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.