# roots of a polynomial inside a circle

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.

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It's a classic situation for Rouché's theorem. –  Daniel Fischer Jul 22 '14 at 17:44
oh sorry forgot to mention. I am supposed to show this without the theorem –  ciao Jul 22 '14 at 17:54

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 $\|z\|=x\leq 1+\frac 1{n-1}$.

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excellent! Could you clarify the $(1 + (x-1))^{n} > 1 + n(x-1)$ inequality? –  ciao Jul 22 '14 at 19:17
oh the Bernoulli inequality...thanks –  ciao Jul 22 '14 at 19:23
Don't really need to think in terms of Bernoulli since $x-1>0$ and $n$ is an integer, and, for $a>0$, binomial theorem shows $(1+a)^n> 1+\binom{n}{1}a = 1+na$. Bernoulli is only "interesting" because it is true for $a\geq -1$... –  Thomas Andrews Jul 22 '14 at 19:28
Thomas, i am trying to apply the same principle to the roots of $1 + nz + z^{n}$ and to show that they lie inside $\|z\| = 1 + \frac{2}{(n-1)}.$ So for a real number outside of that range, i get $x^{n} > x + 2$ for $n > 2.$ Now looking at the equation and setting $\|z\| = x,$ i get $x^{n} \leq 1 + nx.$ My question is how should i relate these two results to get the contradiction? THanks –  ciao Jul 22 '14 at 20:36
Doesn't seem to work in that case. @ciao –  Thomas Andrews Jul 22 '14 at 20:44

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 $$\|z\|\leq 1+\frac{7}{10(n-1)}.$$

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That's not an obvious application of Bernoulli. Bernoulli, at least as defined on Wikipedia, is $(1+x)^n\geq 1+xn$ for $x\geq -1$. –  Thomas Andrews Jul 22 '14 at 18:46
@ThomasAndrews: all right, it is a consequence of the fact that $a_n=\left(1+\frac{1}{n}\right)^{n+1/2}$ decreases towards $e$. I used to call it "Bernoulli inequality" anyway, but probably I am making an improper use of the words, thank you for pointing that out. –  Jack D'Aurizio Jul 22 '14 at 19:06
Ah, yes, I was unaware of that theorem. :) –  Thomas Andrews Jul 22 '14 at 19:11