Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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

2 Answers 2

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

share|cite|improve this answer
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 $$\|z\|\leq 1+\frac{7}{10(n-1)}.$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.