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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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    |z|<1 is not a circle2013-05-07

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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|. The base case $n=1$ is easy, since $|a_0z|=a_0|z|. 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}| 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 wrong2012-09-08