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I'm working on the following exercise:

"Show that if $p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$, then there must be at least on point $z$ with $|z|=1$ such that $|p(z)|\geq 1$."

They provide the following hint.

"If $|p(z)|<1$ for all $z$ with $|z|=1$,how many zeros does $a_{n-1}z^{n-1}+\cdots+a_1z+a_0$?"

The first thing that came to mind was a proof by contradiction. So I have have the following proof.

Proof: Assume, to the contrary, that $|p(z)|<1$ for all $z$ with $|z|=1$. Let $q(z)=p(z)-z^n$ and $f(z)=-z^n$. Then, \begin{align*} |q(z)-f(z)|&=|p(z)-z^n-(-z^n)|\\ &=|p(z)-z^n+z^n|\\ &=|p(z)|\\ &<1\\ &=|f(z)| \end{align*} Thus, by Rouche's theorem, we obtain that $f(z)$ and $q(z)$ have the same number of zeros counting multiplicities, inside the unit disk. But, $f(z)$ has a zero at $z=0$ of order $n$, so $q(z)$ has $n$ zeros also. This is a contradiction because $q(z)$ is a polynomial of degree $n-1$, and by the Fundamental Theorem of Algebra, we know that it must have $n-1$ zeros, counting multiplicity. Therefore, the result holds. $\blacksquare$

I feel like there is something wrong with the proof. One thing that has me worried is that Rouche's theorem requires the closed curve $\gamma$ passes through no zeros or poles. Since polynomials are entire, I am not worried about poles, but what about the zeros? Is it possible that $q(z)$ has a zero on the boundary of the unit circle and I am breaking the conditions for Rouche's theorem? Any insight is appreciated. Thanks in advance.

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    $\begingroup$ If $|z|=1$, then $|q(z)|\geq |z|^n-|p(z)|=1-|p(z)|>0$. $\endgroup$ Commented Apr 10, 2016 at 11:38
  • $\begingroup$ What a re you getting at @ChrisPanos? $\endgroup$
    – iRubeeeeen
    Commented Apr 10, 2016 at 11:48
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    $\begingroup$ In your proof you assume that $|p(z)|<1$ and you are worried if $q(z)$ has a zero on the unit circle. But $q(z)$ does not have a zero on the unit circle, since $|q(z)|>0$ on the unit circle. $\endgroup$ Commented Apr 10, 2016 at 11:54
  • $\begingroup$ Ahhhhh! I got you now. I just need to add that to the proof to make it more precise. Thanks a lot @ChrisPanos $\endgroup$
    – iRubeeeeen
    Commented Apr 10, 2016 at 12:13

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Assume, to the contrary, that $|p(z)|<1$ for all $z$ on the unit circle. Let $q(z)=p(z)-z^n$. Then $q(z)$ is a polynomial of degree at most $n-1$. Let $f(z)=-z^n$ and observe that for all $z$ with $|z|=1$, $|f(z)|=1$ and $$|q(z)|=|p(z)-z^n|\geq 1-|p(z)|>0,$$ meaning that $q(z)$ does not vanish on the unit circle. Also, since $q(z)$ is entire, there are no poles to agonize over. Furthermore, we get that \begin{align*} |q(z)-f(z)|&=|p(z)-z^n-(-z^n)|\\ &=|p(z)|\\ &<|f(z)|. \end{align*} By Rouche's theorem, we get that $f(z)$ and $q(z)$ have the same number zeros inside the unit disk. But since $f(z)$ has a zero of order $n$ at the origin, then $q(z)$ has $n$ zeros inside the unit circle. This contradicts the fundamental theorem of algebra because $q(z)$, being of degree at most $n-1$, has at most $n-1$ zeros. Therefore, the result holds.

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