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I have trouble with the following problem: Let $P(z) = z^n + a_{n-1}z^{n-1}+\ldots +a_0$ and $|P(z)| \leq 1$ if $|z|=1$. Prove that $P(z)=z^n$.

If one tries to apply Rouché's theorem, first look at $g(z)=-z^n$. Then, of course, $|g(z)|=1$ on the unit circle. If $|P(z)| \lt 1$, then $n = |- z^n|_0 = |P(z) - z^n|_0$ by Rouché, and $a_{n-1}z^{n-1}+\ldots +a_0 = 0$ because of the fundamental theorem of algebra.

But now we know that $|tP(z) - g(z)|_0$ with $0 \leq t \lt 1$. I have been trying to apply Rouché's theorem a second time, maybe on another disk.

My Attempt (which is not correct)

Since $|tP(z) - g(z)|_0 = n$ on the unit circle (and everywhere else), we choose $t=\frac{1}{2}$. So $$\tilde{P}(z) = \frac{1}{2}P-z^n =\frac{-z^n}{2}+\frac{a_{n-1}}{2}z^{n-1} + \ldots + \frac{a_0}{2}$$ If we multiply the term, the zeros wont change, so $$2\tilde{P}(z)=-z^n + P(z) - z^n$$ Now, we apply Rouché's theorem: $|2\tilde{P}| \gt |P|$, so $$n = |2\tilde{P}(z)|_0 = |2\tilde{P}(z) + P(z)|_0 = |2(P(z)-z^n)|_0$$ Finally, the $2(P(z)-z^n)$ should have the same zeros as $(P(z)-z^n)$, which was a polynomial of grade $n-1$. But we know that it has $n$ zeros, so because of the fundamental theorem of algebra, it is constant. Therefore zero.

-> There are 2 Erros: firstly: $|2\tilde{P}| \gt |P|$ looks nice but is a baseless claim. secondly, the conclusion. Because Rouche, $(P(z)-z^n)$ should have exactly $n$ zeros, which is just a contradiction

Fabsch
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Aug 03 '24 at 22:09
  • something is off in your approach -- if you are able to apply Rouche then $2(P(z)-z^n)$ has exactly $n$ zeros so it cannot be constant – user8675309 Aug 04 '24 at 03:22
  • my inequality $|2\tilde{P}| \gt |P|$ simply didnt hold – Fabsch Aug 04 '24 at 11:24

2 Answers2

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I am not really sure how to apply Rouche's theorem here, but below I give a simple proof via maximum modulus principle.

Consider another polynomial $Q(z) = 1 + a_{n-1}z + a_{n-2}z^2 + \dots a_{0}z^n$. Clearly, for $z \ne 0$ we have $Q(z) = z^n P(1/z)$. On the unit circle $Q$ inherits the property $|Q(z)| \le 1$ from $P$. But, clearly, $Q(0) = 1$. Thus, by the maximum modulus principle $Q$ is constant, hence, $Q \equiv 1$ and $P(z) = z^n$.

Matsmir
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I suggest observing that if $P(z)\neq z^n$ then say for $z\in \partial B\big(0,2\big)$
$\vert - z^n\vert \gt \vert P(z)\vert\implies \Big(-z^n + P(z)\Big)$ has degree at most $n-1$ and exactly $n$ roots in $B\big(0,2\big)$ per Rouche which is impossible.

The inequality is Suppose $P(z) = a_0+a_1z + \dots + a_nz^n$ is bounded by 1 for $|z|\leq 1$. Show that $|P(z)|\leq |z^n|$ for all $|z^n|\geq 1$

user8675309
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  • i dont quite follow your proof. is $|-z^n|>|P(z)|$ prooven in the link? There, $P(z)$ is bounded by $1$ per definition, which is not the case here – Fabsch Aug 04 '24 at 11:09
  • Your comment "There, () is bounded by 1 per definition, which is not the case here" does not make sense. Please carefully re-read the problem and answers in the link. In the link and this problem we are talking about a degree at most $n$ polynomial that has modulus $\leq 1$ for $z \in\overline B(0,1)$. The link then tells you that $\vert P(z)\vert \leq \vert z\vert^n$ for $\vert z\vert \gt 1$ and the inequality is strict if $P(z)\not\propto z^n$. I proved the result using Rouche; you'll see another answer using $ z^n P(1/z)$ so there are common proof techniques to this page. – user8675309 Aug 04 '24 at 15:44
  • so the linked inequality is: w/ degree at most $n$, if $P(z)$ has modulus at most 1 in the $\overline B(0,1)$ then $\vert P(z)\vert \leq \vert z^n\vert $ for all $z \in \mathbb C-\overline B(0,1)$ and the inequality is strict everywhere- unless $P(z)=\alpha z^n$ for some $\alpha\in S^1$. I show this in the comment to my post in the link; the strictness conditions may be easier to understand via the max modulus in Conrad's answer though I appreciate the enthusiasm for Rouche. Now for this page: in your OP we know that $P(z)$ is monic so $\alpha =1$ is the only case to consider. – user8675309 Aug 05 '24 at 22:24
  • i liked your proof because one applies Rouche twice - earlier in the contradiction for the inequalitiy - ill streamline it and add it to the question when i find the time – Fabsch Aug 06 '24 at 11:30
  • The linked inequality is of considerable independent interest so I would not recommend any combining. Note: the 3rd response in the link used Parseval; you can do a 1 line proof to your question here by direct application of Parseval. So all 3 proof techniques there apply here though Parseval will be the quickest (try it). Also when you ask a question it in the comments and I answer it and you then delete the question afterwards... it makes these comments very hard to follow for 3rd parties. – user8675309 Aug 06 '24 at 16:16
  • i deleted the question because i saw the answer myself, i didnt realize you replied – Fabsch Aug 06 '24 at 18:23