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Let $\omega$ be primitive nineth root of unity. Let $a= \omega^4+\omega^3+\omega^2-\omega-2$. Is there a nice way to show the sum is non zero?

Given some linear combination of powers of primitive $n$-th roots of unity with coefficient from real, Is there a nice way to show whether it is 0 or not?

(One may also assume the coefficients are integers, and maximum power of primitive $n$-th that comes in linear combination $\leq \lfloor{\frac{n}{2}}\rfloor.)$

Cloud JR K
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    Since $\varphi(n)\le\lfloor\frac n2\rfloor$ for lots of $n$, there will be many ways to get $0$ under your conditions. – abiessu Jan 03 '24 at 06:25

3 Answers3

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The degree of the minimal polynomial of $\omega$ over the rationals is $\phi(9)=6$.

In general, for any $f\in\Bbb{Z}[X]$ one has $f(\zeta_n)=0$ if and only if $f$ is divisible by the minimal polynomial of $\zeta_n$, which is the $n$-th cyclotomic polynomial.

Servaes
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Since $$\omega^4+\omega^3+\omega^2-\omega-2=(\omega-1)(\omega+1)(\omega^2+\omega+2),$$ it is enough to show that $\omega^2+\omega+2\neq0$. Since $\cos(\tfrac{2\pi k}{9})>0$ for $k=1,2,$ we are done.

Xander Henderson
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Bob Dobbs
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In this particular case, you can bound the real part of $a$ away from zero by looking at individual terms:

  • the real parts of the terms $\omega^4, \omega^3, -\omega$ are obviously negative, by considering their arguments $8\pi/9, 2\pi/3, 11\pi/9$
  • the real part of $\omega^2$ is less than $+1$;
  • the real part of $-2$ is $-2$

So $Re(a) < 0 + 1 - 2 = -1$ and thus $a \not = 0$.

Michael Lugo
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