# What is the sum of the squares of the 10th roots of unity?

Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?

• The square of a tenth rooot is a fifth root ... – Hagen von Eitzen Jan 9 at 16:19

We have $$q^n-1 =(q-1)(1+q+\ldots+q^{n-1})$$.

Thus if $$\xi$$ is a primitive $$n$$-th root of unity, $$\xi^n=1$$ and so $$1+\xi+\ldots+\xi^{n-1}=0$$ as required.

The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.

Thus you're summing twice the fifth roots.

• I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is? – AnoUser1 Jan 9 at 16:27
• @AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas). – egreg Jan 9 at 16:33
• Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root – AnoUser1 Jan 9 at 16:35
• @AnoUser1 $1=x^{10}=(x^2)^5$. – egreg Jan 9 at 16:36
• I'm still not following sorry. How do we know x^10 is 1? – AnoUser1 Jan 9 at 16:39

square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?

You have $$\sum_{k=0}^9 e^{-ik\pi/10} = 0,$$ and you are asked to compute $$\sum_{k=0}^9 \left(e^{-ik\pi/10}\right)^2 = \sum_{k=0}^9 e^{-ik\pi/5},$$ which is zero as well because it is just traversing the 5-roots twice.

• I don't get you? – AnoUser1 Jan 9 at 16:21
• @AnoUser1 see update – gt6989b Jan 9 at 18:11
• got you. but how are we meant to know what each individual root is, without having an angle? – AnoUser1 Jan 9 at 18:14
• @AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice – gt6989b Jan 9 at 19:59
• I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me – AnoUser1 Jan 10 at 13:25