Geometric Interpretation of Liouville's Theorem? 
The only bounded entire functions in $\mathbb{C}$ are constants.

Could someone please give me a geometric interpretation of the theorem above? I don't intuitively understand why it's true.
Also, aren't periodic functions counterexamples? i.e. $\sin(z)$?
 A: Here is an interpretation based on a proof by Edward Nelson.
If $f : \mathbb{C} \rightarrow \mathbb{C}$ is entire (*), then it verifies the Mean value property: its value at the center of a disk equals to the average of its values on that same disk.
Consider two points $a, b \in \mathbb{C}, a \neq b$, a radius $R > 0$ and the two disks $D_1, D_2$ of radius $R$ centered respectively at $a$ and $b$.
We have (Mean value property):
$f(a) = \frac{1}{\pi R^2} \iint_{D_1}{f}{dS} = \frac{1}{\pi R^2}( \iint_{D_1 \cap D_2}{f}{dS} + \iint_{D_1 \setminus D_2}{f}{dS})$
Doing the same for $b$ and $D_2$ we find:
$f(a) - f(b) = \frac{1}{\pi R^2} (\iint_{D_1 \setminus D_2}{f}{dS} - \iint_{D_2 \setminus D_1}{f}{dS})$
If $f$ is bounded, those two integrals converge to zero when $R \rightarrow \infty$, because the area on which we integrate becomes negligible compared to the whole area of the disks (see picture below) and we find that $f(a) = f(b)$ for all $a,b \in \mathbb{C}$: $f$ is constant.

(*) same proof still holds for harmonic functions, since we only use the mean value property
A: The geometric interpretation is the following : any holomorphic function of the Riemann Sphere/complex projective line is constant.
Edit : I put the proof of the previous fact in the general case of a compact Riemann surface in comments below.
