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Let $\mathbb D = \{z \in \mathbb C : \vert z \vert < 1\}$ and $\mathcal P$ the set of all polynomials with

$$ P(z) = 1 + \sum_{i=1}^n a_i z^i, \ a_n \neq 0,\ n \in \mathbb N.$$

Then for every $P \in \mathcal P$ it holds that $\max_{z \in \partial \mathbb D} \vert P(z) \vert > 1$.

Let $P \in \mathcal P$. Using the maximum principle I achieved

$$ \max_{z \in \partial \mathbb D} \vert P(z) \vert = \max_{z \in \overline{\mathbb D}} \vert P(z) \vert \geq \vert P(0) \vert = 1.$$

Further I know that $$ \left\vert \frac{P(z)}{a_n z^n} \right\vert \to 1 \quad \text{for} \quad \vert z \vert \to \infty.$$

Hence there is a $r > 0$ with $\vert P(z) \vert \geq \frac 1 2 \vert a_n \vert \vert z \vert^n$ for all $\vert z \vert \geq r$. Hence there exists a $R > 0$ with $$\vert P(z) \vert > \vert P(0) \vert = 1 \quad \text{for all} \quad \vert z \vert = R.$$

But how can I deduce that $R = 1$ or get that $\max_{z \in \partial \mathbb D} \vert P(z) \vert > 1$ from the first equation? I think I am missing some details. Thanks for your help :)

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  • $\begingroup$ Do you recall the open mapping theorem? (Or the maximum principle in its local form?) $\endgroup$ – Daniel Fischer Oct 25 '16 at 19:45
  • $\begingroup$ Sorry, there was a "1" missing. I edited it. Thanks for noticing :) $\endgroup$ – Yaddle Oct 25 '16 at 19:45
  • $\begingroup$ Presumably you mean to have $a_i z^i$? $\endgroup$ – copper.hat Oct 25 '16 at 19:46
  • $\begingroup$ Sorry, another typo. @DanielFischer, I really don't see how this could help me :/ $\endgroup$ – Yaddle Oct 25 '16 at 19:57
  • $\begingroup$ Is $P$ a constant function? $\endgroup$ – Daniel Fischer Oct 25 '16 at 19:59
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There are many ways of doing this (open mapping theorem is a good one to have in your toolbox), here is an elementary way:

We can write $P(z) = 1+z^k g(z)$, where $g$ is a polynomial and $g(0) \neq 0$.

Let $g(0) = r e^{i \theta}$.

Let $\phi(t) = |P(t e^{-i { \theta \over k}})|$, with $t \ge 0$ and note that \begin{eqnarray} \phi(t) &=& | 1 + t^k {\overline{g(0)} \over |g(0)|} g(t e^{-i { \theta \over k}}) | \\ &=& | 1 + t^k {\overline{g(0)} \over |g(0)|} (g(0)+g(t e^{-i { \theta \over k}})-g(0) ) | \\ &=& | 1 + t^k |g(0)| + t^k {\overline{g(0)} \over |g(0)|} ( g(t e^{-i { \theta \over k}})-g(0) ) | \\ &\ge& 1 + t^k |g(0)| - t^k |{\overline{g(0)} \over |g(0)|}| |g(t e^{-i { \theta \over k}})-g(0)| \end{eqnarray} By continuity, we can find a $\delta > 0$ such that if $t \in [0,\delta)$ then $|{\overline{g(0)} \over |g(0)|}| |g(t e^{-i { \theta \over k}})-g(0)| < {1 \over 2} |g(0)|$, and so $\phi(t) \ge 1 + {1 \over 2} t^k |g(0)| $. In particular, we can choose $t$ such that $t e^{-i { \theta \over k}}$ lies in the unit disc, and so $|P(t e^{-i { \theta \over k}})| > 1$.

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  • $\begingroup$ Thank you! But now I'm curious how to sie the open mapping theorem. $\endgroup$ – Yaddle Oct 26 '16 at 7:11
  • $\begingroup$ The open mapping theorem says that a non constant analytic function maps open sets to open sets. Hence $P(B(0,1))$ is an open set containing $1$, and from this we see that it must contain $1+ \epsilon$ for some $\epsilon >0$. $\endgroup$ – copper.hat Oct 26 '16 at 7:37
  • $\begingroup$ Oh, that is much easier. Thank you :) $\endgroup$ – Yaddle Oct 26 '16 at 7:39
  • $\begingroup$ The open mapping theorem is a very powerful result. $\endgroup$ – copper.hat Oct 26 '16 at 7:41
  • $\begingroup$ Yes, i see. I will keep it in mind! $\endgroup$ – Yaddle Oct 26 '16 at 7:43

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