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The title says pretty much everything:

We have $z\in\mathbb{C}$ such that $\|z\|=1$ and a monic polynomial $p\in\mathbb{Z}[x]$ such that $p(z)=0$.
Is it true that for some $n\in\mathbb{N}$, $z^n=1$?

My thoughts about it: assume that $z=e^{\pi i\alpha}$ and $z$ is not a root of unity, i.e. $\alpha\not\in\mathbb{Q}$. Then $\alpha$ has to be a trascendental number: otherwise, by the Gelfond-Schneider theorem, $z$ is a trascendental number. But obviously this is far from closing the question: for instance, $\frac{3+4i}{5}$ is a complex number with unit modulus, and $\frac{1}{\pi}\arctan\left(\frac{4}{3}\right)$ is a trascendental number, but there is no monic polynomial with integer coefficients having $\frac{3+4i}{5}$ as a root. So, what is the good way to go for proving the claim above, since I believe it is true? Counter-examples are welcome as well, obviously.

Jack D'Aurizio
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    If $z^\alpha = 1$ where $\alpha$ is irrational and $z \neq 1$, it seems pretty hard to believe that an INTEGER combination of non-negative powers of $z$ could possibly be 0. However proof eludes me. Nice question! – user2566092 Sep 16 '15 at 15:36
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    The answer is no, and the question has been discussed here: http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity – Manny Reyes Sep 16 '15 at 15:47
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    @MannyReyes: thank you so much, I am closing the question. – Jack D'Aurizio Sep 16 '15 at 15:50
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    @Manny: Good job digging up that duplicate (using the link elsewhere)! Jack, here I gave $u=\sqrt2-1+i\sqrt{2\sqrt2-2}$ as a specific example of an algebraic integer on the unit circle that is not a root of unity. Finding such things is not too hard. Pick a real part that is an algebraic integer $\in(0,1)$, and find a matching imaginary part. With any luck the minimal polynomial will have roots not on the unit circle. That then immediately implies that we don't have a root of unity for otherwise all the conjugates would have absolute value $1$. – Jyrki Lahtonen Sep 16 '15 at 19:57
  • @JyrkiLahtonen: thanks for the explicit counter-example, too. – Jack D'Aurizio Sep 16 '15 at 20:02

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