 (R ) 360 (r )
120
120
2
2
360 ((300 ) 360 ((25) )
120
360 (90,000 625)
93,593 ft2
If each person’s seat requires 6 ft2 of space,
2 52 2(1)(5) cos 6
34
1
7
1 2
5 1
0 cos 12
7
26
10
cos
12 93,593
there are 6 or 15,599 seats.
51. The distance formula is symmetric with respect to
(r1, v1) and (r2, v2). That is,
5.35
r22 r12 2r2r1 cos (v1
v2)
r12 r22 2r1r2 cos[
(v2 v1)]
(2.5)2 (1.75)2 2(2.5)(1.75)cos
2
5 8
r12 r22 2r1r2 cos(v2
v1)
21
9.3125
8.75
cos 40
5
3
3
2
Let R 3 100 or 300 and let r 0.25 100
or 25.
44. P1P2 11
6
4
3
41. r 2
or r 2
for any v.
2 2(
42 6
4)(6) c
os (10
5° 170°)
42. P1P2 16
36
48
cos°)
(65
52
48
cos)
(65
5.63
43. P1P2 0
1 2 3 4
7
6
330˚
210˚
6
30˚
180˚
3
5
6
300˚
60˚
150˚
2
2
3
21
6.25 30.0625 8.75 cos 40
52a.
120˚
90˚
60˚
30˚
150˚
3.16
180˚
45. P1P2 1.32 (3.6
)2 2(1.3)(
3.6) co
s (62
° (47°))
1.69
12.966
9.3(62°
cos °)
47
14.65
6
9.3(15°
cos )
4.87
46. r (3)2 42
240˚
5
180° 53° 127°
Sample answer: (5, 127°)
47. There are 360° in a circle. If the circle is cut into 6
360
equal pieces, each slice measures 6 or 60°.
Beginning at the origin, the equation of the first
line is v 0°. The equation of the next line,
rotating counterclockwise, is v 0 60 or 60°.
The equation of the last line is v 60 60 or
120°. Note that lines extend through the origin, so
3 lines create 6 pieces.
3 mph
54. 3, 2, 4
1, 4, 0
(3)(1) (2)(4) (4)(0)
380
11
No, the vectors are not perpendicular because
their inner product is not 0.
r1 r2
Chapter 9
300˚
8 mph
48. P1P2 r12 r22 2r1r2 cos (v v)
r12 r22 2r1r2
2
(r1 r
2)
270˚
52b. P1P2 52 62
2(
5)(6) c
os (34
5° 310°)
25
36
60
cos
(35°)
61
60
cos
(35°)
3.44
No; the planes are 3.44 miles apart.
53. Draw a picture.
sin v 38
Boat
1
sin (sin v) sin138
v 22.0°
4
0˚
330˚
210˚
sin v 5, v 53°
r12 r22 2r1r2 cos 0
2 4 6 8
276
55. Rewrite y 9x 3 as 9x y 3 0.
d
63.
3282
82
56.
1682
41
1 sin2a
sin2a
Distance is always positive.
1
1
sin2a
csc2a 1
cot2a
3
57. Arc cos 2 30°
1
In a 30°-60°-90° right
triangle, the angle opposite
the smallest leg is 30°.
58. y 5 cos 4v
Amplitude 5; Period 24 or 2
2
9.3
sin 30°
18.6 sin 30°
9.3
60°
sin B
90° B
Find c.
c
sin 60°
9-2
Graphs of Polar Equations
Page 565
Check for Understanding
1. Sample answer: r sin 2v
The graph of a polar equation whose form is
r a cos nv or a sin nv, where n is a positive
integer, is a rose.
2. 1 sin v 1 for any value of v. Therefore, the
maximum value of r 3 5 sin v is r 3 5(1)
or 8. Likewise, the minimum value of r 3 5
sin v is r 3 5(1) or 2.
3. The polar coordinates of a point are not unique. A
point of intersection may have one representation
that satisfies one equation in a system, another
representation that satisfies the other equation,
but no representation that satisfies both
simultaneously.
4. Barbara is correct. The interval 0 v is not
always sufficient. For example, the interval 0 v
only generates two of the four petals for the
rose r sin 2v. r sin 2v is an example where
values of v from 0 to 4 would have to be
considered.
C 180° 90° 30°
18.6 sin 30° 9.3 sin B
3
59. b sin A 18.6 sin 30°
9.3
Since a b sin A, there is one solution.
Find B.
Find C.
18.6
sin B
2 4 1
4 0
1 1
0
4
(1)
1 1 0 2 1
3 5
3 4
5
4
3 4 5
2(5) 4(5) 1(1)
11
64. 11 (3) 14
11 (2) 13
11 (1) 12
11 0 11
{(3, 14), (2, 13), (1, 12), (0, 11)}
For each x-value, there ia a unique v-value.
Yes, the relation is a function.
65. Since the two triangles formed are right triangles,
the side opposite the right angles, A
B
, intercept
an arc measuring 180°, or half the circle. AB
is a
diameter.
C d
50 d
50 d
The correct choice is E.
Ax1 By1 C
A2 B2
9(3) (1)(2) (3)
92 (
1)2
32
82
32
82
82
82
9.3
sin 30°
c sin 30° 9.3 sin 60°
9.3 sin 60°
c
sin 30°
c 16.1
60. 3 or 1 positive
f(x) x3 4x2 4x 1
0 negative
P
1
Q
Since there are only positive real zeros, the only
rational real zero is 1.
61.
x 3
x 5x2
2x
3
x2
5
x
3x 3
3x 15
10
10
→ 0. Therefore, the slant
As x → , x5
asymptote is y x 3.
62. y-axis:
For x:
f(x) x4 3x2 2
For x: f(x) (x)4 3(x)2 2
x4 3x2 2
So, in general, point (x, y) is on the graph if and
only if (x, y) is on the graph.
5.
90˚
120˚
30˚
150˚
180˚
1
2
0˚
330˚
210˚
240˚
270˚
cardioid
277
6.
60˚
300˚
120˚
90˚
60˚
30˚
150˚
180˚
2 4 6 8
0˚
330˚
210˚
240˚
270˚
300˚
limaçon
Chapter 9
7.
2
3
2
8.
3
6
5
6
11
6
4
3
11.
3
2
2
5
3
0
4
3
3
2
If or 5
6
equation, r 1. If 3
2
2
3
4
3
3
2
330˚
120˚
270˚
300˚
90˚
60˚
30˚
150˚
1
0˚
2
330˚
210˚
300˚
240˚
rose
17.
270˚
300˚
rose
2
3
2
0
4 8 12 16
4
3
3
2
0
19.
2
3
4
3
cardioid
278
3
2
5
3
0˚
330˚
270˚
300˚
2
3
2
3
6
5
6
0
11
6
7
6
4 6 8
limaçon
20.
1 2 3 4
2
240˚
6
5
6
60˚
30˚
180˚
5
3
3
90˚
210˚
Spiral of Archimedes
2
120˚
150˚
11
6
7
6
5
3
18.
3
6
14
10b. Sample answer: 0 v 3
Begin at the origin and “spiral” twice around it,
or through 4 radians. Move straight up
through 4 2 or 92 radians. Now move to the
left slightly, through approximately 92 6 or
14
radians.
3
Chapter 9
0˚
is substituted in either
11
6
7
6
270˚
60˚
1 2 3 4
0˚ 180˚
2
330˚
5
6
4 8 12 16
16.
1
90˚
30˚
240˚
30˚
240˚
6
300˚
lemniscate
60˚
210˚
3
5
6
90˚
180˚
1, 6, 1, 56, and 2, 32.
10a.
270˚
210˚
5
3
150˚
original equation, r 2. The solutions are
2
3
2
120˚
180˚
0
11
6
120˚
is substituted in either original
330˚
150˚
spiral of Archimedes
6 or 56 or 32
0˚
2 4 6 8
210˚
14.
3
5 10 15 20
15.
30˚
240˚
6
4
3
2 sin 2 cos 2
sin cos 2
sin 1 2 sin2 2 sin2 sin 1 0
(2 sin 1)(sin 1) 0
2 sin 1 0 or sin 1 0
sin 12
sin 1
6
2
3
2
7
6
5
3
60˚
cardioid
11
6
90˚
150˚
300˚
5
6
2
7
6
270˚
circle
13.
1
330˚
240˚
120˚
0˚ 180˚
1 2 3 4
210˚
6
12.
60˚
30˚
180˚
3
5
6
90˚
150˚
spiral of Archimedes
2
3
Exercises
120˚
0
11
6
4
3
rose
9.
3 6 9 12
7
6
5
3
3
2
6
0
Pages 565–567
3
5
6
1
7
6
2
2
3
1 2 3 4
0
11
6
7
6
4
3
lemniscate
3
2
5
3
21.
90˚
120˚
22.
60˚
120˚
30˚
150˚
180˚
330˚
240˚
30˚
0˚
1 2 3 4
330˚
210˚
300˚
270˚
28. (1, 0.5), (1, 1.0), (1, 2.1), (1, 2.6), (1, 3.7), (1, 4.2),
(1, 5.2), (1, 5.8)
60˚
150˚
0˚ 180˚
1
210˚
90˚
240˚
270˚
300˚
rose
cardioid
23. Sample answer: r sin 3v
The graph of a polar equation of the form r a cos
3v or r a sin 3v is a rose with 3 petals.
24. Sample answer: r 2v
4
a2
r 12
1
2
a
r 2
25. 3 2 cos 1 cos 0
The solution is (3, 0)
2
3
2
3
6
5
6
1 2 3 4
4
3
26. 1 cos 1 cos 2 cos 0
cos 0
2
3
Substituting each angle
into either of the original
equations gives r 1, so
the solutions of the system
are 1,
27.
2
3
2
3
6
1
2
and 1, .
0
[4, 4] scl1 by [4, 4] scl1
11
6
7
6
3
2
2
5
3
3
2
5
6
2 or 32
[6, 6] scl:1 by [6, 6] scl1
30. (3.6, 0.6), (2.0, 4.7)
0
11
6
7
6
2
[2, 2] scl1 by [2, 2] scl1
29. (2, 3.5), (2, 5.9)
4
3
3
2
31a. If the lemniscate is 6 units from end to end, then
a 12(6) or 3.
5
3
r2 9 cos 2 or r2 9 sin 2
31b. If the lemniscate is 8 units from end to end, then
a 12(8) or 4.
3
r2 16 cos 2 or r2 16 sin 2
6
5
6
32.
1
2
4
3
3
2
90˚
60˚
0
30˚
150˚
11
6
7
6
120˚
180˚
5
3
2 4 6 8
330˚
210˚
2 sin 2 sin 2
sin sin 2
sin 2 cos sin 0 2 cos sin sin 0 sin (2 cos 1)
sin 0 or 2 cos 1 0
0˚
240˚
270˚
300˚
This microphone will pick up more sounds from
behind than the cardioid microphone.
33. 0 v 4: Begin at the origin and curl around
once, or through 2 radians. Curl around a second
time and go through 2 2 or 4 radians.
34. All screens are [1, 1] scl1 by [1, 1] scl1
cos 12
0 or or 3 or 53
If 0 or is substituted in either original
equation, r 0. If 3 or 53 is substituted in
either original equation, r 3
or r 3
,
respectively. The solutions are (0, 0), (0, ),
3, 3, and 3, 53.
279
Chapter 9
34a. r cos 2
r cos 7
r cos 4
r cos 9
When n 11, the innermost loop will be on the
left and there will be an additional outer ring.
35. Sample answer: r 1 sin v
A heart resembles the shape of a cardioid. The
sine function orients the heart so that the axis of
symmetry is along the y-axis. If a 1, the heart
points in the right direction.
36a. For a limaçon to go back on itself and have an
inner loop, r must change sign. This will happen
if b a.
36b. For the other two cases, a b.
Experimentation shows that the dimple
disappears when a 2b, so there is a
dimple if b a 2b.
36c. For this remaining case, there is neither an
inner loop nor a dimple if a 2b.
37a. Subtracting a from v rotates the graph
counterclockwise by an angle of a.
37b. Multiplying v by 1 reflects the graph about the
polar axis or x-axis.
37c. Multiplying the function by 1 changes r to its
opposite, so the graph is reflected about the
origin.
37d. Multiplying the function by c results in a
dilation by a factor of c. (Points on the graph
move closer to the origin if c 1, or farther
away from the origin if c 1.)
38. Sample answer: (4, 405°), (4, 765°), (4, 135°),
(4, 225°)
(r, 360k°)
→ (4, 45° 360(1)°) → (4, 405°)
→ (4, 45° 360(2)°) → (4, 765°)
(r, (2k 1)180°)
→ (4, 45° (1)180°) → (4, 135°)
→ (4, 45° (1)180°) → (4, 225°)
r cos 6
r cos 8
When n 10, two more outer rings will appear.
34b. r cos 3
r cos 5
Chapter 9
280
i j k
2 3 0
1 2 4
3 0 2 0 2 3 i
j
k
2 4
1 4
1 2
12i 8j 7k
12, 8, 7
2, 3, 0
12, 8, 7
24 (24) 0 or 0
1, 2, 4
12, 8, 7
12 (16) 28 or 0
40. 3.5 cm, 87°
w
39. v
x2
r cos 2
2
r
cos r 2 sec 4. To convert from polar coordinates to rectangular
coordinates, substitute r and v into the equations
x r cos v and y r sin v. To convert from
rectangular coordinates to polar coordinates, use
the equation r x2 y2 to find r. If x 0, v Arctan yx. If x 0, v Arctan yx . If x 0, you
can use 2 or any coterminal angle for v.
3.
2
sin x
2
41. cos4 x cos2 x sin2 x tan x
sin2 x
cos2 x (cos2 x sin2 x)
sin2 x
(cos2 x)(1)
sin2 x
cos2 x
tan2 x
tan2
y
x
tan2 x
r
tan2 x tan2 x
42. Find C.
C 180° 21°15 49°40
109°5
Find b.
b
sin 49°40
O
28.9
sin 109°5
4
or 2
28.9 sin 49°40
sin 109°5
2, 34
b 23.3
6. r (2)2 (5
)2
Find a.
a
sin 21°15
28.9
sin 109°5
a sin 109°5 28.9 sin 21°15
28.9 sin 21°15
7.
a sin 109°5
a 11.1
NY LA Miami
$240 $199 $260
$254 $322 $426
43.
Bus
Train
8.
12
1
3
44. 1 6 1 8
4
8
1
6
8
9.
8
13
4
8
1
3 16
8
So 3
8
3
3
16
16
10.
2
6
3
The correct choice is A.
9-3
Page 571
11.
Polar and Rectangular
Coordinates
12.
Check for Understanding
x
34
2
2
5
v Arctan 2 5.39
4.33
29
(5.39, 4.33)
x 2 cos 43
y 2 sin 43
1
3
(1, 3
)
x 2.5 cos 250°
y 2.5 sin 250°
0.86
2.35
(0.86, 2.35)
y2
r sin v 2
2
r
sin v
r 2 csc v
x2 y2 16
(r cos v)2 (r sin v)2 16
r2(cos2 v sin2 v) 16
r2 16
r 4 or r 4
r6
x2 y2 6
x2 y2 36
r sec v
r
r
1. Sample answer: (22
, 45°)
22 22
r x r cos 5. r (2
)2 (
)2 v Arctan 2
b sin 109°5 28.9 sin 49°40
b
y r sin 1
r cos v
1
1
x
Arctan 22
x 1
8
45°
22
2. The quadrant that the point lies in determines
y
y
whether v is given by Arctan x or Arctan x .
281
Chapter 9
13a.
90˚
120˚
22. x 1 cos 6
60˚
3
1 2 30˚
150˚
180˚
3
1 2 3 4
23 , 12
330˚
210˚
240˚
270˚
23. x 2 cos 270°
0
(0, 2)
24. x 4 cos 210°
300˚
13b. No. The given point is on the negative x-axis,
directly behind the microphone. The polar
pattern indicates that the microphone does not
pick up any sound from this direction.
3
2
4
25.
Pages 572–573
Exercises
2
2
2
14. r 2
(
2)
v Arctan 2
26.
4
8
or 22
Add 2 to obtain v 74.
22, 74
15. r 02 12
27.
1
or 1
Since x 0 when y 1, v 2.
1, 2
16. r 12 (
3
)2
v Arctan
4
or 2
3
2, 3
17. r 1 2
4
28.
3
1
3
2
4
4
16
v Arctan
3
4
14
29.
3
Arctan or 43
1
24 or 12
12, 43
18. r 32 82
73
8.54
(8.54, 1.21)
42 (
7)2
19. r 30.
v Arctan 83
1.21
7
v Arctan 4 31.
65
8.06
1.05
Add 2 to obtain v 5.23.
(8.06, 5.23)
20. x 3 cos 2
0
(0, 3)
y 3 sin 2
3
32.
21. x 12 cos 34
y 12 sin 34
12 2
12 2
4
2
4
2
2
42 , 42 Chapter 9
2
33.
282
112
12
2
0˚
y 1 sin 6
y 2 sin 270°
2
y 4 sin 210°
412
23
2
(23
, 2)
x 14 cos 130°
y 14 sin 130°
9.00
10.72
(9.00, 10.72)
x 7
r cos v 7
7
r cos v
r 7 sec v
y5
r sin v 5
5
r sin v
r 5 csc v
x2 y2 25
(r cos v)2 (r sin v)2 25
r2(cos2 v sin2 v) 25
r2 25
r 5 or r 5
x2 y2 2y
(r cos v)2 (r sin v)2 2r sin v
r2 (cos2 v sin2 v) 2r sin v
r2 2r sin v
r 2 sin v
x2 y2 1
(r cos v)2 (r sin v)2 1
r2 (cos2 v sin2 v) 1
r2 (cos 2v) 1
1
r2 cos 2v
2
r sec 2v
x2 (y 2)2 4
x2 y2 4y 4 4
(r cos v)2 (r sin v)2 4r sin v 0
r2(cos2 v sin2 v) 4r sin v 0
r2 4r sin v 0
r2 4r sin v
r 4 sin v
r2
x2 y2 2
x2 y 2 4
r 3
2
x y2 3
x2 y2 9
v 3
34.
y
43. horizontal distance:
25(4 2 cos 120°) 75 m east
vertical distance:
25(3 2 sin 120°) 118.30 m north
y
y
Arctan x 3
y
3
x
1
2
6
3
y 3
x
1
3
2
1 2y
y2
r 3 cos v
r2 3r cos v
x2 y2 3x
37.
r2sin 2v 8
r22 sin v cos v 8
2r sin v r cos v 8
2yx 8
xy 4
38.
yx
36.
Arctan
6.07
5.47
6.07
5.47
5.47
cos 47.98°
tan v
Arctan 1
r sin v
r2 r sin v
x2 y2 y
40. x 325 cos 70°
111.16
(111.16, 305.40)
r
8.17 r
8.17∠47.98°
44d. 8.17 sin (3.14t 47.98°)
45.
r 2a sin v 2a cos v
r2 2ar sin v ar cos v
x2 y2 2ay 2ax
2
2
x 2ax y 2ay 0
(x a)2 (y a)2 2a2
The graph of the equation is the circle centered at
(a, a) with radius 2
a.
1
39.
4
r sin v
r cos v
47.98 v; 47.98°
5.47 r cos 47.98°
v 4
5
4
y 325 sin 70°
305.40
46.
120˚
90˚
60˚
30˚
150˚
—
41. —
6 6 5
24
4
24
24
180˚
1 2 3 4
0˚
330˚
210˚
0.52 unit
42. Drop a perpendicular from the point with polar
coordinates (r, v) to the x-axis. r is the length of
the hypotenuse in the resulting right triangle.
x is the length of the side adjacent to angle v, so
cos v xr. Solving for x gives x r cos v. y is the
y
length of the side opposite angle v, so sin v r.
Solving for y gives y r sin v. (The figure is drawn
for a point in the first quadrant, but the signs
work out correctly regardless of where in the
plane the point is located.)
240˚
270˚
300˚
47. Sample answer: (2, 405°), (2, 765°), (2, 225°),
(2, 585°)
(r, v 360k°)
→ (2, 45° 360(1)°) → (2, 405°)
→ (2, 45° 360(2)°) → (2, 765°)
(r, v (2k 1)180°)
→ (2, 45° (1)180°) → (2, 225°)
→ (2, 45° (3)180°) → (2, 585°)
48. r2 502 4252 2 50 425 cos 30°
r 382.52 mph
y
(r, )
50
sin v
r
382.52
sin 30°
50 sin 30° 382.52 sin v
r
50 sin 30°
382.52
O
x
44a. x 4 cos 20°
y 4 sin 20°
3.76
1.37
3.76, 1.37
x 5 cos 70°
y 5 sin 70°
1.71
4.70
1.71, 4.70
44b. 3.76, 1.37
1.71, 4.70
3.76 1.71, 1.37 4.70
5.47, 6.07
44c.
5.47 r cos v; 6.07 r sin v
r sin v
y
x
y
x
120˚
O
x
35. r 2 csc v
r
r
2
sin v
30˚ 425 mph
50 mph
3°45 v
The direction is 3°45 west of south.
x
283
Chapter 9
3. The graph of the equation x k is a vertical line.
Since the line is vertical, the x-axis is the normal
line through the origin. Therefore, f 0° or f 180°, depending on whether k is positive or
negative, respectively. The origin is k units
from the given vertical line, so p k. The polar
form of the given line is k r cos (v 0°) if k is
positive or k r cos (v 180°) if k is negative.
Both equations simplify to k r cos v.
4. You can use the extra ordered pairs as a check on
your work. If all the ordered pairs you plot are not
collinear, then you have made a mistake.
sin2 A cos A 1
1 cos2 A cos A 1
0 cos2 A cos A 2
0 (cos A 2)(cos A 1)
cos A 2 0
or cos A 1 0
cos A 2
cos A 1
A 0°
y
50.
49.
2
y 2 cos 1
O
1
90˚ 180˚ 270˚ 360˚ 5. A2 B2 32 (
4)2
5
Since C is negative, use 5.
2
4
f Arctan 43
x
3, 1
2
2
(
53° or 307°
p r cos (v f)
2 r cos (v 307°)
)
6. A2 B2 (2)2 42
25
Since C is negative, use 25
.
52. Enter the x-values in L1 and the f(x)-values in L2
of your graphing calculator. Make a scatter plot.
The data points are in the shape of parabola.
Perform a quadratic regression.
a 0.07, b 0.73, c 1.36
Sample answer:
y 0.07x2 0.73x 1.36
53. 2 1 0 0 3 0 20
2
4
8
1
0
2
0
1 2 4 5 10  0
x4 2x3 4x2 5x 10
625 145
54. m 25 17
45y 2 0
cos f 35, sin f 5, p 2
30˚
3
cos 210° 2
3
x
5
y
51. The terminal side is in the
third quadrant and the
reference angle is
210 180 or 30°.
2
4
9
x y 0
2
2
2
5
5
5
95
25
5
, sin f , p cos f 10
5
5
f Arctan(2)
63°
Since cos f 0, but sin f 0, the normal lies in
the second quadrant.
f 180° 63° or 117°
p r cos (v f)
(y 145) 60(x 17)
95
10
60
y 60x 875
55. x y and y z, so x z.
r cos (v 117°)
7. 3 r cos (v 60°)
0 r cos (v 60°) 3
0 r (cos v cos 60° sin v sin 60°) 3
If x z, then 0 xz 1.
The correct choice is C.
3
r sin v 3
0 12r cos v 2
3
y 3
0 12x 2
9-4
y 6 or
0 x 3
x 3
y 6 0
Polar Form of a Linear Equation
Pages 577–578
8.
Check for Understanding
r cos (v 4) 2
1. The polar equation of a line is p r cos (v f).
r and v are the variables. p is the length of the
normal segment from the line to the origin and
f is the angle the normal makes with the positive
x-axis.
2. For r to be equal to p, we must have cos (v f) 1. The first positive value of v for which this is
true is v f.
Chapter 9
r 2 sec v 4
r cos v cos 4 sin v sin 4 2 0
2
2
2
r sin v 2 0
r cos v 2
2
x
2
2
y 2 0
2
x 2
y 4 0
2
284
9.
2
3
2
10.
3
6
5
6
90˚
30˚
180˚
1 2 3 4
2 4 6 8
7
6
240˚
270˚
4
f Arctan 3
53°
Since cos f 0, but sin f 0, the normal lies in
the fourth quadrant.
f 360° 53° or 307°
p r cos (v f)
2.1 r cos (v 307°)
300˚
Since the shortest distance is along the normal,
the answer is (p, f) or 5, 56.
2
3
2
15. A2 B2 32 22
13
Sinc C is negative, use 13
.
3
6
5
6
3
x
13
0
3
2
5
3
16.
4
x
41
25
Since C is positive, use 25.
cos f
24
25y 4 0
7
24
,
2
5 , sin f 25
5
10
y 0
1
41
4
541
1041
4
41
, p , sin f cos f 41
41
41
f Arctan 4
51°
Since cos f 0, but sin f 0, the normal lies in
the fourth quadrant.
f 360° 51° or 309°
p r cos (v f)
5
p4
f Arctan 274
74°
Since cos f 0, but sin f 0, the normal lies in
the second quadrant.
f 180° 74° or 106°
p r cos (v f)
4 r cos (v 106°)
13. A2 B2 212 202
29
Since C is negative, use 29.
21
x
29
5
13
r cos (v 34°)
13
A2 B2 42 (
5)2
14
Since C is negative, use 41
.
Pages 578–579
Exercises
2B
2 12. A
72 (
24)2
7
2
5x
5
f Arctan 23
34°
p r cos (v f)
11
6
4
3
2
y 0
13
13
3
213
5
13
13
, sin f , p cos f 13
13
13
2 4 6 8
7
6
4
cos f 5, sin f 5, p 2.1
11a. p r cos (v f) → 5 r cos v 56
11b.
1
18
y 21
0
0
0
3
5
3
3
2
6
x
10
0˚
330˚
210˚
11
6
4
3
14. A2 B2 62 (
8)2
10
Since C is negative, use 10.
60˚
150˚
0
120˚
1041
41
r cos (v 309°)
A2 B2 (–1)2 32
17. 10
Since C is negative, use 10
.
1
x
10
cos f 2209y 8279 0
3
10
10
10
7
10
y 0
3
10
7
10
, p sin f 10
10
f Arctan (3)
72°
Since cos f < 0, but sin f > 0, the normal lies in
the second quadrant.
f 180° 72° or 108°
p r cos (v f)
cos f 2219, sin f 2209, p 3
f Arctan 2201
44°
p r cos (v f)
3 r cos (v 44°)
7
10
10
r cos (v 108°)
18. 6 r cos (v 120°)
0 r (cos v cos 120° sin v sin 120°) 6
3
r sin v 6
0 12 r cos v 2
3
y 6
0 12x 2
y 12 or
0 x 3
x 3
y 12 0
285
Chapter 9
28.
19. 4 r cos v 4
0 r cos v cos 4 sin v sin 4 4
x 2
y 8 or
0 2
x 2
y 8 0
2
20. 2 r cos (v )
0 r (cos v cos sin v sin ) 2
0 r cos v 2
0 x 2
x 2
21. 1 r cos (v 330°)
0 r (cos v cos 330° sin v sin 330°) 1
0
r cos v cos
r cos v sin v sin
7
6
1
y
2
30˚
150˚
180˚
3 6 9 12
330˚
210˚
240˚
26.
120˚
270˚
90˚
300˚
0˚
180˚
1 2 3 4
330˚
210˚
240˚
270˚
300˚
3
11
6
7
6
120˚
30˚
2
0
27.
3
2
90˚
5
3
60˚
30˚
150˚
180˚
2 4 6 8
0˚
330˚
210˚
240˚
3
13
5
13
0
3
13
5
13
, p sin f 13
13
r cos (v 56°)
Use a graphing calculator and the INTERJECT
feature to find solutions to the system at (2.25,
0.31) and (5.39, 0.31). Since p, the length of the
normal, must be positive, use f 2.25 and p 0.31.
0.31 r cos (v 2.25)
32a. p r cos (v f) → 6 r cos (v 16°)
Since the shortest distance is along the normal,
the closest the fly came was p or 6 cm.
32b. (p, f) or (6, 15°)
33. Since both normal segments have length 2, p must
be 2 in both equations. Since the two lines
intersect at right angles, their normals also
intersect at right angles. This can be achieved by
having the two f-values differ by 90°. To make
sure neither line is vertical, neither f-value
should be a multiple of 90°. Therefore, a sample
answer is 2 r cos (v 45°) and 2 r cos (v 135°).
34. m 0
(y 4) 0(x 5) → y 4 0
cos f 0, sin f 1, p 4
Since cos f 0 when sin f 1, f 90°.
p r cos (v f)
4 r cos (v 90°)
1 2 3 4
4
3
5
3
→ p 2 cos 76 f
6
3
2
→ p 3 cos 4 f
5
6
60˚
150˚
Chapter 9
0˚
4
3
31. p r cos (v f)
11 0
2
3
13
5
13
11 0
25.
11
6
56°
p r cos (v f)
11
60˚
7
6
5
3
2
13
,
cos f 13
f Arctan 32
y 10 0
x 3
90˚
2 4 6 8
13
Since C is negative, use 13
.
3
1
r cos v r sin v 5 0
2
2
3
1
x y 5 0
2
2
120˚
0
A2 B2 22 32
x y 22 0
3
23.
r 5 sec (v 60°)
r cos (v 60°) 5
r (cos v cos 60° sin v sin 60°) 5 0
24.
6
2
1
3
r cos v r sin v 11 0
2
2
3
x
2
3
(y 1) 3(x 4) → 2x 3y 5 0
r 11 sec v 76
7
6
11
6
3
2
2
4
2
or 30. m 6
3
2
13
22.
7
6
0
4
3
2
3
5
6
2 4 6 8
7
6
sin v 1
x y 2 or
0 3
3
x y 2 0
29.
3
6
2
2
x y 4
0
2
2
0
2
5
6
2
2
r cos v r sin v 4
0
2
2
1
3
r cos v r
2
2
1
3
x y 1
2
2
2
3
270˚
300˚
286
35a.
120˚
90˚
40. x 3y 6
60˚
x 6
y
3
30˚
150˚
y 13x 2
0˚
x t, y 13t 2
12
5
25
0
37
505
0
180˚
N
r2
41. A 360 330˚
210˚
240˚
270˚
65
62
360 300˚
35b. p r cos (v f)
→ p 125 cos (130 f)
→ p 300 cos (70 f)
Use a graphing calculator and the INTERSECT
feature to find the solutions to the system at
(45, 124.43) and (135, 124.43). Sinc p, the
length of the normal, must be positive, use f 135° and p 124.43.
124.43 r cos (v 135°)
36. k r sin (v a)
k r [sin v cos a cos v sin a]
k r sin v cos a r cos v sin a
k y cos a x sin a
This is the equation of a line in rectangular
coordinates. Solving the last equation for y yields
k
. The slope of the line shows
y (tan a)x cos a
that a is the angle the line makes with the x-axis.
To find the length of the normal segment in the
figure, observe that the complementary angle to a
in the right triangle is 90° a, so the v-coordinate
of P in polar coordinates is 180° (90° a) a 90°. Substitute into the original polar
equation to find the r-coordinate of P:
k r sin (a 90° a)
k r sin 90°
kr
Therefore, k is the length of the normal segment.
y
20.42 ft2
42. Since 360° lies on the x-axis
of the unit circle at (1, 0),
sin 360° y or 0.
(1, 0)
x
43. 2x3 5x2 12x 0
x(2x2 5x 12) 0
x(2x 3)(x 4) 0
x 0 or x 32 or x 4
44.
c2 d2 48
(c d)(c d) 48
12(c d) 48
cd4
Page 579
1.
Mid-Chapter Quiz
90˚
120˚
2.
60˚
180˚
2 4 6 8
240˚
120˚
P
270˚
90˚
37. p r cos (v f)
→ p 40 cos (0° f)
→ p 40 cos (72° f)
Use a graphing calculator and the INTERSECT
feature to find the solutions of the system at
(144, 32.36) and (36, 32.36). Since p, the length
of the normal, must be positive, use f 36° and
p 32.36.
32.36 r cos (v 36°)
38.
r6
x2 y2 6
x2 y2 36
39. The graph of a polar equation of the form
r a sin nv is a rose.
0˚
330˚
270˚
300˚
5. r (2
)2 (
2
)2
4
or 2
11
6
7
6
4.
2 4 6 8
240˚
0
1 2 3 4
4
3
60˚
210˚
6
30˚
180˚
x
3
0˚
300˚
150˚
2
5
6
330˚
210˚
2
3
30˚
150˚
3.
O
y
2
3
5
3
3
2
2
3
6
5
6
1
0
11
6
7
6
4
3
5
3
3
2
v Arctan
4
2
Since (2
, 2
) is in the third quadrant,
v 4 or 54.
2, 54
6. r 02 (
4)2
16
or 4
Since x 0 when y 4, v 32.
4, 32
287
Chapter 9
x2 y2 36
x2 y2 36
r 6 or r 6
8.
r 2 csc v
r sin v 2
y2
9. A2 B2 52 (
12)2
13
Since C is positive, use 13.
4. Sample answer: x2 1 0
(x i)(x i) 0, where the solutions are x i.
x2 xi xi i2 0
x2 (1) 0
x2 1 0
6
4
2
5. i (i ) i2
12 (1)
1
6. i10 i2 (i4)2 i2 i2
(1)2i2 i2
1 (1) or 2
7. (2 3i) (6 i) (2 (6)) (3i i)
4 4i
8. (2.3 4.1i) (1.2 6.3i)
(2.3 (1.2)) (4.1i (6.3i))
3.5 10.4i
9. (2 4i) (1 5i) (2 (1)) (4i 5i)
1 9i
10. (2 i)2 (2 i)(2 i)
4 4i i2
3 4i
7.
5
12
3
y 0
1
3x 13
13
5
12
3
, p cos f 1
3 , sin f 13
13
2
f Arctan 15
67°
Since cos f 0, but sin f 0, the normal lies in
the second quadrant.
f 180° 67° or 113°
p r cos ( f)
3
13
r cos (v 113°)
A2 B2 (2)2 (6
)2
10. i
i
1 2i
11. 1 2i
1 2i 1 2i
210
Since C is negative, use 210
.
6
2
10
2
i 2i2
1 4i2
2
210
y 0
210
x
cos f 10
1
,
0
f Arctan sin f 310
10,
p
i2
5
10
10
25 15i
3
1
12. (2.5 3.1i) (6.2 4.3i)
(2.5 (6.2)) (3.1i 4.3i)
3.7 7.4i N
72°
Since cos f 0 and sin f 0, the normal lies in
the third quadrant.
f 180° 72° or 252°
p r cos (v f)
10
10
Pages 583–585
r cos (v 252°)
Simplifying Complex Numbers
9-5
Page 583
Check for Understanding
1. Find the (positive) remainder when the exponent
is divided by 4. If the remainder is 0, the answer
is 1; if the remainder is 1, the answer is i; if the
remainder is 2, the answer is 1; and if the
remainder is 3, the answer is i.
2.
Complex Numbers (a bi )
Reals
(b 0)
Imaginary
(b 0)
19. 12 i (2 i) 12 (2) (i (i))
Pure
Imaginary
(a 0)
32 2i
20. (3 i) (4 5i) (3 (4)) (i (5i))
7 4i
21. (2 i)(4 3i) 8 10i 3i2
5 10i
22. (1 4i)2 (1 4i)(1 4i)
1 8i 16i2
15 8i
3. When you multiply the denominators, you will be
multiplying a complex number and its conjugate.
This makes the denominator of the product a real
number, so you can then write the answer in the
form a bi.
Chapter 9
Exercises
13. i6 i4 i2
1 1
1
14. i19 (i4)4 i3
14 i
i
15. i1776 (i4)444
1444
1
16. i9 i5 (i4)2 i (i4)2 i3
12 i 12 i
i (i) or 0
17. (3 2i) (4 6i) (3 (4)) (2i 6i)
1 8i
18. (7 4i) (2 3i) (7 2) (4i 3i)
9 7i
288
3i
3i
34. (2 i)2
(2 i)(2 i)
23. (1 7
i)(2 5
i)
2 5
i 27
i 35
i2
(2 35
) (27
5
)i
24. (2 3
)(1 12
) (2 3
i)(1 12
i)
2 212
i 3
i
36
i2
2 43
i 3
i 6
8 33
i
3i
4 4i i2
3i
3 4i
3 4i
3i
3 4i 3 4i
9 9i 4i2
9 16i2
13 9i
25
1 2i
2i
2i
25. 1 2i
1 2i 1 2i
2 3i 2i2
1 4i2
1235 295i
4 3i
5
35.
(1 i)2
(3 2i)2
45 35i
3 2i
i
3 2i
4 26. 4 i 4 i
4 i
5 12i
2i
5 12i 5 12i
10i 24i2
25 144i2
11
0
11
17i
7
27.
5i
5i
24 10i
169
5i
5i
24
10
i
169
169
25 10i i2
25 i2
36a. Z R (XL XC )j
→ Z 10 (1 2)j → Z 10 j ohms
→ Z 3 (1 1)j → Z 3 0j ohms
36b. (10 j) (3 0j) (10 3) (1j 0j)
13 j ohms
24 10i
26
12
5
i
13
13
28. (x i)(x i) 0
x2 i2 0
x2 1 0
29.
(x (2 i))(x (2 i)) 0
(x 2 i)(x 2 i) 0
x2 2x xi 2x 4 2i xi 2i i2 0
x2 4x 4 1 0
x2 4x 5 0
30. (2 i)(3 2i)(1 4i) (6 i 2i2)(1 4i)
(8 i)(1 4i)
8 31i 4i2
12 31i
31. (1 3i)(2 2i)(1 2i) (2 8i 6i2)(1 2i)
(4 8i)(1 2i)
4 16i 16i2
12 16i
32.
1
3
i
2
1 2
i
1
1
36c. S Z → S 6 3j
1
6 3j
6 3j 6 3j
6 3j
36 9j2
6 3j
45
0.13 0.07j siemens
37a. x 3 4i
37b. No
37c. The solutions need not be complex conjugates
because the coefficients in the equation are not
all real.
37d. (3 4i)2 8i(3 4i) 25 0
7 24i 24i 32 25 0
00
(3 4i)2 8i(3 4i) 25 0
7 24i 24i 32 25 0
00
38. f(x yi) (x yi)2
x2 2xyi y2
(x2 y2) 2xyi
39a. z0 2 i
z1 i(2 i) i2 or 1 2i
z2 i(2i 1) 2i2 i or 2 i
z3 i(2 i) 2i i2 or 1 2i
z4 i(1 2i) i 2i2 or 2 i
z5 i(2 i) 2i i2 or 1 2i
1
3
i
2
i
1 2
i
1 2
1 2
i
1
2
i 3i 6i2
2
2
1 2i
12 6 22 3i
3
6
3
2
i
16 3 3
6
2 2
i
3 6i
2 2
i
3 6
i
3 6
i
3 6
i
6 2
6i 3
2i 12i2
9 6i2
(6 2
3) (2
6 3
2)i
15
23
8i (8i)2 4(1)(
25)
2(1)
8i 36
2
2
33.
1 2i i2
9 12i 4i2
2i
5 12i
12 11i 2i2
16 i2
10 11i
17
5i
5i
(1 i)(1 i)
(3 2i)(3 2i)
26
2
i
25 15 15
5
289
Chapter 9
39b. z0 1 0i
z1 (0.5 0.866i)(1 0i) 0.5 0.866i
z2 (0.5 0.866i)(0.5 0.866i)
0.25 0.866i 0.75
0.500 0.866i
z3 (0.5 0.866i)(0.500 0.866i)
0.250 0.750
1.000 0.000i
z4 (0.5 0.866i)(1.000) 0.500 0.866i
z5 (0.5 0.866i)(0.500 0.866i)
0.250 0.866i 0.75
0.500 0.866i
46. tan a 43
tan2 a 1 sec2 a
4 2
3
1
25
9
9
25
3
5
sin2
2
48. h x3
x3
tan 52° x 45
x tan 52° 45 tan 52° x3
x tan 52° x3
45 tan 52°
45 tan 52°
tan 52° 3
x x 127.40
h x3
127.40(3
) 221 ft
3
10
p
20
8˚
30˚
120˚
52˚
45 ft
h
60˚
x
49. Enter the x-values in L1 and the f(x)-values in L2
of your graphing calculator. Make a scatter plot.
The data points are in the shape of a parabola, so
a quadratic function would best model the set of
data.
50. Let d depth of the original pool.
The second pool’s width 5d 4, the length 10d 6, and the depth d 2.
(5d 4)(10d 6)(d 2) 3420
(50d2 70d 24)(d 2) 3420
50d3 100d2 70d2 140d 24d 48 3420
50d3 170d2 164d 3372 0
25d3 85d2 82d 1686 0
Use a graphing calculator to find the solution
d 3.
The dimensions of the original pool are 15 ft by
30 ft by 3 ft.
51. 80 k(5)(8)
2k
y 2(16)(2)
64
0
11
6
5
3
44. x (3), y 6
t1, 4
x 3, y 6
t1, 4
18, 6, 4
22, 6, 3
45. u
4
2, 32, 1 4, 12, 6
6, 227, 5
Chapter 9
3
3
2
12
y 3.5 cos 6t
6
4
3
4
5
period 1
2 or 6
r cos (v 162°)
7
6
5
cos B 1
3
1
18°
Since cos f 0, but sin f 0, the normal lies in
the second quadrant.
f 180° 18° or 162°
p r cos (v f)
7 14 21 28
25
cos2 B 169
47. amplitude 2(7) or 3.5
6
210
cos2 B 1
33
x y 0
5
6
2
1123
65
42. A2 b2 62 (
2)2
210
Since C is positive, use 210
.
2
sin B
B cos2 B 1
16
25
4
5
3
41. c1(cos 2t i sin 2t) c2(cos 2t i sin 2t)
c1 cos 2t c1i sin 2t c2 cos 2t c2i sin 2t
(c1 c2)(cos 2t) (c1 c2)(i sin 2t)
(c1 c2)(cos 2t) only if c1 c2
2
3
sin2 B
51
3 5 13 11
43.
sin2
csc2 B
cos (a B) cos a cos B sin a sin B
2
i
12 5 125
3
10
20
169
144
144
169
12
13
cos2 a
sin a 11 2i
11 2i
5
sec2 a
sin2 a 2
1 1
csc2 B
2
a
3 2
1
(3 4i)(1 2i)
2
3
210
210
310
10
,
cos f 10 , sin f 10
1
f Arctan 3
1 cot2 B csc2 B
sin2 a 5 1
1
1
11 2i
1
11 2i
11 2i
125
sec2
cos a
a cos2 a 1
40. (1 2i)3 (1 2i)3
cot B 152
290
52.
4. The conjugate of a bi is a bi.
(a
b
i)(a
bi) a2 b2, so the friend’s
method gives the same answer.
Sample answer: The absolute value of 2 3i is
22 32 13
. Using the friend’s method, the
absolute value is (2
3
i)(2
3i) 4
9
13
.
5. 2x y (x y)i 5 4i
2x y 5
xy4
2x (x 4) 5
y x 4
x1
y (1) 4 or 3
6.
7.
y 7 x2
x 7 y2
x 7 y2
x 7 y2
x
7
y
f1(x) 7
x
y
53.
(6, 8)
(6, 1)
O (1, 1)
x
f(x, y) 2x y
f(1, 1) 2(1) 1 or 3
f(6, 1) 2(6) 1 or 11
f(6, 8) 2(6) 8 or 4
The maximum value is 3 and the minimum value
is 11.
54. x 2y 7z 14
x 3y 5z 21
y 2z 7
x 3y 5z 21 → 5x 15y 25z 105
5x y 2z 7
5x y 2z 7
16y 27z 112
y 2z 7
→
16y 32z 112
16y 27z 112
6y 27z 112
1
59z 0
z0
y 2(0) 7
→ y7
x 2(7) 7(0) 14 → x 0
(0, 7, 0)
55. Since BC BD, m∠BDC m ∠DCB x
m ∠DBC 180 120 or 60.
x x 60 180
2x 120
x 60
x 40 60 40 or 100
The correct choice is A.
i
2
2
1
1
O
2 1
(2, 1)
i
1
1
2
2
1
2
z 12 (
2
)2
3
2
v Arctan 2 2
z (22 (1)2
5
2
8. r 2 (
2)2
7
or 22
8
4
v is in the fourth quadrant.
7
7
2 2i 22
cos 4 i sin 4
5
9. r 42 52
v Arctan 4
41
0.90
4 5i 41
(cos 0.90 i sin 0.90)
0
(2)2 02
10. r v Arctan 2 or 2
4
is on the x-axis at 2.
2 2 (cos i sin )
11.
The Complex and Polar Form
of Complex Numbers
Pages 589–590
2 1 O
1
2
3
2
3
(4, 3 )
i
5
6
9-6
2
(1, 2 )
11
6
4
3
1. To find the absolute value of a bi, square a and
b, add the squares, then take the square root of
the sum.
2. i 0 i; cos 2 0 and sin 2 1
i cos 2 i sin 2
3. Sample answer: z1 i, z2 i
3
2
5
3
4cos 3 i sin 3
3
——
2
4 i
1
2
i
2 23
2
3
6
2 4 6 8
7
6
Check for Understanding
12.
0
i
2
3
6
5
6
(2, 3)
0
1 2 3 4
7
6
11
6
4
3
3
2
5
3
2(cos 3 i sin 3)
2(0.99 i(0.14))
1.98 0.28i
z1 z2 z1 z2
i (i) i i
0 i i
0 2i
291
Chapter 9
13.
2
3
2
i
19.
3
20.
i
i
6
5
6
(2, 3)
3
( 2 , 2) 0
1 2 3 4
7
6
O
5
3
3
2
3
(cos 2 i sin
2
3
2 (1 i(0))
3
2
14.
O
11
6
4
3
(3, 4)
2)
z 22 32
13
z 32 (
4)2
25
or 5
21.
i
22.
i
1
i
0.63
1
0.90 0.36 1 1.30
O
0.38i
O
O
(0, 3)
1
(1, 5)
102 152
15a. magnitude 325
18.03 N
(1)2 (5
)2 z 02 (
3)2
z 26
9
or 3
23.
24.
15
15b. Arctan 1
0
56.31°
(1, 5 )
i
i
2
Pages 590–591
(4, 2 )
1
Exercises
16. 2x 5yi 12 15i
2x 12
5y 15
x6
y 3
17. 1 (x y)i y 3xi
1y
x y 3x
x (1) 3x
1 2x
1
x
2
18. 4x (y 5)i 2x y (x 7)i
y5x7
4x 2x y
y x 12
4x 2x (x 12)
3x 12
x 4
y (4) 12 or 8
2 1 O
1
2
O
1
2
z (1)2 (5
)2
6
25. r (4)2 62
52
or 213
z 42 (
2
)2
18
or 32
v Arctan 3
3
26. r 32 32
18
or 32
4
3 3i 32
cos 4 i sin 4
27. r (1)2 (
)2
3
3
v Arctan 1 4
or 2
4
3
v is in the third quadrant.
4
4
i 2cos 3 i sin 3
1 3
28. r 62 (
8)2
or 10
5.36
100
v is in the fourth quadrant.
6 8i 10(cos 5.36 i sin 5.36)
Chapter 9
292
8
v Arctan 6 2
v Arctan 4 1
29. r (4)2 12
2.90
17
v is in the second quadrant.
4 i 17
(cos 2.90 i sin 2.90)
39.
5
2
0
9
or 3
v is on the x-axis at 3.
3 3(cos 0 i sin 0)
41.
0
42
or 42
32
v is on the x-axis at 42
.
42
42
(cos i sin )
2
34. r 0 (2)2
4
or 2
3
Since x 0 when y 2, v 2.
i
3
2
3
(3, 4 ) 6
5
6
1 2 3 4
4
3
5
3
3
2
2
32
3
2
3
7
6
11
6
4
3
3
2
5
3
3(cos i sin )
3(1 0)
3
1
O
0.50 0.39i
0.60 0.39i
0.25 i 1
44.
i i
1
1
2 2i
32
2
3
i
2
38.
3
2
3
6
5
6
0
1 2 3 4
7
6
0
1 2 3 4
5
3
2 2 i
37.
6
0.44 0.44i
5
3
2
3
i
1
cos 6 i sin 6
32 i 2
2
6
4
3
3cos 4 i sin 4
i
(3, )
11
6
(1, )
6 11
7
6
2
3
1
0
1 2 3 4
5
3
0.44 0.5i
6
11
6
7
6
i
3
5
6
0
(5, 0) 0
43.
2
3
2
5
6
5(cos 0 i sin 0)
5(1 0)
5
3
36.
42.
3
2 4 6 8
3
2
11
6
2.5(0.54 i(0.84))
6
4
3
0
1.35 2.10i
7
6
2i 2cos 2 i sin 2
2
3
i
v Arctan 3
5
6
0
33. r (4
)2 2
02
2
2
3
2
6
2.5(cos 1 i sin 1)
2
2
2
i
2
v Arctan 3
32 02
32. r (2.5, 1) 4
3
5
i
3
7
6
2cos 4 i sin 4
2
2
2
1 2 3 4
5
3
3
2
i
11
6
4
or 25
2.03
20
v is in the second quadrant.
2 4i 25
(cos 2.03 i sin 2.03)
2
3
5
6
0
1 2 3 4
(2, 5 )
4
7
6
4
3
v Arctan 2 31. r (2)2 42
40.
3
6
21
or 29
5.47
841
v is in the fourth quadrant.
20 21i 29(cos 5.47 i sin 5.47)
35.
2
i
5
6
v Arctan 2
0 2
202 (21
)2
30. r 2
3
(2, 4 )
3
4
3
3
2
4
11
6
5
3
4
2cos 3 i sin 3
3
i
2
1
3
6
5
6
1
0
3 6 9 12
(10, 6)
7
6
45.
4
3
3
2
0.5 0.5i
5
3
1
10(cos 6 i sin 6)
10(0.960 i(0.279))
i
1 3
9.60 2.79i
i
1
11
6
22 i 2
1
1
O
O
1
0.5 0.5i
1
293
Chapter 9
46a. 40∠30° 40(cos 30° j sin 30°)
3
2
40
j
1
2
51. (6 2i)(2 3i) 12 22i 6i2
6 22i
52. x 3 cos 135°
y 3 sin 135°
34.64 20j
60∠60° 60(cos 60° j sin 60°)
1
3
602 j 2
2
49a.
49b.
49c.
49d.
50a.
71.96
v Arctan 64.64
tan 60° tan 45°
1 tan 60° tan 45°
3
1
1 (3
)(1)
3
1 1 3
1 3
1 3
3
3 1 3
4 23
2
2 3
v
55. w t
12(2)
1 or 24
v rq
18(24) or 432 cm/s
432 cm/s 4.32 m/s
13.57 m/s
14.11
Arctan 21.69
12
56. sin A 1
8
A sin1 3
2
A 41.8°
47. 2a
3a
1
5
2a 1 3a 5
4a
58. as x → , y → ; as x → , y → y 2x2 2
x
1000
10
0
10
1000
z2 5(cos 0.93 i sin 0.93)
z1z2 52
(cos 1.71 i sin 1.71)
7.07(cos 1.71 i sin 1.71)
50c. Sample answer: Let z1 2 4i and
z2 1 3i. Then
z1 25
(cos 5.18 i sin 5.18)
4.47(cos 5.18 i sin 5.18),
z2 10
(cos 1.89 i sin 1.89)
3.16(cos 1.89 i sin 1.89), and
z1z2 (2 4i)(1 3i)
10 10i
y
2 106
202
2
202
2 106
59. In the fourth month, the person will have received
3 pay raises.
$500(1.10)3 $665.50
The correct choice is D.
cos 4 i sin 4
102
14.14(cos 0.79 i sin 0.79).
50d. To multiply two complex numbers in polar form,
multiply the moduli and add the amplitudes. (In
the sample answer for 50c, note that 5.18 1.89
7.07, which is coterminal with 0.79.)
Chapter 9
32
50b. z1 2 cos 4 i sin 4 1.41(cos 0.79 i sin 0.79)
2
53. magnitude (3)2 72
58
u 7j
u
3, 7
3i
54. tan 105° tan (60° 45°)
25.88
0.58
21.69 14.11j 25.88 (cos 0.58 j sin 0.58) ohms
Translate 2 units to the right and down 3 units.
Rotate 90° counterclockwise about the origin.
Dilate by a factor of 3.
Reflect about the real axis.
Sample answer: let z1 1 i and z2 3 4i.
z1z2 (1 i)(3 4i)
1 7i
2
32 32
2 , 2 96.73
48°
v(t) 96.73 sin (250t 48°)
47. The graph of the conjugate of a complex number is
obtained by reflecting the original number about
the real axis. This reflection does not change the
modulus. Since the amplitude is reflected, we can
write the amplitude of the conjugate as the
opposite of the original amplitude. In other words,
the conjugate of r(cos v i sin v) can be written as
r(cos (v) i sin (v)), or r(cos v i sin v).
48a. 10(cos 0.7 j sin 0.7) 7.65 6.44j
16(cos 0.5 j sin 0.5) 14.04 7.67j
48b. (7.65 6.44j) (14.04 7.67j)
(7.65 14.04) (6.44j 7.67j)
21.69 14.11j ohms
2
48c. r 21.69
14
.112
32
32
30 51.96j
46b. (34.64 20j) (30 51.96j)
(34.64 30) (20j 51.96j)
64.64 71.96j
46c. v(t) r sin (250t v°)
2 71
r 64.64
.962
2
32
294
3
9-6B Graphing Calculator Exploration:
5. r 4
Geometry in the Complex Plane
2
v 6 3
3
6 or 2
cos 2 i sin 2 34(0 (1)i)
3
4
Page 592
3
4i
1. They are collinear.
9
v 4 2
4
6. r 2 or 2
9
2
11
4 4 or 4
11
11
2
2
2cos 4 i sin 4 22 i2
2
2
i
7. r 1
(6)
2
5
v 3 6
or 3
2
5
7
6 6 or 6
2. Yes. M is the point obtained when T 0, and N is
the point obtained when T 1.
3. The points are again collinear, but closer together.
7
7
3
3cos 6 i sin 6 32 i2
1
33
3
2 2i
22 (
23
)2
8. r1 16
or 4
r 4(23
) or 83
r2 (3)2 (3
)2
12
or 23
23
3
v1 Arctan 2
v2 Arctan 3
5
6
3
5
v 3 6
2
9-7
7
3
1
11
11
2cos 6 j sin 6 3cos 3 j sin 3
r 2(3) or 6
11
11
2
13
v 6 3
6 6
6 or 6
V 6cos 6 j sin 6 volts
Pages 596–598
Exercises
10. r 4(7) or 28
2
v 3 3
3
3 or Check for Understanding
7
12 43
i
9. E IZ
28(cos i sin ) 28(1 i(0))
28
1. The modulus of the quotient is the quotient of the
moduli of the two complex numbers. The
amplitude of the quotient is the difference of the
amplitudes of the two complex numbers.
2. Square the modulus of the given complex number
and double its amplitude.
3. Addition and subtraction are easier in rectangular
form. Multiplication and division are easier in
polar form. See students’ work for examples.
4. r 2 2 or 4
7
83
cos 6 i sin 6 83
2 i2
Products and Quotients of
Complex Numbers in Polar Form
Pages 596
5
6 6 or 6
4. The points are on the line through M and N.
5. If one of a, b, or c equals 0, then aK bM cN is
on KMN. If none of a, b, or c equals 0, then
aK bM cN lies on or inside KMN.
6. M is the point obtained when T 0 and N is the
point obtained when T 1. Thus, a point between
M and N is obtained when 0 T 1.
7. The distance between z and 1 i is 5. This
defines a circle of radius 5 centered at 1 i.
8. The distance between a point z and a point at
2 3i is 2.
z (2 3i) 2
6
3
2
v 4 4
11. r 2 or 3
4 or 2
3cos 2 i sin 2 3(0 i(1))
3i
12. r 1
2
3
1
v 3 6
or 6
2
3
6 6 or 6
v 2 2
cos 6 i sin 6 1623 i12
1
6
4
2 or 2
3
4(cos 2 i sin 2) 4(1 i(0))
4
1
12 12 i
295
Chapter 9
3
13. r 5(2) or 10
24. r1 22 (
2)2 r2 (3)2 32
8
or 22
18
or 32
r 22
(32
) or 12
v 4
4
3
7
4 4 or 4
7
7
2
10cos 4 i sin 4 102 i2
2
v1 Arctan 2 2
7
3
7
10
11
3
3
12i
25. r1 (2
)2
(2
)2 r2 (3
)2 (
2
32
)2
4
or 2
36
or 6
r 2 6 or 12
11
v1 Arctan 2 v2 Arctan 3cos 6 i sin 6 32 i 2
33
1
7
17. r 2
2
2
22
2
7
12(cos i sin ) 12(1 i(0))
12
(3
)2
(
1)2
r2 22 (23
)2
26. r1 4
or 2
16
or 4
3
v 4 4
4
4 or 2
11
7
4
cos
4
3
2
i sin
2
4
(0
3
i sin
2
2
2
2
or 3
4
2
v1 Arctan 42
3
4
3
v 4 4
2
4 or 2
2
2
cos i sin 2
2
3
3
4
2
5
4 or 4
2
2
2 i 2
2
6 6 or 6
3
23. r 1
5
v 3 3
or 8
4
3
4
4
3
8cos 3 i sin 3 82 i2
1
4 43
i
Chapter 9
v2 Arctan 6
6
4
22
3 (0 i(1))
2
2
3i
12cos 6 i sin 6 122 i2
63
6i
8
62
4
32
42
6
r 2 62 62
72
or 62
r 1
)2 2
(42
)2
27. r1 (4
64
or 8
i (1))
22
4 4i
v 4 2
5
4
3
v 3 6
22. r 2(6) or 12
4
1
2
10
cos 6 i sin 6 1223 i12
7
22
6
6
15
6 or 2
2 2i
11
1
2
4
3i
22
cos
5
6 6 or 6
11
3
) or 22
21. r 2(2
5
4
11
v 6 3
5
3
v 6 3
4[cos (5.6) i sin (5.6)] 3.10 2.53i
20
v2 Arctan 2
1
3
6
v 2 3.6 or 5.6
20. r 15 or 3
23
v1 Arctan 2
2(cos i sin ) 2(1 i(0))
2
18. r 3(0.5) or 1.5
v 4 2.5 or 6.5
1.5(cos 6.5 i sin 6.5) 1.46 0.32i
4
1
r 4 or 2
or 2
19. r 1 or 4
5
12
3 33
i
7
4
4 or 3
2
6(cos 300° i sin 300°) 6 i
5
3
2
32
v 4 4
v 240° 60°
300°
1
2
2
2
4
3
2 2i
16. r 2(3) or 6
12cos 2 i sin 2 12(0 i(1))
6 6 or 6
11
4 or 2
14
3
v 4 4
v 3 2
15. r 1 or 3
3
4
4
5
v 3 6
2
5
6 6
3
6 or 2
18cos 2 i sin 2 18(0 i(1))
18i
3
7
52
52
i
14. r 6(3) or 18
v2 Arctan 3 2
296
E
28. I Z
5
34.
13
5
3 2j
r1 13
13
13
r cos v 6 5
r 2 32 (
2)2
13
rcos v cos 6 sin v sin 6 5 0
v2 Arctan 3
2x 2y 5 0
5
5
3
1
2r cos v 2r sin v 5 0
r or 13
3
2
v1 0
35.
v 0 (0.59) or 0.59
I 13
(cos 0.59 j sin 0.59) 3 2j amps
29. Z 130˚
x lb
x lb
23 lb
x lb
23 lb
50˚
r 2 42 (
3)2
25
or 5
r1 100
x lb
Prop
Since the triangle is isosceles, the base angles are
100
r 5 or 20
180 50
congruent. Each measures 2 or 65°.
3
v2 Arctan 4
v1 0
23
sin 50°
0.64
v 0 (0.64) or 0.64
z 20(cos 0.64 j sin 0.64)
16 12j ohms
30. Start at z1 in the complex
plane. Since the modulus
z1
5
of z2 is 1, z1z2 and z will
6
2
both have the same
modulus
as z1. Then z1z2 z1
and z can be located by
2
rotating z1 by 6
7
6
counterclockwise and
clockwise, respectively.
23 sin 65°
sin 50°
2
3
i
z1z2
2
z1
z2
3
36.
6
0
1 2 3 4
3
2
5
x
27.21 x; 27.21 lb
cos 2x sin x 1
1 2 sin2 x sin x 1
2 sin2 x sin x 0
sin x (2 sin x 1) 0
sin x 0 or 2 sin x 1 0
x 0°
11
6
4
3
x
sin 65°
23 sin 65° x sin 50°
1
sin x 2
x 30°
y cos x
x cos y
arccos x y
38. BC ED BE AF CD 3
AB FE 2
AC AB BC
2 3 or 5
FD FE ED
2 3 or 5
perimeter of rectangle ACDF 3 5 3 5
or 16
perimeter of square BCDE 4(3) or 12
16 12 4
The correct choice is C.
5
3
37.
31a. The point is rotated counterclockwise about the
origin by an angle of v.
31b. The point is rotated 60° counterclockwise about
the origin.
32. Since a 1, the equation will be the form z2 bz
c 0. The coefficient c is the product of the
7
7
solutions, which is 6cos 6 i sin 6, or
33
3i in rectangular form. The coefficient b
is the opposite of the sum of the solutions, so
convert the solutions to rectangular form to do the
addition.
5
b 3cos 3 i sin 3 2cos 6 i sin 6
3
3
2 2i (3
i)
3
2
33
2 3
2i
3
Therefore, the equation is z2 2
33
2
1
x y 10 0
3
0.59
E
I
100
4 3j
r 5 secv 6
3
2
iz (33 3i) 0.
52 (
12)2
33. r 9-8
3
Page 602
12
Arctan 5 2
Powers and Roots of Complex
Numbers
Graphing Calculator Exploration
1. Rewrite 1 in polar form as 1(cos 0 i sin 0).
Follow the keystrokes to find the roots at 1,
0.5 0.87i, and 0.5 0.87i.
or 13
5.11
169
5 12i 13(cos 5.11 i sin 5.11)
2. Rewrite i in polar form as 1cos 2 i sin 2.
Follow the keystrokes to find the roots at
0.92 0.38i, 0.38 0.92i, 0.92 0.38i, and
0.38 0.92i.
297
Chapter 9
3. Rewrite 1 i in polar form as 2
cos 4 i sin
4
.
v 2
1
6
cos 162 i sin 162
1cos 1
2 i sin 12 1
Follow the keystrokes to find the roots at 1.06 0.17i, 0.17 1.06i, 0.95 0.49i, 0.76 0.76i, and 0.49 0.95i.
4. equilateral triangle
5. regular pentagon
6. If a 0 and b 0, then a bi a. The principal
roots of a positive real number is a positive real
number which would lie on the real axis in a
complex plane.
Pages 604-605
0.97 0.26i
1
3
5
cos () i sin (3)()
02 (
1)2 1
r 1
4
3
3 4n
3
3
7
7
x2 cos 8 i sin 8 0.92 0.38i
11
11
15
15
x3 cos 8 i sin 8 0.38 0.92i
x4 cos 8 i sin 8 0.92 0.38i
i
1
i
1
1
O
0.92 0.38i
0.38 0.92i
1
a ai
a
0.38 0.92i
0.92 0.38i
4 4i
2. Finding a reciprocal is the same as raising a
number to the 1 power, so take the reciprocal of
the modulus and multiply the amplitude by 1.
a ai
3 4n
x1 cos 8 i sin 8 0.38 0.92i
10. 2x3 4 2i 0 → x3 2 i
Find the third roots of 2 i.
(2)2 (1
)2 5
r a
a ai
O
a
1
1
3
)
x2 (5
4
. By De Moivre’s Theorem,
the polar form of (a ai)2 is 2a2cos 2 i sin 2.
i sin
)
x3 (5
23cos (3)6 i sin (3)6
1
3
8 cos 2 i sin 2
8(0 i (1))
8i
v 2
v 2
i sin 1.29 0.20i
cos 3
3
v 4
v 4
cos
i
sin
3
3 0.81 1.02i
O
1
1.29 0.20i
1
5
32 (
5)2 or 34
v Arctan 3
6. r 1.030376827
34
4 (cos (4)() i sin (4)(v))
644 960i
Chapter 9
v
0.47 1.22i
1
v Arctan or 6
(3
)2
(
1)2 or 2
5. r 1
3
v
i
Since cos 2 0, this is a pure imaginary number.
v 2n
x1 (5
) 3 cos 3 i sin 3 0.47 1.22i
4. Shembala is correct. The polar form of a ai is
a2
cos
v 2n
(5
) 3 cos 3 i sin 3
1
4
1
v Arctan 2 3.605240263 1
1
(2 i) 3 [5
(cos (v 2i) i sin (v 2n))] 3
a ai
a
1
3
cos 8 i sin 8
5
5
cos 4 i sin 4
2
2
2 i 2
42
3
v 2
(i) 1 (cos 2 2n i sin 2 2n 4
(2
)5 cos (5)4 i sin (5)4
42
2.677945045
1
3
0.82 1.02i
9. x4 i 0 → x4 i
Find the fourth roots of i.
Check for Understanding
1
v Arctan 2 (2)2 (1
)2 or 5
8. r 1. Same results, 4 4i; answers may vary.
(1 i)(1 i)(1 i)(1 i)(1 i)
(1 2i i2)(1 2i i2)(1 i)
(2i)(2i)(1 i)
4(1 i)
4 4i
(1 i)5
→ r 2
, v 4
3.
02 12 or 1
7. r 298
1
0.82 1.02i
11. For w1, the modulus (
0.82 (0.7
)2)2 or 1.13.
For w2, the modulus 1.132 or 1.28.
For w3, the modulus 1.282 or 1.64.
This moduli will approach infinity as the number
of iterations increases. Thus, it is an escape set.
21. r (2)2 12 5
)
( 5
1
4
v Arctan 2 1
2.677945045
cos (v) i sin (v)
1
4
1
4
0.96 0.76i
1
v Arctan 4
42 (
1)2 17
22. r Pages 605–606
12.
33
Exercises
cos (3) 6 i sin
27 cos 2 i sin 2
6
(3)
13.
)
(17
(22
)
162
i sin
2
2
i
2
2
7
( 2
)
7
4
1
2
1
4
16 cos 3 i sin 3
16 i
3
2
8 83
i
v 2
cos 122 i sin 122
2n
2n
cos 3 i sin 3
x1 cos 0 i sin 0 1
6
3
v Arctan 32 (
6)2 35
16. r 4
4
1
3
1
2
0.9827937232
(13
)2 (cos (2)(v) i sin (2)(v))
0.03 0.07i
2
O
1
v Arctan 2
4
1
1 3 i
2
1.107148718
4
(25
) (cos (4)(v) i sin (4)(v))
112 384i
2
3
1 3 i
3
2
1
i
v Arctan 2
22 42 25
18. r 2
x3 cos 3 i sin 3 2 2i
(35
)4 (cos (4)(v) i sin (4)(v))
567 1944i
22 32 13
17. r 2
x2 cos 3 i sin 3 2 2i
1.107148718
1
cos 1434 i sin 1434
0.71 0.71i
26. x3 1 0 → x3 1
Find the third roots of 1.
r 1 12 02 1
v0
1
3
1 [1 (cos (0 2n) i sin (0 2n))] 3
v Arctan 1 3
3
24 cos (4)3 i sin (4)3
1
2
1
v Arctan 1 0.91 0.61i
12 (
3
)2 2
15. r 1
4
02 12 1
25. r 21
4
16 16i
cos 134 i sin 134
3
v Arctan 4
2
4
7
2
2
v Arctan 2 4
1.37 0.37i
(22
)3cos (3)9 i sin (3)4
162
cos
1
3
(1)2 (1
)2 2
24. r 162
162
i
2
2
14. r (2) 2 22
21
4
cos (v) i sin (v)
1
3
22 22 22
23. r cos (5) 4 i sin (5) 4
5
5
32 cos 4 i sin 4
2
2
32 2 i 2
0.2449786631
1
3
1.60 0.13i
27(0 i(1))
27i
25
1
3
2
1
19. 32 5 cos 53 i sin 53
1
2cos
2
15
1
i sin
2
15
1.83 0.81i
20. r (1)2 02 1
1
4
1
v
cos () i sin ()
1
4
1
4
cos 4 i sin 4
0.71 0.71i
299
Chapter 9
27. x5 1 → x5 1
Find the fifth roots of 1.
r (1)2 02 1
i
2 2i
v
1
5
2 2i
1
1
5
(1) [1 (cos ( 2n) i sin ( 2n))]
2n
2n
cos 5 i sin 5
x1 cos
x2 cos
x3 cos
x4 cos
x5 cos
O
1
i sin 0.81 0.59i
5
5
3
3
i sin 0.31 0.95i
5
5
5
5
i sin 1
5
5
7
7
i sin 0.31 0.95i
5
5
9
9
i sin 0.81 0.59i
5
5
1
1
2i
2 2i
2 30. x4 (1 i) 0 → x4 1 i
Find the fourth roots of 1 i.
1
v Arctan 1 4
1
r 12 12 2
1
(1 i) 4 2
cos 4 2n i sin 4 2n 4
i
1
0.31 0.95i
1
8n
8n
(2
) 4 cos 16 i sin 16
0.81 0.59i
i
1
O
1
1.07 0.21i
0.81 0.59i
0.31 0.95i
1
3
1
v0
2i
2
1
9
9
1
)
x3 (2
1
x4 (2
)
3
2
3
17
4
1
(1 3
i) 4
4
64 [64 (cos (0 2n) i sin (0 2n))]
n
n
i sin
2
1
x1 2
1
x2 2 4
(cos i sin ) 22
x3 22
1
x3 2 4
22i
Find the fourth roots of 16.
2 02 16
r (16)
v
1
4
1
1
i sin
4
28
0.59 1.03i
1.03 0.59i
2n
i sin
28
1
4
x1 2 cos 4 i sin 4 2
2
i
i sin
4
cos 12 i sin 12 0.59 1.03i
10
10
cos 1
2 i sin 12 1.03 0.59i
22
22
cos 12 i sin 12 0.59 1.03i
i
2 cos 4 i sin 4
3
4
5
4
7
4
4 6n
x4 2 4 cos 12 i sin 12 1.03 0.59i
(16) [16 (cos ( 2n) i sin ( 2n))]
2n
4 6n
1
4
22i
3
3
x4 22
cos 2 i sin 2
3x4 48 0 → x4 16
1
2 4 cos 12 i sin 12
x1 22
(cos 0 i sin 0) 22
2
4
2 cos 3 2n i sin 3 2n 4
22
cos 2 i sin 2
3
4
5
4
7
4
17
cos 1
6 i sin 16 1.07 0.21i
25
25
cos 16 i sin 1
6 0.21 1.07i
3
1
4
1
4
or v Arctan 3
3
1 2i
2
1
4
x2 22
cos
1
4
i 0 → x4 1 3
i.
31. 2x4 2 23
Find the fourth roots of 1 3
i.
r (1)2 (
)2 2
3
2
Chapter 9
) 4 cos 16 i sin 16 0.21 1.07i
x2 (2
3 2 1 O
1
x4 2 cos
1
2
2
2
x3 2 cos
0.21 1.07i
x1 (2
) 4 cos 16 i sin 1
6 1.07 0.21i
i
2
2
x2 2 cos
1
O
1
1.07 0.21i
28. 2x4 128 0 → x4 64
Find the fourth roots of 64.
r 642 02 64
29.
1
0.21 1.07i
1
2 2i
2 2i
2 2i
O
0.59 1.03i
1
300
1
1.03 0.59i
32. Rewrite 10 9i in polar form as
9
10
cos tan1 181
38a. The point at (2, 2) becomes the point at (0, 2).
From the origin, the point at (2, 2) had a length
of 22
and the new point at (0, 2) has a length
2
of 2. The dilation factor is 2.
9
10
i sin tan1 .
Use a graphing calculator to find the fifth roots at
0.75 1.51i, 1.20 1.18i, 1.49 0.78i,
0.28 1.66i, and 1.66 0.25i.
33. Rewrite 2 4i in polar form as
25
[cos (tan1 (2)) i sin (tan1 (2))].
Use a graphing calculator to find the sixth roots at
1.26 0.24i, 0.43 1.21i, 0.83 0.97i,
1.26 0.24i, 0.43 1.21i, and 0.83 0.97i.
34. Rewrite 36 20i in polar form as
2 y
(0, 2)
1
x
2
2
2
3
4
38b.
For w2, the modulus (0.81)2 or 0.66.
For w3, the modulus (0.66)2 or 0.44.
This moduli will approach 0 as the number of
iterations increase. Thus, it is a prisoner set.
36a. In polar form the 31st roots of 1 are given by
2n
x3 cos
x4 cos
x5 cos
x6 cos
i sin
i sin
i sin
i sin
2
3
3
3
4
3
5
3
1
2(cos 90° i sin 90°)
5
10
v 6 3
11
3
11
6 6 or 6
1
45°
3
1
2
3
i
2
1 cos 45°
2
1
2
43. cos 22.5° cos 2
x2 cos 3 i sin 3 2 2i
2
3
3
3
4
3
5
3
33
3i
41. (2 5i) (3 6i) (6 2i)
(2 (3) (6)) (5i 6i 2i)
5i
42. x t, y 2t 7
n
1
22 (cos 45° i sin 45°)
11
cos 3 i sin 3
2
i sin 2
6cos 6 i sin 6 6 2 i 2
1
40. r 2(3) or 6
1 6 [1 (cos (0 2n) i sin (0 2n))] 6
x1 cos 0 i sin 0 1
(cos 45° i sin 45°)
2
2
2
2
The square is rotated 90° counterclockwise and
dilated by a factor of 0.5.
39. The roots are the vertices of a regular polygon.
Since one of the roots must be a positive real
number, a vertex of the polygon lies on the
positive real axis and the polygon is symmetric
about the real axis. This means that the non-real
complex roots occur in conjugate pairs. Since the
imaginary part of the sum of two complex
conjugates is 0, the imaginary part of the sum of
all the roots must be 0.
cos 3
1 i sin 31 , n 0, 1, . . . , 30. Then
2n
a cos 31 . The maximum value of a cosine
expression is 1, and it is achieved in this
situation when n 0.
36b. From the polar form in the solution to part a, we
2n
2n
get b sin 3
1 . b will be maximized when 31 is
as close to 2 as possible. This occurs when n 8,
16
so the maximum value of b is sin 3
1 , or about
0.9987.
37. x6 1 0 → x6 1
Find the sixth roots of 1.
r 12 02 1
v0
n
2
0.5 0.5i
2
122 2 or 0.81.
1
1
2
5
35. For w1, the modulus 2n
1 O
1
4106
cos tan1 9 i sin tan1 9.
Use a graphing calculator to find the eighth roots
at 1.59 0.10i, 1.05 1.19i, 0.10 1.59i,
1.19 1.05i, 1.59 0.10i, 1.05 1.19i,
0.10 1.59i, and 1.19 1.05i.
5
(2, 2)
2 2
1
3
2 2i
1
3
i
2
2
2
1
2
2
2 2
4 2 2
2
44. Find B.
B 180° 90° 81°15
8°45
Find a.
a
tan 81°15 2
8
28 tan 81°15 a
181.9 a
301
Chapter 9
Find c.
15. Sample answer: (4, 585°), (4, 945°), (4, 45°),
(4, 405°)
(r, v 360k°)
→ (4, 225° 360(1)°) → (4, 585°)
→ (4, 225° 360(2)°) → (4, 945°)
(r, v (2k 1)180°)
→ (4, 225° (1)180°) → (4, 45°)
→ (4, 225° (1)180°) → (4, 405°)
90˚
16.
17.
2
60
120
2
28
cos 81°15 c
28
c
cos 81°15
c 184.1
45. Let x the number of large bears produced.
Let y the number of small bears produced.
x 300
y
1200
y 400
(300, 900)
1000
x y 1200
800
f(x, y) 9x 5y
(800, 400)
600
f(300, 400) 9(300) 5(400) 400
4700
200 (300, 400)
x
f(300, 900) 9(300) 5(900)
O 200 600 1000
7200
f(800, 400) 9(800) 5(400)
9200
Producing 800 large bears and 400 small bears
yields the maximum profit.
46. 0.20(6) 1.2 quarts of alcohol
0.60(4) 2.4 quarts of alcohol
1.2 2.4
64
˚
˚
3
180˚
1 2 3 4
240˚
270˚
18.
90˚
120˚
3.6
1
or 36% alcohol
0
19.
1.
3.
5.
7.
9.
Check for Understanding
2.
4.
6.
8.
10.
absolute value
prisoner
pure imaginary
rectangular
Argand
120˚
Polar
iteration
cardioid
spiral of Archimedes
modulus
11.
120˚
12.
180˚
0˚
1 2 3 4
13.
2
3
270˚
2
240˚
300˚
14.
3
5
6
C
6
1 2 3 4
7
6
11
6
4
3
Chapter 9
3
2
5
3
2
4
3
3
2
270˚
22 32
(32
, 32
)
25. x 2 cos 330°
5
3
2
3
2
3
(3
, 1)
302
8 16 24 32
0
11
6
7
6
2
3
3
2
5
3
2
3
6
5
6
2 4 6 8
0
11
6
7
6
4
3
300˚
6
11
6
0˚
330˚
limaçon
24. x 6 cos 45°
D
7
6
23.
2 4 6 8
3
0
3
6
4
3
30˚
240˚
1 2 3 4
5
3
5
6
0˚
60˚
210˚
6
90˚
300˚
5
6
0
2
3
270˚
3
2
2
Spiral of Archimedes
180˚
330˚
210˚
2
3
300˚
150˚
0˚
1 2 3 4
B
330˚
240˚
30˚
180˚
A
210˚
120˚
11
6
21.
330˚
22.
60˚
150˚
30˚
150˚
7
6
60˚
210˚
270˚
1 2 3 4
4
3
2 4 6 8
3
0
circle
90˚
120˚
60˚
0˚
30˚
180˚
Skills and Concepts
90˚
90˚
2
5
3
6
300˚
150˚
240˚
Pages 608–610
270˚
3
2
5
6
330˚
20.
Pages 607
2
3
60˚
210˚
240˚
11
6
4
3
1 2 3 4
Chapter 9 Study Guide and Assessment
0
1 2 3 4
7
6
30˚
180˚
6
300˚
150˚
The correct choice is A.
0˚
330˚
210˚
3
5
6
30˚
150˚
3
2
rose
y 6 sin 45°
22 6
32
y 2 sin 330°
1
2 2
1
5
3
34
2
22
34
2
22
26. x 2 cos
2
(2
, 2
)
27. x 1 cos y 1 sin 0
(0, 1)
y 6 or
0 x 3
x 3
y 6 0
35. 4 r cos v 2
(3
)2 (
3)2
v Arctan
or 23
12
4
3
3
3
5
50
or 52
4
52, 4
36.
v Arctan 3 1
30. r (3)2 12
3.16
10
(3.16, 2.82)
37.
2.82
38.
v Arctan 4
2
31. r 42 22
20
4.47
(4.47, 0.46)
0.46
39.
A2 B2 22 12
32. 5
Since C is positive, use 5
.
2
5
1
5
25
cos f 5,
1
f Arctan 2
40.
3
5
x y 0
41.
5
35
sin f 5, p 5
4 2i
5 2i
5 2i 5 2i
20 18i 4i2
16 18i
16
18
29 29i
42.
r cos (v 207°)
A2 B2 32 12
33. 10
Since C is positive, use 10
.
5i
1 2i
1 2i
1 2
i
5i
1 2i
5 52
i i 2
i2
1 2i2
(5 2
) (52
1)i
3
1 52
5 2
3 3i
v Arctan 2
2
22 22
43. r 4
10
x y 0
8
or 22
10
22
cos 4 i sin 4
2
10
sin f 1
, p 5
0
4
44. r 12 (
3)2
v Arctan 1 2
3
5.03
10
10
(cos 5.03 i sin 5.03)
18°
Since cos f 0 and sin f 0, the normal lies in
the third quadrant.
f 180° 18° or 198°
p r cos (v f)
210
5
4 2i
5 2i
29
27°
Since cos f 0 and sin f 0, the normal lies in
the third quadrant.
f 180° 27° or 207°
p r cos (v f)
0 0 r sin v 4
0 y 4
0 y 4 or
y40
i10 i25 (i4)2 i2 (i4)6 i
(1)2 (1) (1)6 i
1 i
(2 3i) (4 4i) (2 (4)) (3i (4i))
2 7i
(2 7i) (3 i) (2 (3)) (7i (i))
1 6i
3
3
i (4 3i) 4i 3i4
4(i) 3(1)
3 4i
(i 7)(i 7) i2 14i 49
1 14i 49
48 14i
25 4i2
1
10
310
cos f 10,
1
f Arctan 3
0 r cos v cos 2 r sin v sin 2 4
v Arctan 5
29. r 52 52
3
10
3
0 2x 2 y 3
1
23, 43
35
5
3
1
0 2r cos v 2 r sin v 3
1
2
0 r cos v cos 3 r sin v sin 3 3
2
2
28. r 34. 3 r cos v 3
y 2 sin
45. r (1)2 (3
)2
2cos 3 i sin 3
r cos (v 198°)
3
2
or 2
4
2
v Arctan 1 3
2
46. r (6)2 (4
)2
4
v Arctan 6 or 213
3.73
52
213
(cos 3.73 i sin 3.73)
47. r (4)2 (1
)2
1
v Arctan 4 3.39
17
17
(cos 3.39 i sin 3.39)
303
Chapter 9
48. r 42 02
16
or 4
4(cos 0 i sin 0)
(2
)2 2
02
49. r 8
or 22
22
(cos i sin )
v0
02 32
50. r v
60. r (
3
)2 (1
)2 2
27 cos
v
2
3
5
6
128 2
6
2 cos 6 i sin 6
1
3
2 2 i 2
( 3, 5
3)
4
3
2
2
12 cos 3 i sin 3
1
3
12 2 i 2
3
5
11
6
4
5
15
2
16 16i
1
4
1
1
0.92 0.38i
2
1
3
3
3
1
1
1.24 0.22i
162
162
i
55. r 2(5) or 10
10 (cos 2.5 i sin 2.5)
8.01 5.98i
v 2 0.5 or 2.5
Page 611
Applications and Problem Solving
65. lemniscate
8
56. r 2 or 4
5
7
10
v
3
cos i sin 3 1
3
2 2 i 2
2
3
6
r cos v 2 5 0
r cos cos 2 r sin v sin 2 5 0
6
6
or
r sin v 5 0
y50
y 5
3
E
68. I Z
50 180j
32
4 5j
4 4i
2.2
50 180j
0.5 (cos 0.9 i sin 0.9)
0.31 0.39i
59. r 22 22 22
v Arctan 2 4
2
4 5j
4 5j 4 5j
v 1.5 0.6 or 0.9
58. r 4.4 or 0.5
59.04°
67.
3
2
125
21,25
0
145.77
(145.77, 59.04°)
4cos 2 i sin 2 6 6 or 2
4(0 i(1))
4i
v Arctan 7
5 752 1252
66. r 7
v 6 3
3
1
2 3 cos 18 i sin 18
4 4 or 4
2
1
200 470j 900j2
16 25j2
1100 470j2
41
26.83 11.46j amps
(2
)8 cos (8)4 i sin (8)4
4096 (cos 2 i sin 2)
4096
304
v Arctan 3
6
2 cos 36 i sin 36
v 4 2
2
Chapter 9
1 cos 42 i sin 42
322 i 2
3
v 2
(3
)2
12 2
64. r 3
or
02 12 1
63. r cos 8 i sin 8
32cos 4 i sin 4
3
2
2
162
2 i 2
54. r 8(4) or 32
57. r 15
162
cos 4 i sin 4
6 63
i
6
4
5
(22
)3 cos (3)4 i sin (3)4
2
v Arctan 2 (2)2 (2
)2 22
62. r 5
3
3
2
3
4(cos 3 i sin 3)
4
5
5
3 cos 3 i sin 3
1
3
3 2 i 2
3
33
2 2i
2
v 3 3 or 3
53. r 4(3) or 12
1
(2
)4 cos (4)4 i sin (4)4
1 2 3 4
7
6
3
i
v Arctan 1 3
0
5
3
3
2
4
3
6
11
6
4
3
i 2
5
6
1 2 3 4 7
6
2
3
0
643
64i
52.
( 2, 6 )
(1)2 12 2
61. r 3
i 2
128 9
or 3
3cos 2 i sin 2
51.
1
6
3
(7) 6 i sin (7) 6
7
7
cos 6 i sin 6
1
3
2 i 2
v Arctan
Page 611
180 (c b) 180 x
xcb
The correct choice is E.
4. Volume wh
16,500 75 w 10
16,500 750w
22 w
Open-Ended Assessment
1a. Sample answer: 4 6i and 3 2i
(4 6i) (3 2i) (4 3) (6i 2i)
7 4i
1b. No. Sample explanation: 2 3i and 5 i also
have this sum.
(2 3i) (5 i) (2 5) (3i (i))
7 4i
2a. Sample answer: 4 i
z 42 12
17
2b. No. Sample explanation: 1 4i also has this
absolute value.
z 12 42
17
w
5.
8
h
C
The answer choices include sin x. Write an
expression for the height, using the sine of x.
h
1
sin x 8
A 2bh
1
8 sin x h
2 (10)(8 sin x)
40 sin x
The correct choice is B.
3. Since PQRS is a parallelogram, sides PQ and SR
are parallel and m∠Q m∠S b.
M
c˚
P
Q
b˚
T
x˚
R
a˚
S
1
10
The correct choice is A.
6. Consider the three unmarked angles at the
intersection point. One of these angles, say the top
one, is the supplement of the other two unmarked
angles, because of vertical angles. So the sum of
the measures of the unmarked angles is 180°.
The sum of the measures of the marked angles
and the three unmarked angles is 3(180), since
these angles are the interior angles of three
triangles.
m(sum of marked angles) m(sum of unmarked angles) 3(180)
m(sum of marked angles) 180 3(180)
m(sum of marked angles) 360
The correct choice is C.
7. Subtract the second equation from the first.
5x2 6x
70
5x2
6y 10
6x 6y 60
x y 10, so 10x 10y 100.
The correct choice is E.
8. Since ∠B is a right angle, ∠C is a right angle also,
because they are alternate interior angles.
In the triangle containing ∠C, 90 x y 180 or
x y 90.
The straight angle at D is made up of 3 angles.
120 x x 180
2x 60 or x 30
x y 90
(30) y 90
y 60
The correct choice is B.
9. In the slope-intercept form of a line, y mx b,
m represents the slope of the line, and b
represents the y-intercept. Since the slope is
3
given as 2, the slope-intercept form of the line is
3
y 2x b.
Since (–3, 0) is on the line, it satisfies the
3
9
equation. 0 2(–3) b. So b 2.
The correct choice is D.
10. Note that consecutive interior angles are
supplementary.
110 2x 180
y x 180
2x 70
y (35) 180
x 35
y 145
The answer is 145.
B
10
1
1099 100100 10100
9
SAT and ACT Practice
x˚
1
100100
10100
1. ∠a and ∠b form a linear pair, so ∠b is
supplementary to ∠a. Since ∠b and ∠d are
vertical angles, they are equal in measure. So ∠d
is also supplementary to ∠a. Since ∠d and ∠f are
alternate interior angles, they are equal. So ∠f is
supplementary to ∠a. And since ∠f and ∠h are
vertical angles, ∠h is supplementary to ∠a. The
angles supplementary to ∠a are angles b, d, f, and
h. The correct choice is A.
2. Draw the given triangle and draw the height h
from point B.
A
75 ft
The correct choice is A.
Chapter 9 SAT & ACT Preparation
Page 613
h 10 ft
O
N
In SMO, c b a 180 or a 180 (c b).
Also, x a 180 or a 180 x since consecutive
interior angles are supplementary.
305
Chapter 9

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