10: Change of Variables in Multiple Integrals
Tải bản đầy đủ - 0trang
97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 1194
1194
CHAPTER 17
SECOND-ORDER DIFFERENTIAL EQUATIONS
20. A spring with a mass of 2 kg has damping constant 16, and a
force of 12.8 N keeps the spring stretched 0.2 m beyond its
natural length. Find the position of the mass at time t if it
starts at the equilibrium position with a velocity of 2.4 m͞s.
21. Assume that the earth is a solid sphere of uniform density with
mass M and radius R 3960 mi. For a particle of mass m
within the earth at a distance r from the earth’s center, the
gravitational force attracting the particle to the center is
Fr
ϪGMr m
r2
where G is the gravitational constant and Mr is the mass of the
earth within the sphere of radius r.
ϪGMm
r.
R3
(b) Suppose a hole is drilled through the earth along a diameter. Show that if a particle of mass m is dropped from rest
at the surface, into the hole, then the distance y y͑t͒ of
the particle from the center of the earth at time t is given by
(a) Show that Fr
yЉ͑t͒ Ϫk 2 y͑t͒
where k 2 GM͞R 3 t͞R.
(c) Conclude from part (b) that the particle undergoes simple
harmonic motion. Find the period T.
(d) With what speed does the particle pass through the center
of the earth?
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA1.qk_97879_Apdx7eMV_Apdx7eMV_pA1 11/9/10 4:54 PM Page A1
Appendixes
F Proofs of Theorems
G Complex Numbers
H Answers to Odd-Numbered Exercises
A1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A2
A2
APPENDIX F
F
PROOFS OF THEOREMS
Proofs of Theorems
In this appendix we present proofs of several theorems that are stated in the main body of
the text. The sections in which they occur are indicated in the margin.
Section 11.8
In order to prove Theorem 11.8.3, we first need the following results.
Theorem
1. If a power series
c n x n converges when x b (where b
Խ Խ Խ Խ
whenever x Ͻ b .
2. If a power series c n x n diverges when x d (where d
whenever x Ͼ d .
Խ Խ Խ Խ
0), then it converges
0 ), then it diverges
c n b n converges. Then, by Theorem 11.2.6, we have
PROOF OF 1 Suppose that
lim n l ϱ c n b 0. According to Definition 11.1.2 with 1, there is a positive integer
N such that cn b n Ͻ 1 whenever n ജ N. Thus, for n ജ N, we have
n
Խ
Խ
Խ
Խ
cn x n
Ϳ Ϳ
ͿͿ ͿͿ
cn b nx n
x
cn b n
bn
b
Խ
Խ
n
x
Ͻ
b
n
If x Ͻ b , then x͞b Ͻ 1, so x͞b n is a convergent geometric series. Therefore,
by the Comparison Test, the series ϱnN c n x n is convergent. Thus the series c n x n is
absolutely convergent and therefore convergent.
Խ Խ Խ Խ
Խ
Խ
Խ
Խ
Խ
Խ
c n d n diverges. If x is any number such that Խ x Խ Ͼ Խ d Խ, then
c n x cannot converge because, by part 1, the convergence of c n x n would imply the
convergence of c n d n. Therefore c n x n diverges whenever Խ x Խ Ͼ Խ d Խ.
PROOF OF 2 Suppose that
n
Theorem For a power series
c n x n there are only three possibilities:
1. The series converges only when x 0.
2. The series converges for all x.
Խ Խ
3. There is a positive number R such that the series converges if x Ͻ R and
Խ Խ
diverges if x Ͼ R.
PROOF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers b
and d such that c n x n converges for x b and diverges for x d. Therefore the set
S ͕x c n x n converges͖ is not empty. By the preceding theorem, the series diverges if
x Ͼ d , so x ഛ d for all x ʦ S. This says that d is an upper bound for the set S.
Thus, by the Completeness Axiom (see Section 11.1), S has a least upper bound R. If
x Ͼ R, then x S, so c n x n diverges. If x Ͻ R, then x is not an upper bound for
S and so there exists b ʦ S such that b Ͼ x . Since b ʦ S, c n b n converges, so by the
preceding theorem c n x n converges.
Խ
Խ Խ Խ Խ
Խ Խ
Խ Խ Խ Խ
Խ Խ
Խ Խ
Խ Խ
Խ Խ
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A3
APPENDIX F
3
Theorem For a power series
PROOFS OF THEOREMS
A3
cn͑x Ϫ a͒n there are only three possibilities:
1. The series converges only when x a.
2. The series converges for all x.
3. There is a positive number R such that the series converges if
Խ
Խ
diverges if x Ϫ a Ͼ R.
Խ x Ϫ a Խ Ͻ R and
PROOF If we make the change of variable u x Ϫ a, then the power series becomes
c n u n and we can apply the preceding theorem to this series. In case 3 we have con-
Խ Խ
Խ Խ
vergence for u Ͻ R and divergence for u Ͼ R. Thus we have convergence for
x Ϫ a Ͻ R and divergence for x Ϫ a Ͼ R.
Խ
Section 14.3
Խ
Խ
Խ
Clairaut’s Theorem Suppose f is defined on a disk D that contains the point ͑a, b͒.
If the functions fxy and fyx are both continuous on D, then fxy͑a, b͒ fyx͑a, b͒.
PROOF For small values of h, h
0, consider the difference
⌬͑h͒ ͓ f ͑a ϩ h, b ϩ h͒ Ϫ f ͑a ϩ h, b͔͒ Ϫ ͓ f ͑a, b ϩ h͒ Ϫ f ͑a, b͔͒
Notice that if we let t͑x͒ f ͑x, b ϩ h͒ Ϫ f ͑x, b͒, then
⌬͑h͒ t͑a ϩ h͒ Ϫ t͑a͒
By the Mean Value Theorem, there is a number c between a and a ϩ h such that
t͑a ϩ h͒ Ϫ t͑a͒ tЈ͑c͒h h͓ fx͑c, b ϩ h͒ Ϫ fx͑c, b͔͒
Applying the Mean Value Theorem again, this time to fx , we get a number d between b
and b ϩ h such that
fx͑c, b ϩ h͒ Ϫ fx͑c, b͒ fxy͑c, d͒h
Combining these equations, we obtain
⌬͑h͒ h 2 fxy͑c, d͒
If h l 0, then ͑c, d ͒ l ͑a, b͒, so the continuity of fxy at ͑a, b͒ gives
lim
hl0
⌬͑h͒
lim fxy͑c, d͒ fxy͑a, b͒
͑c, d͒ l ͑a, b͒
h2
Similarly, by writing
⌬͑h͒ ͓ f ͑a ϩ h, b ϩ h͒ Ϫ f ͑a, b ϩ h͔͒ Ϫ ͓ f ͑a ϩ h, b͒ Ϫ f ͑a, b͔͒
and using the Mean Value Theorem twice and the continuity of fyx at ͑a, b͒, we obtain
lim
hl0
⌬͑h͒
fyx͑a, b͒
h2
It follows that fxy͑a, b͒ fyx͑a, b͒.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A4
A4
APPENDIX F
PROOFS OF THEOREMS
8 Theorem If the partial derivatives fx and fy exist near ͑a, b͒ and are continuous at ͑a, b͒, then f is differentiable at ͑a, b͒.
Section 14.4
PROOF Let
⌬z f ͑a ϩ ⌬x, b ϩ ⌬y͒ Ϫ f ͑a, b͒
According to (14.4.7), to prove that f is differentiable at ͑a, b͒ we have to show that we
can write ⌬z in the form
⌬z fx͑a, b͒ ⌬x ϩ fy͑a, b͒ ⌬y ϩ 1 ⌬x ϩ 2 ⌬y
where 1 and 2 l 0 as ͑⌬x, ⌬y͒ l ͑0, 0͒.
Referring to Figure 1, we write
y
(a+Ỵx, b+Ỵy)
(u, b+Ỵy)
1
(a, b+Ỵy)
⌬z ͓ f ͑a ϩ ⌬x, b ϩ ⌬y͒ Ϫ f ͑a, b ϩ ⌬y͔͒ ϩ ͓ f ͑a, b ϩ ⌬y͒ Ϫ f ͑a, b͔͒
Observe that the function of a single variable
(a, √)
R
t͑x͒ f ͑x, b ϩ ⌬y͒
(a, b)
0
x
is defined on the interval ͓a, a ϩ ⌬x͔ and tЈ͑x͒ fx͑x, b ϩ ⌬y͒. If we apply the Mean
Value Theorem to t, we get
FIGURE 1
t͑a ϩ ⌬x͒ Ϫ t͑a͒ tЈ͑u͒ ⌬x
where u is some number between a and a ϩ ⌬x. In terms of f, this equation becomes
f ͑a ϩ ⌬x, b ϩ ⌬y͒ Ϫ f ͑a, b ϩ ⌬y͒ fx͑u, b ϩ ⌬y͒ ⌬x
This gives us an expression for the first part of the right side of Equation 1. For the
second part we let h͑y͒ f ͑a, y͒. Then h is a function of a single variable defined on
the interval ͓b, b ϩ ⌬y͔ and hЈ͑y͒ fy͑a, y͒. A second application of the Mean Value
Theorem then gives
h͑b ϩ ⌬y͒ Ϫ h͑b͒ hЈ͑v͒ ⌬y
where v is some number between b and b ϩ ⌬y. In terms of f, this becomes
f ͑a, b ϩ ⌬y͒ Ϫ f ͑a, b͒ fy͑a, v͒ ⌬y
We now substitute these expressions into Equation 1 and obtain
⌬z fx͑u, b ϩ ⌬y͒ ⌬x ϩ fy͑a, v͒ ⌬y
fx͑a, b͒ ⌬x ϩ ͓ fx͑u, b ϩ ⌬y͒ Ϫ fx͑a, b͔͒ ⌬x ϩ fy͑a, b͒ ⌬y
ϩ ͓ fy͑a, v͒ Ϫ fy͑a, b͔͒ ⌬y
fx͑a, b͒ ⌬x ϩ fy͑a, b͒ ⌬y ϩ 1 ⌬x ϩ 2 ⌬y
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A5
APPENDIX G
COMPLEX NUMBERS
A5
1 fx͑u, b ϩ ⌬y͒ Ϫ fx͑a, b͒
where
2 fy͑a, v͒ Ϫ fy͑a, b͒
Since ͑u, b ϩ ⌬y͒ l ͑a, b͒ and ͑a, v͒ l ͑a, b͒ as ͑⌬x, ⌬y͒ l ͑0, 0͒ and since fx and fy
are continuous at ͑a, b͒, we see that 1 l 0 and 2 l 0 as ͑⌬x, ⌬y͒ l ͑0, 0͒.
Therefore f is differentiable at ͑a, b͒.
Complex Numbers
G
Im
2+3i
_4+2i
i
0
_i
_2-2i
Re
1
3-2i
FIGURE 1
Complex numbers as points in
the Argand plane
A complex number can be represented by an expression of the form a ϩ bi, where a and
b are real numbers and i is a symbol with the property that i 2 Ϫ1. The complex number a ϩ bi can also be represented by the ordered pair ͑a, b͒ and plotted as a point in a
plane (called the Argand plane) as in Figure 1. Thus the complex number i 0 ϩ 1 ؒ i is
identified with the point ͑0, 1͒.
The real part of the complex number a ϩ bi is the real number a and the imaginary
part is the real number b. Thus the real part of 4 Ϫ 3i is 4 and the imaginary part is Ϫ3.
Two complex numbers a ϩ bi and c ϩ di are equal if a c and b d, that is, their real
parts are equal and their imaginary parts are equal. In the Argand plane the horizontal axis
is called the real axis and the vertical axis is called the imaginary axis.
The sum and difference of two complex numbers are defined by adding or subtracting
their real parts and their imaginary parts:
͑a ϩ bi͒ ϩ ͑c ϩ di͒ ͑a ϩ c͒ ϩ ͑b ϩ d͒i
͑a ϩ bi͒ Ϫ ͑c ϩ di͒ ͑a Ϫ c͒ ϩ ͑b Ϫ d͒i
For instance,
͑1 Ϫ i ͒ ϩ ͑4 ϩ 7i ͒ ͑1 ϩ 4͒ ϩ ͑Ϫ1 ϩ 7͒i 5 ϩ 6i
The product of complex numbers is defined so that the usual commutative and distributive
laws hold:
͑a ϩ bi͒͑c ϩ di͒ a͑c ϩ di͒ ϩ ͑bi͒͑c ϩ di͒
ac ϩ adi ϩ bci ϩ bdi 2
Since i 2 Ϫ1, this becomes
͑a ϩ bi͒͑c ϩ di ͒ ͑ac Ϫ bd͒ ϩ ͑ad ϩ bc͒i
EXAMPLE 1
͑Ϫ1 ϩ 3i͒͑2 Ϫ 5i͒ ͑Ϫ1͒͑2 Ϫ 5i͒ ϩ 3i͑2 Ϫ 5i͒
Ϫ2 ϩ 5i ϩ 6i Ϫ 15͑Ϫ1͒ 13 ϩ 11i
Division of complex numbers is much like rationalizing the denominator of a rational
expression. For the complex number z a ϩ bi, we define its complex conjugate to be
z a Ϫ bi. To find the quotient of two complex numbers we multiply numerator and
denominator by the complex conjugate of the denominator.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A6
A6
COMPLEX NUMBERS
APPENDIX G
EXAMPLE 2 Express the number
Ϫ1 ϩ 3i
in the form a ϩ bi.
2 ϩ 5i
SOLUTION We multiply numerator and denominator by the complex conjugate of 2 ϩ 5i,
namely 2 Ϫ 5i, and we take advantage of the result of Example 1:
Ϫ1 ϩ 3i
Ϫ1 ϩ 3i 2 Ϫ 5i
13 ϩ 11i
13
11
ؒ
2
ϩ
i
2
2 ϩ 5i
2 ϩ 5i
2 Ϫ 5i
2 ϩ5
29
29
Im
The geometric interpretation of the complex conjugate is shown in Figure 2: z is the
reflection of z in the real axis. We list some of the properties of the complex conjugate in
the following box. The proofs follow from the definition and are requested in Exercise 18.
z=a+bi
i
0
Re
Properties of Conjugates
_i
zϩwzϩw
z=a-bi
–
zw z w
zn zn
Խ Խ
FIGURE 2
The modulus, or absolute value, z of a complex number z a ϩ bi is its distance
from the origin. From Figure 3 we see that if z a ϩ bi, then
Im
|z|
0
FIGURE 3
Խ z Խ sa
z=a+bi
b„@
„„
+
@
„
œ „a
bi
=
b
a
2
ϩ b2
Notice that
zz ͑a ϩ bi͒͑a Ϫ bi ͒ a 2 ϩ abi Ϫ abi Ϫ b 2i 2 a 2 ϩ b 2
Re
Խ Խ
zz z
and so
2
This explains why the division procedure in Example 2 works in general:
z
w
zw
ww
zw
ԽwԽ
2
Since i 2 Ϫ1, we can think of i as a square root of Ϫ1. But notice that we also have
͑Ϫi͒2 i 2 Ϫ1 and so Ϫi is also a square root of Ϫ1. We say that i is the principal
square root of Ϫ1 and write sϪ1 i. In general, if c is any positive number, we write
sϪc sc i
With this convention, the usual derivation and formula for the roots of the quadratic equation ax 2 ϩ bx ϩ c 0 are valid even when b 2 Ϫ 4ac Ͻ 0:
x
Ϫb Ϯ sb 2 Ϫ 4ac
2a
EXAMPLE 3 Find the roots of the equation x 2 ϩ x ϩ 1 0.
SOLUTION Using the quadratic formula, we have
x
Ϫ1 Ϯ s1 2 Ϫ 4 ؒ 1
Ϫ1 Ϯ sϪ3
Ϫ1 Ϯ s3 i
2
2
2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A7
APPENDIX G
COMPLEX NUMBERS
A7
We observe that the solutions of the equation in Example 3 are complex conjugates of
each other. In general, the solutions of any quadratic equation ax 2 ϩ bx ϩ c 0 with real
coefficients a, b, and c are always complex conjugates. (If z is real, z z, so z is its own
conjugate.)
We have seen that if we allow complex numbers as solutions, then every quadratic
equation has a solution. More generally, it is true that every polynomial equation
a n x n ϩ a nϪ1 x nϪ1 ϩ и и и ϩ a 1 x ϩ a 0 0
of degree at least one has a solution among the complex numbers. This fact is known as
the Fundamental Theorem of Algebra and was proved by Gauss.
Polar Form
We know that any complex number z a ϩ bi can be considered as a point ͑a, b͒ and that
any such point can be represented by polar coordinates ͑r, ͒ with r ജ 0. In fact,
Im
a+bi
r
a r cos
b
b r sin
ă
0
a
Re
as in Figure 4. Therefore we have
z a ϩ bi ͑r cos ͒ ϩ ͑r sin ͒i
FIGURE 4
Thus we can write any complex number z in the form
z r͑cos ϩ i sin ͒
Խ Խ
r z sa 2 ϩ b 2
where
and
tan
b
a
The angle is called the argument of z and we write arg͑z͒. Note that arg͑z͒ is not
unique; any two arguments of z differ by an integer multiple of 2.
EXAMPLE 4 Write the following numbers in polar form.
(a) z 1 ϩ i
(b) w s3 Ϫ i
SOLUTION
Խ Խ
(a) We have r z s12 ϩ 12 s2 and tan 1, so we can take ͞4.
Therefore the polar form is
Im
z s2
1+i
ͩ
cos
ϩ i sin
4
4
ͪ
2
œ„
Խ Խ
π
4
0
π
_
6
Re
2
œ„
3-i
FIGURE 5
(b) Here we have r w s3 ϩ 1 2 and tan Ϫ1͞s3 . Since w lies in the
fourth quadrant, we take Ϫ͞6 and
ͫ ͩ ͪ ͩ ͪͬ
w 2 cos Ϫ
6
ϩ i sin Ϫ
6
The numbers z and w are shown in Figure 5.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 2:17 PM Page A8
A8
APPENDIX G
COMPLEX NUMBERS
The polar form of complex numbers gives insight into multiplication and division. Let
z1 r1͑cos 1 ϩ i sin 1 ͒
z2 r2͑cos 2 ϩ i sin 2 ͒
be two complex numbers written in polar form. Then
Im
z™
z1 z2 r1r2͑cos 1 ϩ i sin 1 ͒͑cos 2 ϩ i sin 2 ͒
z¡
r1r2 ͓͑cos 1 cos 2 Ϫ sin 1 sin 2 ͒ ϩ i͑sin 1 cos 2 ϩ cos 1 sin 2
ă
Therefore, using the addition formulas for cosine and sine, we have
ăĂ
Re
ăĂ+ă
z1z2 r1r2 cos1 2 ϩ i sin͑1 ϩ 2 ͔͒
1
z¡z™
This formula says that to multiply two complex numbers we multiply the moduli and add the
arguments. (See Figure 6.)
A similar argument using the subtraction formulas for sine and cosine shows that to
divide two complex numbers we divide the moduli and subtract the arguments.
FIGURE 6
Im
z
z1
r1
͓cos͑1 Ϫ 2 ͒ ϩ i sin͑1 Ϫ 2
z2
r2
r
ă
0
_ă
Re
1
r
z2
0
In particular, taking z1 1 and z2 z (and therefore 1 0 and 2 ), we have the
following, which is illustrated in Figure 7.
1
z
If
FIGURE 7
z r͑cos ϩ i sin ͒, then
1
1
͑cos Ϫ i sin ͒.
z
r
EXAMPLE 5 Find the product of the complex numbers 1 ϩ i and s3 Ϫ i in polar form.
SOLUTION From Example 4 we have
1 ϩ i s2
z=1+i
2
œ„
6
ϩ i sin Ϫ
6
So, by Equation 1,
zw
͑1 ϩ i͒(s3 Ϫ i) 2s2
π
12
Re
2s2
2
3-i
w=œ„
FIGURE 8
ͪ
ͩ ͪͬ
ϩ i sin
4
4
ͫ ͩ ͪ
2 œ„2
0
cos
s3 Ϫ i 2 cos Ϫ
and
Im
ͩ
ͫ ͩ
ͩ
cos
cos
Ϫ
4
6
ͪ ͩ
ͪ
ϩ i sin
Ϫ
4
6
ͪͬ
ϩ i sin
12
12
This is illustrated in Figure 8.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A9
APPENDIX G
COMPLEX NUMBERS
A9
Repeated use of Formula 1 shows how to compute powers of a complex number. If
z r ͑cos ϩ i sin ͒
then
z 2 r 2͑cos 2 ϩ i sin 2 ͒
and
z 3 zz 2 r 3͑cos 3 ϩ i sin 3 ͒
In general, we obtain the following result, which is named after the French mathematician
Abraham De Moivre (1667–1754).
2
De Moivre’s Theorem If z r͑cos ϩ i sin ͒ and n is a positive integer, then
z n ͓r ͑cos ϩ i sin ͔͒ n r n͑cos n ϩ i sin n ͒
This says that to take the nth power of a complex number we take the nth power of the
modulus and multiply the argument by n.
EXAMPLE 6 Find
SOLUTION Since
1
2
( 12 ϩ 12 i)10.
ϩ 12 i 12 ͑1 ϩ i͒, it follows from Example 4(a) that 12 ϩ 12 i has the
polar form
1
1
s2
ϩ i
2
2
2
ͩ
cos
ϩ i sin
4
4
ͪ
So by De Moivre’s Theorem,
ͩ
ͪ ͩ ͪͩ
ͩ
1
1
ϩ i
2
2
10
s2
2
25
2 10
10
cos
cos
10
10
ϩ i sin
4
4
5
5
ϩ i sin
2
2
ͪ
ͪ
1
i
32
De Moivre’s Theorem can also be used to find the n th roots of complex numbers. An
n th root of the complex number z is a complex number w such that
wn z
Writing these two numbers in trigonometric form as
w s͑cos ϩ i sin ͒
and
z r ͑cos ϩ i sin ͒
and using De Moivre’s Theorem, we get
s n͑cos n ϩ i sin n͒ r͑cos ϩ i sin ͒
The equality of these two complex numbers shows that
sn r
and
cos n cos
or
and
s r 1͞n
sin n sin
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:05 PM Page A10
A10
APPENDIX G
COMPLEX NUMBERS
From the fact that sine and cosine have period 2 it follows that
n ϩ 2k
Thus
ͫ ͩ
w r 1͞n cos
ϩ 2k
n
ͪ ͩ
ϩ 2k
n
or
ϩ 2k
n
ϩ i sin
ͪͬ
Since this expression gives a different value of w for k 0, 1, 2, . . . , n Ϫ 1, we have the
following.
3 Roots of a Complex Number Let z r͑cos ϩ i sin ͒ and let n be a positive
integer. Then z has the n distinct n th roots
ͫ ͩ
wk r 1͞n cos
ϩ 2k
n
ͪ ͩ
ϩ i sin
ϩ 2k
n
ͪͬ
where k 0, 1, 2, . . . , n Ϫ 1.
Խ Խ
Notice that each of the nth roots of z has modulus wk r 1͞n. Thus all the nth roots of
z lie on the circle of radius r 1͞n in the complex plane. Also, since the argument of each successive nth root exceeds the argument of the previous root by 2͞n , we see that the
n th roots of z are equally spaced on this circle.
EXAMPLE 7 Find the six sixth roots of z Ϫ8 and graph these roots in the complex
plane.
SOLUTION In trigonometric form, z 8͑cos ϩ i sin ͒. Applying Equation 3 with
n 6, we get
ͩ
wk 8 1͞6 cos
ϩ 2k
ϩ 2k
ϩ i sin
6
6
ͪ
We get the six sixth roots of Ϫ8 by taking k 0, 1, 2, 3, 4, 5 in this formula:
Im
2 i wĂ
w
2
_
wá
0
2 Re
wÊ
w
_
2i
wÂ
FIGURE 9
The six sixth roots of z=_8
ͩ
ͩ
ͩ
ͩ
ͩ
ͩ
ͪ ͩ ͪ
ͪ
ͪ ͩ
ͪ
ͪ ͩ
ͪ
ͪ
ͪ ͩ ͪ
w0 8 1͞6 cos
ϩ i sin
6
6
w1 8 1͞6 cos
ϩ i sin
2
2
w2 8 1͞6 cos
5
5
ϩ i sin
6
6
s2
Ϫ
1
s3
ϩ i
2
2
w3 8 1͞6 cos
7
7
ϩ i sin
6
6
s2
Ϫ
1
s3
Ϫ i
2
2
w4 8 1͞6 cos
3
3
ϩ i sin
2
2
Ϫs2 i
w5 8 1͞6 cos
11
11
ϩ i sin
6
6
s2
1
s3
ϩ i
2
2
s2 i
s2
1
s3
Ϫ i
2
2
All these points lie on the circle of radius s2 as shown in Figure 9.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.