SOLUTION: If 1 and w are 2 of the 5 roots of (w)^5 = 1, then prove the following: a) (w)^2, (w)^3, and (w)^4 are the remaining roots of (w)^5 = 1 b) 1 + w + (w)^2 + (w)^3 + (w)^4 = 0

Algebra ->  Exponents -> SOLUTION: If 1 and w are 2 of the 5 roots of (w)^5 = 1, then prove the following: a) (w)^2, (w)^3, and (w)^4 are the remaining roots of (w)^5 = 1 b) 1 + w + (w)^2 + (w)^3 + (w)^4 = 0       Log On


   

Question 1181163: If 1 and w are 2 of the 5 roots of (w)^5 = 1, then prove the following:
a) (w)^2, (w)^3, and (w)^4 are the remaining roots of (w)^5 = 1
b) 1 + w + (w)^2 + (w)^3 + (w)^4 = 0
c) (w)^(5n) = 1 for any integer n
Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
If 1 and w are 2 of the 5 roots of (w)^5 = 1, then prove the following:
a) (w)^2, (w)^3, and (w)^4 are the remaining roots of (w)^5 = 1
where cis(θ) = cos(θ)+isin(θ),
The 5 fifth roots of 1 are 



%28cis%2872%5Eo%29%5E%22%22%29%5E2=cis%282%2A72%5Eo%29=cis%28144%5Eo%29
%28cis%28144%5Eo%29%5E%22%22%29%5E2=cis%282%2A144%5Eo%29=cis%28288%5Eo%29


%28cis%2872%5Eo%29%5E%22%22%29%5E3=cis%283%2A72%5Eo%29=cis%28216%5Eo%29



%28cis%2872%5Eo%29%5E%22%22%29%5E4=cis%284%2A72%5Eo%29=cis%28288%5Eo%29
b) 1 + w + (w)^2 + (w)^3 + (w)^4 = 0, provided w ≠ 1
w%5E5=1
w%5E5-1=0
%28w-1%29%28w%5E4%2Bw%5E3%2Bw%5E2%2Bw%2B1%29=0
Since w ≠ 1, w-1 ≠ 0, so
w%5E4%2Bw%5E3%2Bw%5E2%2Bw%2B1+=+0
1%2Bw%2Bw%5E2%2Bw%5E3%2Bw%5E4=0
c) (w)^5n = 1 for any integer n
w%5E5+=+1
Raise both sides to the 5th power:
%28w%5E5%29%5En=1%5En
w%5E%285n%29=1

Edwin

Answer by mccravyedwin(407) About Me  (Show Source):