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

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