Simple proof that holomorphic functions maps dense sets to dense sets As stated in the title I'm searching a simple way to prove that:  


*

*$\Omega \subset \mathbb{C}$ dense in $\mathbb{C}$  

*$f$ a holomorphic map $f$ from $\Omega $ to $\mathbb{C}$ 

*f is not constant 


$\implies f(\Omega)$ is also dense in $\mathbb{C}$.  
Thanks in advance !
 A: It is not sufficient that $\Omega$ is dense in $\mathbb{C}$. A simple counterexample is the principal branch of the square root, which is holomorphic
on $\mathbb{C} \setminus (-\infty, 0]$.
However, if we add the condition that $\mathbb{C} \setminus \Omega$ is closed and discrete, I think the following should work.
We can proceed by contraposition. Suppose that $\overline{f(\Omega)} \neq \mathbb{C}$. Then there are $w \in \mathbb{C}, r > 0$
such that $|f(z) - w| \ge r$ for all $z \in \Omega$, which implies that the function $g(z) = 1/(f(z) - w)$ is a bounded holomorphic function on $\Omega$. Because the singularities of $g$ are isolated, it follows by Riemann's theorem
that $g$ can be extended to an bounded entire function, which by Liouville's theorem
must be constant. Then of course $f(z) = w + 1/g(z)$ is also constant, which contradicts the hypothesis.

Edit: I noticed only when rereading the proof that there was no real need to treat essential singularities separately, so I removed that part.
