(a) Answer Thecube roots of 1000 are z 0 = 2 ( cos 0 + i sin 0 ) , z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) and z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) Explanation Given info: 625 i 4 Formula used: The trigonometric form of the complex number z = a + i b is z = r ( cos θ + i sin θ ) Where r = a 2 + b 2 and θ = tan − 1 ( b a ) For a positive integer n the complex number z = r ( cos θ + i sin θ ) has exactly n distinct nth roots given by z k = r n ( cos ( θ + 2 π k n ) + i sin ( θ + 2 π k n ) ) where k = 0, 1, 2, . . . , ,n -1. Calculation: We have a = 8 , b = 0 r = a 2 + b 2 = 8 2 + 0 2 = 8 θ = tan − 1 ( 0 8 ) = 0 8 = 8 ( cos 0 + i sin 0 ) z k = r n ( cos ( θ + 2 π k n ) + i sin ( θ + 2 π k n ) ) n = 3 , r = 8 , θ = 0 z 0 = 8 3 ( cos ( 0 + 2 π × 0 3 ) + i sin ( 0 + 2 π × 0 3 ) ) z 0 = 2 ( cos 0 + i sin 0 ) z 1 = 8 3 ( cos ( 0 + 2 π × 1 3 ) + i sin ( 0 + 2 π × 1 3 ) ) z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) z 2 = 8 3 ( cos ( 0 + 2 π × 2 3 ) + i sin ( 0 + 2 π × 2 3 ) ) z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) Thecube roots of 1000 are z 0 = 2 ( cos 0 + i sin 0 ) , z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) and z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) Conclusion: Thus,the cube roots of 1000 are z 0 = 2 ( cos 0 + i sin 0 ) , z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) and z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) (b) To determine The standard form of each of the roots. Answer The standard formof the roots are z 0 = 1 , z 1 = − 1 + 1.73 i and z 2 = − 1 − 1.73 i Explanation Given info: Thecube roots of 8 are z 0 = 2 ( cos 0 + i sin 0 ) , z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) and z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) Formula used: The trigonometric form of the complex number z = a + i b is z = r ( cos θ + i sin θ ) Where r = a 2 + b 2 and θ = tan − 1 ( b a ) Calculation: The roots from part a are z 0 = 2 ( cos 0 + i sin 0 ) , z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) and z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) We have z 0 = 2 ( cos 0 + i sin 0 ) = 2 z 1 = 2 ( cos 2 π 3 + i sin 2 π 3 ) = − 1 + 1.73 i z 2 = 2 ( cos 4 π 3 + i sin 4 π 3 ) = − 1 − 1.73 i The standard formof the roots are z 0 = 1 , z 1 = − 1 + 1.73 i and z 2 = − 1 − 1.73 i Conclusion: Thus,the standard formof the roots are z 0 = 1 , z 1 = − 1 + 1.73 i and z 2 = − 1 − 1.73 i . (c) To determine The roots graphically. Answer Please refer calculation part. Explanation Given info: Thecube roots of 8 are z 0 = 1 , z 1 = − 1 + 1.73 i and z 2 = − 1 − 1.73 i Formula used: The trigonometric form of the complex number z = a + i b is z = r ( cos θ + i sin θ ) Where r = a 2 + b 2 and θ = tan − 1 ( b a ) Calculation: Graphical representation of roots are given by Conclusion: Thus,the graphical representation of roots are given.

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 6, Problem 127RE

(a)

To determine

Thecube roots of 8 using formula for roots.

(a)

Expert Solution

Answer to Problem 127RE

Thecube roots of 1000 are z0=2(cos0+isin0),z1=2(cos2π3+isin2π3) and z2=2(cos4π3+isin4π3)

Explanation of Solution

Given info:

625i4

Formula used:

The trigonometric form of the complex number z=a+ib is

z=r(cosθ+isinθ)

Where r=a2+b2 and θ=tan−1(ba)

For a positive integer n the complex number z=r(cosθ+isinθ) has exactly n distinct nth roots given by

zk=rn(cos(θ+2πkn)+isin(θ+2πkn))

where k = 0, 1, 2, . . . , ,n -1.

Calculation:

We have

a=8,b=0r=a2+b2=82+02=8θ=tan−1(08)=08=8(cos0+isin0)zk=rn(cos(θ+2πkn)+isin(θ+2πkn))n=3,r=8,θ=0z0=83(cos(0+2π×03)+isin(0+2π×03))z0=2(cos0+isin0)z1=83(cos(0+2π×13)+isin(0+2π×13))z1=2(cos2π3+isin2π3)z2=83(cos(0+2π×23)+isin(0+2π×23))z2=2(cos4π3+isin4π3)

Thecube roots of 1000 are z0=2(cos0+isin0),z1=2(cos2π3+isin2π3) and z2=2(cos4π3+isin4π3)

Conclusion:

Thus,the cube roots of 1000 are z0=2(cos0+isin0),z1=2(cos2π3+isin2π3) and z2=2(cos4π3+isin4π3)

(b)

To determine

The standard form of each of the roots.

(b)

Expert Solution

Answer to Problem 127RE

The standard formof the roots are z0=1,z1=−1+1.73i and z2=−1−1.73i

Explanation of Solution

Given info:

Thecube roots of 8 are z0=2(cos0+isin0),z1=2(cos2π3+isin2π3) and z2=2(cos4π3+isin4π3)

Formula used:

The trigonometric form of the complex number z=a+ib is

z=r(cosθ+isinθ)

Where r=a2+b2 and θ=tan−1(ba)

Calculation:

The roots from part a are z0=2(cos0+isin0),z1=2(cos2π3+isin2π3) and z2=2(cos4π3+isin4π3)

We have

z0=2(cos0+isin0)=2z1=2(cos2π3+isin2π3)=−1+1.73iz2=2(cos4π3+isin4π3)=−1−1.73i

The standard formof the roots are

z0=1,z1=−1+1.73i and z2=−1−1.73i

Conclusion:

Thus,the standard formof the roots are z0=1,z1=−1+1.73i and z2=−1−1.73i .

(c)

To determine

The roots graphically.

(c)

Expert Solution

Answer to Problem 127RE

Please refer calculation part.

Explanation of Solution

Given info:

Thecube roots of 8 are z0=1,z1=−1+1.73i and z2=−1−1.73i

Formula used:

The trigonometric form of the complex number z=a+ib is

z=r(cosθ+isinθ)

Where r=a2+b2 and θ=tan−1(ba)

Calculation:

Graphical representation of roots are given by

EBK PRECALCULUS W/LIMITS, Chapter 6, Problem 127RE

Conclusion:

Thus,the graphical representation of roots are given.

Chapter 6, Problem 126RE

Chapter 6, Problem 128RE

Chapter 6 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E

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