I am self-studying Stein's Complex Analysis text, in which he has as proof for the Fundamental Theorem of Algebra. For some reason, something about it is just not clicking. Here is his proof itself (here $P$ is a non-constant polynomial):
If $P$ has no roots, then $1/P(z)$ is a bounded holomorphic function. To see this, we can of course assume $a_n \neq 0$ and write
$$\frac{P(z)}{z^n} = a_n + \Bigl(\frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n} \Bigr)$$
whenever $z \neq 0$. Since each term in the parentheses goes to 0 as $|z| \rightarrow \infty$ we conclude that there exists $R > 0$ so that if $c = |a_n|/2$, then
$$|P(z)| \geq c|z|^n, \quad \text{whenever } |z| > R$$
In particular, $P$ is bounded from below when $|z| > R$. Since $P$ is continuous and has no roots in the disc $|z| \leq R$, it is bounded from below in that disk as well, thereby proving our claim. By Liouville's theorem we then conclude that $1/P$ is constant. This contradicts our assumption that $P$ is non-constant and proves the corollary.
How does he conclude that there exists $R > 0 $ so that if $c = |a_n|/2$ then $|P(z)| \geq c|z|^n$ for $|z| > R$? Why is he choosing $c$ as he did? Also, how does he use this to deduce that $P$ is bounded from below?
Best Answer
From the expression for $P(z)/z^n$ it follows that $$ \lim\limits_{z\to\infty}\frac{P(z)}{z^n}=a_n\neq 0 $$ Therefore, by definition of limit, by choosing $|z|$ large enough, say $|z|>R$ we can make $\left|\frac{P(z)}{z^n}\right|$ as close to $|a_n|$ as we want, say between $|a_n/2|$ and $|3a_n/2|$, so $$ \left|\frac{P(z)}{z^n}\right|\ge c:=\frac{a_n}{2},\quad |z|>R $$ and, consequently, $|P(z)|\ge cR^n$ if $|z|>R$, so $P$ is bounded below by the positive constant $cR^n$ outside of the ball of radius $R$. Inside, $|P|$ is clearly bounded below, being a positive continuous function on a compact set.