1 |
21.
=
22. | z| = | z|
23. | z 1 z 2| = | z 1|| z 2|
24.
=
z
z
| z
2
z 2
2
2 |
For Exercises 25-30, put the given number in trigonometric form.
25. 2 + 3 i
26. −3 − 2 i
27. 1 − i
28. − i
29. 1
30. −1
31. Verify that De Moivre’s Theorem holds for the power n = 0.
For Exercises 32-35, calculate the given number.
32. 3 (cos 14◦ + i sin 14◦) · 2(cos 121◦ + i sin 121◦)
3 (cos 14◦ + i sin 14◦)
33. [3 (cos 14◦ + i sin 14◦)]4
34. [3 (cos 14◦ + i sin 14◦)]−4
35. 2(cos 121◦ + i sin 121◦)
36. Find the three cube roots of − i.
37. Find the three cube roots of 1 + i.
38. Find the three cube roots of 1.
39. Find the three cube roots of −1.
40. Find the five fifth roots of 1.
41. Find the five fifth roots of −1.
42. Find the two square roots of −2 + 2 3 i.
43. Prove that if z is an nth root of a real number a, then so is z. ( Hint: Use Exercise 20. ) 146
Chapter 6 • Additional Topics
§6.4
6.4 Polar Coordinates
Suppose that from the point (1, 0) in the
y
x y-coordinate plane we draw a spiral around
the origin, such that the distance between
any two points separated by 360◦ along the
← 1
spiral is always 1, as in Figure 6.4.1. We
←
→ 1
can not express this spiral as y = f ( x) for
→
some function f in Cartesian coordinates,
since its graph violates the vertical rule.
x
0
1
2
3
However, this spiral would be simple to
describe using the polar coordinate system.
Recall that any point P distinct from the
origin (denoted by O) in the x y-coordinate
plane is a distance r > 0 from the origin,
−−→
and the ray OP makes an angle θ with the
positive x-axis, as in Figure 6.4.2. We call
the pair ( r, θ) the polar coordinates of P,
Figure 6.4.1
and the positive x-axis is called the polar
axis of this coordinate system. Note that ( r, θ) = ( r, θ +360◦ k) for k = 0, ±1, ±2, ..., so (unlike for Cartesian coordinates) the polar coordinates of a point are not unique.
y
y
P( r, θ)
r
θ
θ
x
x
O
O
− r
P(− r, θ)
Figure 6.4.2
Polar coordinates ( r, θ)
Figure 6.4.3
Negative r: (− r, θ)
In polar coordinates we adopt the convention that r can be negative, by defining (− r, θ) =
−−→
( r, θ + 180◦) for any angle θ. That is, the ray OP is drawn in the opposite direction from the
angle θ, as in Figure 6.4.3. When r = 0, the point ( r, θ) = (0, θ) is the origin O, regardless of the value of θ.
You may be familiar with graphing paper, for plotting points or functions given in Carte-
sian coordinates (sometimes also called rectangular coordinates). Such paper consists of a
rectangular grid. Similar graphing paper exists for plotting points and functions in polar
coordinates, similar to Figure 6.4.4.
Polar Coordinates • Section 6.4
147
90◦
105◦
75◦
120◦
60◦
135◦
45◦
150◦
30◦
165◦
15◦
180◦
O
0◦
195◦
345◦
210◦
330◦
225◦
315◦
240◦
300◦
255◦
285◦
270◦
Figure 6.4.4
Polar coordinate graph
The angle θ can be given in either degrees or radians, whichever is more convenient.
Radians are often preferred when graphing functions in polar coordinates. The reason is
that, unlike degrees, radians can be considered “unitless” (as we mentioned in Chapter 4).
This is desirable when a function given in polar coordinates is expressed as r as a function
of θ (similar to how, in Cartesian coordinates ( x, y), functions are usually expressed as y as
a function of x). For example, if a function in polar coordinates is written as r = 2 θ, then r
would have the same units as θ. But r should be a unitless quantity, hence using radians for
θ makes more sense in this case.
148
Chapter 6 • Additional Topics
§6.4
Example 6.14
Express the spiral from Figure 6.4.1 in polar coordinates.
Solution: We will use radians for θ. The goal is to find some equation involving r and θ that describes
the spiral. We see that
θ = 0
⇒
r = 1
θ = 2 π
⇒
r = 2
θ = 4 π
⇒
r = 3
...
θ = 2 π k
⇒
r = 1 + k
for k = 0,1,2,.... In fact, that last relation holds for any nonnegative real number k (why?). So for
any θ ≥ 0,
θ
θ
θ = 2 π k
⇒
k =
⇒
r = 1 + k = 1 +
.
2 π
2 π
θ
Hence, the spiral can be written as r = 1 +
for θ ≥ 0. The graph is shown in Figure 6.4.5, along
2 π
with the Gnuplot commands to create the graph.
4
set polar
set size square
3
set samples 2000
unset key
2
set zeroaxis
set xlabel "x"
1
set ylabel "y"
plot [0:6*pi] 1 + t/(2*pi)
y
0
1
2
3
4
4
3
2
1
0
1
2
3
4
x
Figure 6.4.5
r = 1 + θ
2 π
Note that when using the set polar command, Gnuplot will assume that the function being plotted
is r as a function of θ (represented by the variable t in Gnuplot).
Polar Coordinates • Section 6.4
149
y
Figure 6.4.6 shows how to convert between polar coordinates and
( x, y) ( r, θ)
Cartesian coordinates. For a point with polar coordinates ( r, θ) and
Cartesian coordinates ( x, y):
r
y
Polar to Cartesian:
θ
x
x
O
= r cos θ
y = r sin θ
(6.9)
x
Cartesian to Polar:
Figure 6.4.6
y
r = ±
x 2 + y 2
tan θ =
if x = 0
(6.10)
x
Note that in formula (6.10), if x = 0 then θ = π/2 or θ = 3 π/2. Also, if x = 0 and y = 0 then the two possible solutions for θ in the equation tan θ
y
=
are in opposite quadrants (for
x
0 ≤ θ < 2 π). If the angle θ is in the same quadrant as the point ( x, y), then r =
x 2 + y 2 (i.e.
r is positive); otherwise r = − x 2 + y 2 (i.e. r is negative).
Example 6.15
Convert the following points from polar coordinates to Cartesian coordinates:
(a) (2, 30◦); (b) (3, 3 π/4); (c) (−1,5 π/3)
Solution: (a) Using formula (6.9) with r = 2 and θ = 30◦, we get:
( x, y) = ( r cos θ, r sin θ) = (2 cos 30◦,2 sin 30◦) = 2 · 3 ,2
2
· 12
⇒
( x, y) =
3, 1
(b) Using formula (6.9) with r = 3 and θ = 3 π/4, we get:
( x, y) = ( r cos θ, r sin θ) = 3 cos 3 π ,3 sin 3 π
, 3
, 3
4
4
= 3 · −1
· 1
⇒
( x, y) = −3
2
2
2
2
(c) Using formula (6.9) with r = −1 and θ = 5 π/3, we get:
( x, y) = ( r cos θ, r sin θ) = −1 cos 5 π ,
,
, 3
3
−1 sin 5 π
3
= −1 · 12 −1 · − 3
2
⇒
( x, y) = − 12 2
Example 6.16
Convert the following points from Cartesian coordinates to polar coordinates:
(a) (3, 4); (b) (−5,−5)
Solution: (a) Using formula (6.10) with x = 3 and y = 4, we get:
y
4
tan θ =
=
⇒
θ = 53.13◦ or θ = 233.13◦
x
3
Since θ = 53.13◦ is in the same quadrant (QI) as the point ( x, y) = (3,4), we can take
r =
x 2 + y 2 = 32 + 42 = 5. Thus, ( r, θ) = (5,53.13◦) .
Note that if we had used θ = 233.13◦, then we would have ( r, θ) = (−5,233.13◦).
150
Chapter 6 • Additional Topics
§6.4
(b) Using formula (6.10) with x = −5 and y = −5, we get:
y
−5
tan θ =
=
= 1
⇒
θ = 45◦ or θ = 225◦
x
−5
Since θ = 225◦ is in the same quadrant (QIII) as the point ( x, y) = (−5,−5), we can take
r =
x 2 + y 2 =
(−5)2 + (−5)2 = 5 2. Thus, ( r, θ) = (5 2,225◦) .
Note that if we had used θ = 45◦, then we would have ( r, θ) = (−5 2,45◦).
Example 6.17
Write the equation x 2 + y 2 = 9 in polar coordinates.
Solution: This is just the equation of a circle of radius 3 centered at the origin. Since r = ± x 2 + y 2 =
± 9, in polar coordinates the equation can be written as simply r = 3 .
Example 6.18
Write the equation x 2 + ( y − 4)2 = 16 in polar coordinates.
Solution: This is the equation of a circle of radius 4 centered at the point (0, 4). Expanding the
equation, we get:
x 2 + ( y − 4)2 = 16
x 2 + y 2 − 8 y + 16 = 16
x 2 + y 2 = 8 y
r 2 = 8 r sin θ
r = 8 sin θ
Why could we cancel r from both sides in the last step? Because we know that the point (0, 0) is on
the circle, so canceling r does not eliminate r = 0 as a potential solution of the equation (since θ = 0◦
would make r = 8 sin θ = 8 sin 0◦ = 0). Thus, the equation is r = 8 sin θ .
Example 6.19
Write the equation y = x in polar coordinates.
Solution: This is the equation of a line through the origin. So when x = 0, we know that y = 0. When
x = 0, we get:
y = x
y = 1
x
tan θ = 1
θ = 45◦
Since there is no restriction on r, we could have r = 0 and θ = 45◦, which would take care of the case
x = 0 (since then ( x, y) = (0,0), which is the same as ( r, θ) = (0,45◦)). Thus, the equation is θ = 45◦ .
Polar Coordinates • Section 6.4
151
Example 6.20
Prove that the distance d between two points ( r 1, θ 1) and ( r 2, θ 2) in polar coordinates is
d =
r 21 + r 22 − 2 r 1 r 2 cos ( θ 1 − θ 2) .
(6.11)
Solution: The idea here is to use the distance formula in Cartesian coordinates, then convert that
to polar coordinates. So write
x 1 = r 1 cos θ 1
y 1 = r 1 sin θ 1
x 2 = r 2 cos θ 2
y 2 = r 2 sin θ 2 .
Then ( x 1, y 1) and ( x 2, y 2) are the Cartesian equivalents of ( r 1, θ 1) and ( r 2, θ 2), respectively. Thus, by the Cartesian coordinate distance formula,
d 2 = ( x 1 − x 2)2 + ( y 1 − y 2)2
= ( r 1 cos θ 1 − r 2 cos θ 2)2 + ( r 1 sin θ 1 − r 2 sin θ 2)2
= r 2 cos2
cos2
sin2
sin2
1
θ 1 − 2 r 1 r 2 cos θ 1 cos θ 2 + r 22
θ 2 + r 21
θ 1 − 2 r 1 r 2 sin θ 1 sin θ 2 + r 22
θ 2
= r 2(cos2
(cos2
1
θ 1 + sin2 θ 1) + r 22
θ 2 + sin2 θ 2) − 2 r 1 r 2(cos θ 1 cos θ 2 + sin θ 1 sin θ 2)
d 2 = r 21 + r 22 − 2 r 1 r 2 cos ( θ 1 − θ 2) ,
so the result follows by taking square roots of both sides.
In Example 6.17 we saw that the equation x 2 + y 2 = 9 in Cartesian coordinates could be
expressed as r = 3 in polar coordinates. This equation describes a circle centered at the
origin, so the circle is symmetric about the origin. In general, polar coordinates are useful
in situations when there is symmetry about the origin (though there are other situations),
which arise in many physical applications.
Exercises
For Exercises 1-5, convert the given point from polar coordinates to Cartesian coordinates.
1. (6, 210◦)
2. (−4,3 π)
3. (2, 11 π/6)
4. (6, 90◦)
5. (−1,405◦)
For Exercises 6-10, convert the given point from Cartesian coordinates to polar coordinates.
6. (3, 1)
7. (−1,−3)
8. (0, 2)
9. (4, −2)
10. (−2,0)
For Exercises 11-18, write the given equation in polar coordinates.
11. ( x − 3)2 + y 2 = 9
12. y = − x
13. x 2 − y 2 = 1
14. 3 x 2 + 4 y 2 − 6 x = 9
15. Graph the function r = 1 + 2 cos θ in polar coordinates.
Appendix A
Answ