TITLE: how to show image of a non constant entire function is dense in $\mathbb{C}$?
QUESTION [24 upvotes]: how to show image of a non constant entire function is dense in $\mathbb{C}$? is there any smallest proof? I have seen this as a theorem in some books but I want some elementary proof.

REPLY [39 votes]: It can be done with the help of the $\color{blue}{Casorati-Weierstrass\ theorem}$ or by the $\color{blue} {Liouville's\ theorem, }$
With the help of $\color{blue}{Casorati-Weierstrass\ theorem}$
Suppose $f$ is an entire function whose image is not dense in $\mathbb{C}$. Then there exists a complex number $\alpha$ and a number $s> 0$  such that |$f(z)-\alpha|>s$ $\forall  
z \in \mathbb{C}$.
Write $f(z) =\sum_{n=0}^{\infty} a_nz^n$ and suppose that there are infinitely many nonzero 
terms in this expansion. Then, for all $z \not=0$, let $g(z)=f(1/z)$ .We see that g has 
an essential singularity at $0$ so by the Casorati-Weierstrass theorem for some $z\ 
near\ 0$ we have $|g(z)-\alpha|<s $ then$|f(1/z)-\alpha|<s$ This contradiction implies 
that the power series expansion of $f$ can have only finitely many terms. Then the 
fundamental theorem of algebra guarantees that $f\ is\ constant$. This contradicts the 
hypothesis. 
With the help of $\color{blue}{Liouville's\ theorem, }$
Suppose there exists a complex number $\alpha$ and $s\in \mathbb{R^+}$ such 
that $|f(z) — \alpha|>s, \forall z \in \mathbb{C}$. Then the function $g(z) = 1/(f(z) — \alpha)$ is 
entire and bounded, so by Liouville's theorem, g is constant. Hence $f$ is constant, 
again contradicting the hypothesis.