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Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.

I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can't find a pattern.

Jean Marie
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    Hint: if $z^{13}=1$, then $z^{-1}=z^{12}$, $z^{-3}=z^{10}$, and ... – Arturo Magidin Oct 24 '20 at 03:57
  • @ArturoMagidin So since $z$ is on the unit circle we have $z + z^{-1} = 2 Re(z),$ and the problem becomes $2(Re(z) + Re(z^3) + Re(z^4))$. But can it be simplified further than this, or do we have to plug in the 13 values of $z$ here? – Blue Oct 24 '20 at 04:37

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As stated in comments, for a fixed $z=e^{i\theta}$ the given expression is equal to $2(\cos\theta+\cos3\theta+\cos4\theta)$. As $z$ runs through its $13$ possible values, so does each term ($1,3,4$ are all coprime to $13$) and the sum over all possibilities is $$6\sum_{k=0}^{12}\cos\frac{2\pi k}{13}$$ but the summation is well-known to evaluate to zero, so the final answer is $0$.

Parcly Taxel
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Since $13$ is prime, the map $z\mapsto z^k$ is a permutation of the roots of $z^{13}-1$, for $k=1,2,3,\dots,12$. Thus, the answer is $6$ times the sum of the roots , and this is $0$ by Vieta.

saulspatz
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If $z^{13} = 1$ and $z \not= 1$, then $z = e^{2\pi i \frac{k}{13}}$ where $(k, 13) = 1$.

Since $z^{12} = z^{-1}$, $z^{10} = z^{-3}$ and $z^{9} = z^{-4}$ we have $z+z^3 + z^4 + z^9 + z^{10}+z^{12} = 2(\cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}))$ where $z = e^{2\pi i \frac{k}{13}}$.

The total is $2\sum_{k=1}^{12} \cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}) = -6$.

If $z = 1$ is included, the total is $0$.

DavidW
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