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10: Change of Variables in Multiple Integrals

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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?



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Appendixes

F Proofs of Theorems

G Complex Numbers

H Answers to Odd-Numbered Exercises



A1



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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 ͸ϱn෇N 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



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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.



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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



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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.



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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



zϩw෇zϩw



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



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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 ␪ ෇ Ϫ␲͞6 and



ͫ ͩ ͪ ͩ ͪͬ



w ෇ 2 cos Ϫ





6



ϩ i sin Ϫ





6



The numbers z and w are shown in Figure 5.



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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 ͒







෇ 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 sin͑␪1 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



͓cos͑␪1 Ϫ ␪ 2 ͒ ϩ i sin͑␪1 Ϫ ␪ 2

z2

r2



r

ă

0







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 n␾͒ ෇ 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 2␲͞n , 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

_





0





2 Re







w

_

2i







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.



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