Let $n\geq 2$, and $z=e^{\frac{2\pi i}{n}}$. Then, for $1\leq m\leq n-1$ we have the identity: $$ \sum_{j=0}^{n-1}z^{jm}=0. $$ Considering a proof of the identity, we can use an idea about periodicity from the unit circle. I am wondering if there is another way to prove it.
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3Please explain the proof you have in mind in more detail. We can't give you an alternative proof if we don't know what your original proof is. – Ben Grossmann Jan 20 '20 at 00:22
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Yes, you're right. Briefly speaking, the proof is in this way. For $gcd(m,n)=1$, we can verify that the left-hand side is $z^0+z^1+\cdots+z^{n-1}$. Let $gcd(m,n)=d$. Then, $z^0+z^m+\cdots+z^{\frac{n}{d}m}$ appears $d$ times in the left-hand side. – kswim Jan 20 '20 at 01:37
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1Does this answer your question? Proof that sum of complex unit roots is zero – DanielV Jan 20 '20 at 01:37
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Not exactly the same, I think. – kswim Jan 20 '20 at 01:42
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Note that the sum is a geometric sum; we get $$(1-z^m)\sum_{j=0}^{n-1}z^{jm}=\sum_{j=0}^{n-1}z^{jm}-\sum_{j=1}^nz^{jm} \stackrel{\ast}{=}z^0-z^n=0,$$ where $\ast$ holds because most terms of the two sums cancel out. Clearly $1-z^m\neq0$ so then the sum itself must equal $0$.
KnatDaaz
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