Understanding Rudin's reasoning on why $|P(re^{i\theta})| > |P(0)|$

Question

I am taking a stroll through Chapter 10 of Big Rudin, and I am stuck on the fundamental theorem of algebra:

10.25 Theorem If $n$ is a positive integer and $$ P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0, $$ where $a_0, \ldots , a_{n-1}$ are complex numbers, then $P$ has precisely $n$ zeros in the plane.

Proof. Choose $r > 1 + 2|a_0| + |a_1| + \cdots + |a_{n - 1}|$. Then $|P(re^{i\theta})| > |P(0)|$ for $0\le \theta\le 2\pi$. ...

Now, using the fact that $P(z)\sim z^n$ for large enough $z$'s (more precisely, $P(z)/z^n\to 1$ as $|z|\to\infty$), I can show an existence of an $R > 0$ so that $|P(z)| > |P(0)|$ whenever $|z| > R$, and this suffices for the rest of the proof. However, I wish to understand Rudin's reasoning as well on why $|P(re^{i\theta})| > |P(0)|$. Any hints?

Answer 1

We have $$\vert P(z)\vert \geq \vert z\vert^n - (\vert a_{n-1} \vert z\vert^{n-1} +\dots \vert a_0\vert).$$ For $\vert z\vert\geq 1$ we have $\vert z\vert^{n-1}\geq \vert z\vert^k$ for $0\leq k \leq n-1$. Thus, we get for $\vert z\vert \geq 1$ \begin{align*} \vert P(z)\vert &\geq (\vert z\vert -\vert a_{n-1} \vert - \dots - \vert a_0\vert)\vert z\vert^{n-1}. \end{align*} Now if the first factor is positive, i.e. if we assume $\vert z\vert >\vert a_{n-1}\vert+\dots+\vert a_0\vert$, then the later expression becomes smaller when making $\vert z\vert$ smaller. Thus, for $\vert z\vert\geq 1+\vert a_{n-1}\vert +\dots + \vert a_0\vert$ (note that $\vert z\vert \geq \max\{1, \vert a_{n-1}\vert +\dots +\vert a_0\vert$ would be good enough)

\begin{align*} \vert P(z)\vert &\geq (\vert z\vert -\vert a_{n-1} \vert - \dots - \vert a_0\vert)\vert z\vert^{n-1}\\ &\geq \vert z\vert -\vert a_{n-1} \vert - \dots - \vert a_0\vert. \end{align*} Note that $P(0)=a_0$ and conclude.

Answer 2

Suppose that $z=re^{i\theta}$, for some $r>0$ and some $\theta\in\Bbb R$. Then $|z|=r$ and\begin{align}\left|z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\right|&\geqslant|z^n|-|a_{n-1}||z^{n-1}|-\cdots-|a_1||z|-|a_0|\\&=r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r-|a_0|.\end{align}So, it is enough to prove that$$r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r-|a_0|>|a_0|=|P(0)|.$$In other words, it is enough to prove that$$r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r-2|a_0|>0.$$But\begin{align}r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r-2|a_0|&>r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r-2|a_0|-r^{n-1}\\&>r^n-|a_{n-1}|r^{n-1}-\cdots-|a_1|r^{n-1}-2|a_0|r^{n-1}-r^{n-1}\\&=r^{n-1}\left(r-|a_{n-1}|-|a_{n-2}|-\cdots-2|a_0|-1\right)\\&>0.\end{align}