Solving 6th Roots of Unity Problems

[bosox097]

How do you do these two problems?

1. Find the sum of the 6th roots of unity.
2. Find the product of the 6th roots of unity.

 

[Data]

The real 2nth roots of unity, for any natural n, are -1 and 1. If a complex number is an mth root of unity (for any m) then its complex conjugate is as well. If z \in \mathbb{C} then z\overline{z} = |z|^2. Thus the product of the 2nth roots of unity (for any n) is -1.

Furthermore, -z is a 2nth root of unity whenever z is. Thus the sum of the 2nth roots of unity (for any n) is 0.

I hope this wasn't a homework problem!

 

[robert Ihnot]

You should look at symetric functions. Take the cubic: (x-a)(x-b)(x-c)=0. Then if this is multiplied out, we get

X^3-X^2(a+b+c)+X(ab+ac+bc)-(abc) = 0.

 

[mathwonk]

look at the vertices of a regular hexgon and think of vector addition, and then use group theory.

 

[HallsofIvy]

(x-a)(x-b)= x²- (a+ b)x+ ab
(x-a)(x-b)(x-c)= x³- (a+ b+ c)x²+ (ab+bc+ ac)x- abc
(x-a)(x-b)(x-c)(x-d)= x⁴- (a+ b+ c+ d)x³+ (ab+ac+ ad+ bc+ bd+ cd)x²- (abc+ acd+ bcd)x+ abcd

Do you see the pattern?

Even more simply: the n^th roots of unity are equally spaced around the unit circle in the complex plane. What does symmetry tell you about their sum?