roots of a polynomial inside a circle

Question

I am asked to show that for $n$ larger or equal to $2,$ the roots of $1 + z + z^{n}$ lie inside the circle $\|z\| = 1 + \frac{1}{n-1}$

Attempt1: Induction for the case $n = 2,$ the roots of $1 + z + z^{2}$ lie inside the circle of radius $2.$ Now we consider $1 + z + z^{n+1},$ if i can factor this into $(1 + z + z^{n})q(x),$ then i assume the answer will follow from the roots of $q,$ but i can't see the factorization.

Thanks in advance.

Answer 1

For real $x>1+\frac{1}{n-1}$ then $$x^n = (1+(x-1))^n> 1+n(x-1) = x+(n-1)(x-1)> x+1$$

Now if $z^n+z+1=0$ then $\|z\|^n = \|z+1\| \leq \|z\|+1$. So, with $x=\|z\|$, we have $x^n\leq x+1$, so $\|z\|=x\leq 1+\frac 1{n-1}$.

Answer 2

Assume that a root $w$ of your polynomial lies outside the disk $\|z\|\leq \frac{n}{n-1}$. Then: $$ \|w^n\| = \|w+1\|, \tag{1}$$ but, due to the following stronger version of the Bernoulli inequality, the LHS is greater than: $$\left(1+\frac{1}{n-1}\right)^n \geq e\left(1+\frac{1}{2n-1}\right),$$ while the RHS is at most $2+\frac{1}{n-1}$, so the equality in $(1)$ cannot hold, contradiction. By the same way you can also prove a little tighter bound, i.e. that all the roots of your polynomial lies inside the disk $$\|z\|\leq 1+\frac{7}{10(n-1)}.$$