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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?

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

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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.

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  • $\begingroup$ 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? $\endgroup$ – AnoUser1 Jan 9 at 16:27
  • $\begingroup$ @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). $\endgroup$ – egreg Jan 9 at 16:33
  • $\begingroup$ 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 $\endgroup$ – AnoUser1 Jan 9 at 16:35
  • $\begingroup$ @AnoUser1 $1=x^{10}=(x^2)^5$. $\endgroup$ – egreg Jan 9 at 16:36
  • $\begingroup$ I'm still not following sorry. How do we know x^10 is 1? $\endgroup$ – AnoUser1 Jan 9 at 16:39
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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.

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  • $\begingroup$ I don't get you? $\endgroup$ – AnoUser1 Jan 9 at 16:21
  • $\begingroup$ @AnoUser1 see update $\endgroup$ – gt6989b Jan 9 at 18:11
  • $\begingroup$ got you. but how are we meant to know what each individual root is, without having an angle? $\endgroup$ – AnoUser1 Jan 9 at 18:14
  • $\begingroup$ @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 $\endgroup$ – gt6989b Jan 9 at 19:59
  • $\begingroup$ 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 $\endgroup$ – AnoUser1 Jan 10 at 13:25

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