## Position of roots

The following result is quite useful in irreductibility problems for polynomials.

Prove that if $a_0 \geq a_1\geq ... \geq a_n >0$ then all the roots $z$ of $p(x)=a_nx^n+...+a_1x+a_0$ satisfy $|z| >1$ or $z$ a root of unity of some order.

Solution:

If all $a_i$ are equal, then the polynomial is a scalar multiplied with $1+x+...+x^n$, and all roots satisfy $x^{n+1}=1$. In the following, suppose there exist two numbers $a_i$ which are not equal.
Suppose $p(z)=0$ and $|z|\leq 1$ (obviously $z \neq 0$ since $p(0)=a_0 >0$). Then we have $-a_0=a_1z+...+a_n z^n$. Multiplying this last relation with $z-1$ which is not zero, since 1 is not a root of $p$, we get $a_0=z(a_0-a_1)+...+z^n(a_{n-1}-a_n)+a_{n+1}z^{n+1}$. Taking modules and using the modulus inequality, we get $a_0\leq |z|(a_0-a_1)+...+|z|^n(a_{n-1}-a_n)+a_{n+1}|z|^{n+1} \leq a_0$, so we get the equality. Therefore, any two non-zero terms are positive real scalar multiple of each other. Since there exists $k$ such that $a_k \neq a_{k+1}$ and $a_{n+1}\neq 0$, we can find $\mu>0$ such that $(a_k-a_{k+1})z^{k+1}=\mu a_{n+1}z^{n+1}$ which yelds $z\in \left\{\displaystyle \varepsilon^q\sqrt[n-k]{\frac{a_k-a_{k+1}}{\mu a_{n+1}}}: \ q=0..n-k-1,\ \varepsilon=\cos\frac{2\pi}{n-k}+i\sin \frac{2 \pi }{n-k}\right\}$. Therefore, a root $z$ can be of modulus strictly greater than 1, or a positive scalar multiple of a root of unity. Since assuming $|z|<1$ we get a contradiction ( we couldn’t have equality in the triangle inequality ), we see that if $|z|=1$ then $z$ is a root of unity of some order. $\Box$