Complex Analysis: Liouville's Theorem and Unbounded Entire Functions I'm working on a problem from Complex Variables by Taylor, in the section about Liouville's Theorem: Show that if $f$ is an entire function and $\lim_{z \rightarrow \infty} f(z) = \infty$, then $f$ must have a zero somewhere in $\mathbb{C}$. 
I can see this amounts to showing that $f(z)$ is a polynomial and then use the Fundamental Theorem of Algebra, and the problem also states to use the previous exercise, the result of which is that if $\lim_{z \rightarrow \infty} f(z) = \infty$, then $\lim_{z \rightarrow \infty} \frac{1}{f(z)} = 0$. I can't really see how to get started except for maybe taking the power series expansion of $f$? 
Any advice?
 A: By contradiction, assume that $f$ does not have a zero in $\mathbb{C}$. Since $\lim_{z\to\infty}f(z)=\infty$, 
$$|f(z)|\geq C_1\mbox{ in }\mathbb{C}-B_R(0)$$
for some constant $C_1>0$. On the other hand, since $f$ does not have a zero in $\mathbb{C}$, 
$f$ does not have a zero in $\overline{B_R(0)}$. Since $\overline{B_R(0)}$ is compact, we have 
$$|f(z)|\geq C_2\mbox{ in }\overline{B_R(0)}$$
for some constant $C_2>0$.
Combining this, we have $\frac{1}{f}$ is an entire function such that 
$$\left|\frac{1}{f}\right|\leq\frac{1}{\min\{C_1,C_2\}}$$
which is bounded. By Liouville's Theore, $\frac{1}{f}$ is constant, which contradicts to the fact that $\lim_{z\to\infty}f(z)=\infty$.
A: From the viewpoint of Riemannian surfaces one can answer your question as follows - and prove some more than what you have asked: Because $f$ has a pole at $z = \infty$, the entire function $f$ extends to a non-constant holomorphic map between compact Riemannian surfaces 
$$F:\mathbb P^1 \longrightarrow \mathbb P^1, \mathbb P^1 := \mathbb C \cup \{\infty\},$$
with $F(\infty) = \infty$. Because $F$ is non-constant, it is an open map. Because $F$ is a continous map from a compact Riemannian surface, the image $F(\mathbb P^1)$ is compact. Hence $F(\mathbb P^1) \subset \mathbb P^1$ is a non-empty open and closed subset, hence $F(\mathbb P^1) = \mathbb P^1$, i.e. $F$ assumes every value.
Note. The map $F$ is finite, hence has finite degree $d \in \mathbb N$ and assumes every value $d$-times counted with multiplicity. Hence the degree $d$ equals the pole order of f at $z = \infty$. A good textbook is "Forster, O.: Lectures on Riemann surfaces".
A: Since ${1\over f}$ is bounded, it cannot be entire, since that would imply it is a constant by Liouville. So it must be that ${1\over f}$ is not analytic at some point, i.e. ThThere are three cases
(i) $f(\omega)=0$ in which case we're done, and this explains why ${1\over f}$ is not analytic at $\omega$.
(ii) $$\lim_{z\to\omega}{1\over {z-\omega}}\left({1\over f(z)}-{1\over f(\omega)}\right)$$ doesn't exist because ${1\over f(z)}$ doesn't have a finite limit as $z\to\omega$, but then this is impossible since ${1\over f(z)}$ is the composition of continuous functions with a non-zero denominator. But then, we can use the partial Taylor series to see that this limit must be infinity which puts us in case (iii)
(iii) $$\lim_{z\to\omega} {1\over f(z)}-{1\over f(\omega)}=\infty$$ which means, after getting a common denominator, that $f(z)\to 0$ as $z\to\omega$, which puts us back in case (i) as
$$\lim_{z\to\omega}{1\over f(z)}=\infty$$
but then by definition, $f(\omega)=0$ since $f$ is, in particular, continuous.
