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Find the largest $c\in\mathbb{R}$ (or prove that doesn't exist) such that if $\alpha$ is zero of polynomial $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$ where $a_0,a_1,\dots,a_n\in\mathbb{C}$ and $n\in\mathbb{N}$ and $|a_0|=|a_1|=\dots=|a_n|>0$ then $|\alpha|>c$.

I first tried to factor out some constant in this polynomial. Because of $|a_0|=|a_1|=\dots=|a_n|>0$ we have $$P(x)=k\left({e^{ib_n}x^n+e^{ib_{n-1}}x^{n-1}+\dots+e^{ib_1}x+e^{ib_0}}\right)$$ for some $k\in\mathbb{R}$ and $b_0,b_1,\dots,b_{n-1},b_n\in\mathbb{R}$. Then $$|P(x)|=|k|\cdot\left|{e^{ib_n}x^n+e^{ib_{n-1}}x^{n-1}+\dots+e^{ib_1}x+e^{ib_0}}\right|\le|k|\left({\left|{e^{ib_n}x^n}\right|+\left|{e^{ib_{n-1}}x^{n-1}}\right|+\dots+\left|{e^{ib_1}x}\right|+\left|{e^{ib_0}}\right|}\right)=|k|\dfrac{|x|^{n+1}-1}{|x|-1}$$ What next? Can this inequality help me to solve this? What is the easiest way to solve this problem?

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  • $\begingroup$ If the inequality is strict there can not exist such $c$. $\endgroup$ – pointer Jan 3 '15 at 20:00
  • $\begingroup$ take $n = 1, a_0 = 0.0001, a_1 = -1,$ then $c = 0.0001$ so $c$ can be as small as you want. $\endgroup$ – abel Jan 3 '15 at 20:11
  • $\begingroup$ @abel. If $a_0=0.0001$ then $a_1$ cannot be $-1$ because $|a_0|=|a_1|$ $\endgroup$ – user164524 Jan 3 '15 at 20:25
  • $\begingroup$ ok. i don't know how i made that error. it is clear now. $\endgroup$ – abel Jan 3 '15 at 21:30
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Without loss of generality one can assume that $$ P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+ 1 $$ and $$ |a_1|=\dots=|a_n| = 1 \quad . $$ If $P(\alpha) = 0$ then with $r := |\alpha|$ $$ 1 = | a_n \alpha^n+a_{n-1} \alpha^{n-1}+\dots+a_1 \alpha | \le r^n+ r^{n-1}+\dots + r $$

Define $q(x) := x^n + x^{n-1} + \dots + x - 1$. It is easy to see that $q(1/2) < 0$.

$q$ is strictly increasing on $[0, \infty)$ and $q(r) \ge 0$. It follows that $r > 1/2$, so $P(\alpha) = 0$ implies $ |\alpha| > 1/2$, i.e. $c = 1/2$ is a solution to your problem.

Choosing $$ P(x) = q(x) = \frac{2x - 1 - x^{n+1}}{1-x} $$ shows that for large $n$ a zero arbitrarily close to $1/2$ exists, so $c=1/2$ is the largest possible solution.

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