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Suppose that $0 < a_0 \le a_1 \le \dots \le a_n$. Prove that the equation $$P(z) = a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0$$ has no root in the circle $|z| < 1$.

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See also: math.stackexchange.com/q/188039/5363 for a related result. –  t.b. Sep 8 '12 at 10:14
$|z|<1$ is not a circle –  El Angel Exterminador May 7 '13 at 15:30

1 Answer 1

We want to show that $a_0z^n+\cdots+a_{n-1}z\neq -a_n$ for $|z|<1$. In fact, by induction we can prove something stronger: that $|a_0z^n+\cdots+a_{n-1}z|<a_n$. The base case $n=1$ is easy, since $|a_0z|=a_0|z|<a_0\leq a_1$. If we have the $n-1$ case, then $$|a_0z^n+\cdots+a_{n-1}z|=|a_0z^{n-1}+\cdots+a_{n-1}||z|<|a_0z^{n-1}+\cdots+a_{n-1}|<a_{n-1}\leq a_n$$ and so the theorem follows by induction.

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I think $| a_0z^{n-1} + \cdots + a_{n-1} | < a_{n-1}$ might be wrong –  user39775 Sep 8 '12 at 9:59

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