# Polynomial with no roots of unity among its roots

I am reading a paper which states without proof that, for $$r \ge 2$$, the polynomial $$f(X) = X^r - X^{r-1}- X^{r-2} - \ldots - 1$$ has no roots (in $$\Bbb{C}$$) that are roots of unity. After playing around with elementary algebra and getting nowhere, I tried to apply the methods of this note by Keith Conrad on polynomials with roots on the unit circle , but with no success. I suspect there is something very simple that I am missing. Any clues about how to prove this will be very gratefully received.

• Have you seen this MO-post? It says "Proving this implies that $P(x)$ has $n$ roots in $\{z\in \mathbb{C}: |z|<1\}$". GH form MO gives a proof for your claim using Rouché's theorem. Commented Aug 18, 2021 at 12:46
• @DietrichBurde: thanks for the pointer: I did a lot of hunting for posts involving Rouché's theorem, but didn't find that one $\ddot{\frown}$. Commented Aug 18, 2021 at 15:28

Notice that $$g(x)=(x-1)f(x)=x^{r+1}-2x^{r}+1$$ and look at its zeroes on the unit circle (we know $$\alpha=1$$ is a root of $$g$$ but clearly not of $$f$$, we will show there is no other). So let $$g(\alpha)=0$$ and $$|\alpha|=1$$, then $$1=|\alpha^{r+1}-2\alpha^r|=|\alpha|^r|\alpha-2|=|\alpha-2|.$$ Only complex number that satisfies both $$|\alpha|=1$$ and $$|\alpha-2|=1$$ is $$\alpha=1$$ (can be seen either as an intersection of two circles in the complex plane, or solved fully algebraically if needed). So $$g(x)$$ has no other roots on the unit circle than the $$\alpha=1$$, and by the construction and the fact that $$f(1) \neq 0$$, $$f(x)$$ has no roots on the unit circle.