Let $|f(z)| \to \infty$ as $|z| \to \infty$, prove that $f(\mathbb{C})= \mathbb{C}$? I've been on this for a while and would appreciate some help. I need to prove that if a holomorphic function $f:\mathbb{C \to C}$ satisfies $|f(z)| \to \infty$ as $|z|\to \infty$, then $f(\mathbb{C})= \mathbb{C}$.
Here's what I have so far. I know that if I can show that $f(\mathbb{C})$ is open closed and nonempty then $f(\mathbb{C})= \mathbb{C}$. It's obviously not empty since it has at least one value since its the image of a function. It is also open by open mapping theorem since $\mathbb{C}$ is open and $f$ is a non-constant function, $f$ sends open sets to open sets. Finally to show $f(\mathbb{C})$ is closed I've tried to show that $\mathbb{C}\setminus f(\mathbb{C})$ is open but to no avail. I've also tried to show that $f$ contains all its limit points but again unable. Could someone help? Thanks.
 A: Note that this proposition is a generalization of the Fundamental Theorem of Algebra (every complex polynomial has a complex root). A well known proof of this theorem makes use of Liouville's theorem. The proof is as follows:
Suppose $f$ is not surjective. Then $\exists a \in \mathbb{C} \setminus f(\mathbb{C})$. Consider the analytic function $g(z) = \frac{1}{f(z)-a}$, defined over the whole $\mathbb{C}$. Then, by your hypothesis on $f$, you have that $g(z) \to 0$ for $|z| \to \infty$. Hence $g$ is bounded, hence constant by Liouville. And this is a contradiction.
A: I guess you are making the hypothesis that $f$ is holomorphic. Suppose $f$ misses $a\in\Bbb C$. By hypothesis on $f$, there exists $R>0$ such that 
$$\forall z\in\Bbb C,\;|z|\geq R\Longrightarrow |f(z)|\geq |a|+1$$
And $f\left(\overline{D(0,R)}\right)$ is compact and doesn't contain $a$ so that there exists $r>0$ with $D(a,r)\subset\Bbb C\setminus f\left(\overline{D(0,R)}\right)$. Then 
$$g(z)=\frac1{f(z)-a}$$
is bounded holomorphic on $\Bbb C$, thus constant which contradicts the hypothesis on $f$.
A: The condition that $|f(z)| \to \infty$ as $|z| \to \infty$ says that $\infty$ is a pole of $f$. It is not hard to prove that if $f$ is an entire function with a pole at $\infty$, then $f$ is a polynomial. Since any polynomial is surjective it follows that $f(\mathbb{C}) = \mathbb{C}$.
A: Assuming $f$ is entire, suppose $f$ doesn't attain all values, for example suppose $f(z)\neq a$. Then $h(z):=f(z)-a$ has no zeros. This means that $h(z)$ has an analytic logorithm, so that we can write be that $h(z)=e^{g(z)}$, for $g$ analytic. However $|h(z)|\rightarrow\infty$ for $|z|\rightarrow\infty$, which means that $g(z)$ has to be bounded below, meaning it's constant. This is a contradiction.
