Roots of $6z^5+5z^4+4z^3+3z^2+2z+1$

Question

By All the zeroes of $p(z)$ lie inside the unit disk, I know there all roots are inside the unit disk. How can I remove the boundary. Wolfram alpha tells me there all there are no solutions on the unit circle. Furthermore, how can I prove the e.g. all roots are inside the circle with radius 0.9? (Basically, this is a transition function of a system, I want to prove this system is stable.)

Answer 1

For this particular polynomial, you can simply note that $6z^5+5z^4+4z^3+3z^2+2z+1=0$ implies $6\overline z^5+5\overline z^4+4\overline z^3+3\overline z^2+2\overline z+1=0$. Now if $z\overline z=1$, we would have $6+5z+4z^2+3z^3+2z^4+z^5=0$ as well, hence

$$7(z^5+z^4+z^3+z^2+z+1)=0$$

meaning any root on the unit circle would necessarily be a (nontrivial) sixth root of unity. But those roots clearly are not roots of the original polynomial.

Answer 2

Multiply the given polynomial by $(1-z)^2$ to get $f(z)=1-7z^6+6z^7$. This is the characteristic polynomial of the matrix $$P=\begin{pmatrix}0&0&0&0&0&0&-\tfrac{1}{6}\\ 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&\tfrac{7}{6}\\\end{pmatrix}$$ Let $M$ be the maximum modulus of the roots of your polynomial. We have $$ 5M^n\geq \left|\text{Tr}(P^n)-2\right| $$ so $M\geq \frac{2}{3}$ by considering $n=13$.
Let us consider $g(z)=z^7-7z+6$, characteristic polynomial of the matrix $$Q=\begin{pmatrix}0&0&0&0&0&0&-6\\ 1&0&0&0&0&0&7\\ 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&0\\\end{pmatrix}$$ and let $N$ be the maximum modulus of the roots of $g(z)$. By a similar argument, $N\geq\frac{4}{3}$, so all the roots of the given polynomial lie in the annulus $$\frac{2}{3}\leq |z|\leq\frac{3}{4}.$$