Question about proof of FTA I read a short proof of the Fundamental Theorem of Algebra, as follows:
Assume $p(z)$ is a nonconstant polynomial with no roots. Then $1/p$ is an analytic function on $\mathbb{C}$. Also, $1/p \to 0$ as $z \to \infty$, so $p$ is bounded. By Liouville's theorem, any bounded analytic function is constant, which is a contradiction.
My issue is I don't understand why $1/p$ is bounded. Could someone please explain this? 

Absolute value of polynomial tends to infinity for $\left|z\right|\to\infty$. That is, for each $M>0$, there exists $R>0$ such that for $\left|z\right|>R$ we have $\left|p(z)\right|>M$. Take sufficiently large closed disk, so that $\left|p(z)\right|>1$ for $z$ outside the disk. The disk is compact, so it's image by $\left|p(z)\right|$ is compact, hence closed. It does not contain 0, so it's bounded away from 0, say, by $a>0$. Thus, 
$$
\left|1/p(z)\right|<1/a
$$
inside the disk, and $\left|1/p(z)\right|<1$ outside.
