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I am trying to prove that if $p(z)=z^n+a_{n-1}z^{n-1}+\dots+a_0$ then all the zeros lie in a circle of radius $R= \max\{1,|a_0|+|a_1|+|a_2|+\dots+|a_{n-1}|\}$

I'm trying to use induction and perhaps Liouville's theorem, but I'm stuck. Could anyone help me please?

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Suppose that $|z_0|$ is larger than $\max(1, |a_0|+\cdots +|a_{n-1}|)$. Estimate the size of $a_{n-1}z_0^{n-1}+ \cdots + a_0$ and compare that to $|z_0^n|$. You should be able to show that $z_0^n$ is greater in absolute value, so $p(z_0)$ can't be zero.

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  • $\begingroup$ why would that imply that p(z_0) can't be zero? $\endgroup$ Commented May 1, 2016 at 0:16
  • $\begingroup$ $p(z_0)$ is the sum of two pieces, $z_0^n$ and $a_{n−1}z_0^{n−1}+⋯+a_0$. If one of these is bigger (in absolute value) than the other, then they can't possibly add to 0. $\endgroup$ Commented May 1, 2016 at 0:18
  • $\begingroup$ OK, I think I get it, I can apply the inductive hypothesis to the second piece and that does the trick, right? $\endgroup$ Commented May 1, 2016 at 0:24
  • $\begingroup$ You don't need induction. $\endgroup$ Commented May 1, 2016 at 0:33
  • $\begingroup$ Right, since $|z_0|$ is larger than R then $|z_0^n|$ is larger than R^n. On the other hand I can also write the size of the second part in terms of R and then some rewriting will yield the inequalities. $\endgroup$ Commented May 1, 2016 at 0:46

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