SOLUTION: (a) Find the sixth roots of 1. (b) Find the fifth roots of 32.

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Question 1183762: (a) Find the sixth roots of 1.
(b) Find the fifth roots of 32.

Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the first ia -1 and 1 if you want the sixth root. If you want the six rootS, they are -1 with multiplicity 3 and 1 with multiplicity 3. I think you want -1, 1 with this question, but the wording matters.
the second is 2.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The answer from the other tutor is incomplete/incorrect. There are 6 6th roots of 1 (or any other number); there are 5 5th roots of 32 (or any other number). The response from the other tutor only shows the real roots -- 2 of the 6 6th roots of 1, and 1 of the 5 5th roots of 32.

De Moivre's Theorem solves this kind of problem easily.

Given a number represented in trigonometric form

N=a%2Acis%28theta%29 or N=a%2Ae%5E%28i%2Atheta%29,

de Moivre's Theorem says that the "primary" n-th root is

a%5E%281%2Fn%29%2Acis%28theta%2Fn%29

and the other n-th roots have the same magnitude as the primary root and are distributed around the Argand plane at intervals of 2pi%2Fn

(a) The 6 6th roots of 1

1+=+1%2Acis%280%29

The primary root is

1%5E%281%2F6%29%2Acis%280%2F6%29+=+1%2Acis%280%29+=+1

The other 6th roots of 1 are distributed about the Argand plane at intervals of 2pi%2F6=pi%2F3; so the 6 6th roots of 1 are

cis(0) = 1
cis(pi/3) = 1/2+i*sqrt(3)/2
cis(2pi/3) = -1/2+i*sqrt(3)/2
cis(pi) = -1
cis(4pi/3) = -1/2-i*sqrt(3)/2
cis(5pi/3) = 1/2-i*sqrt(3)/2

(b) The 5 5th roots of 32

The primary root is

32%5E%281%2F5%29%2Acis%280%2F5%29+=+2%2Acis%280%29+=+2

The other 5th roots of 1 are distributed about the Argand plane at intervals of 2*cis(pi/5); so the 5 5th roots of 32 are:

2*cis(0) = 2
2*cis(pi/5)
2*cis(2pi/5)
2*cis(3pi/5)
2*cis(4pi/5)