About holomorphic function which avoids $\mathbb R$ I was asking myself whether or not the following statement is true, and I think it is.

The conjecture
Let $f$ be an entire function, i.e. $f$ holomorphic and $f:\mathbb C\to \mathbb C$. We assume that 
$$f(\mathbb C)\cap \mathbb R=\emptyset.$$
Then $f$ is constant.

What I tried
This statement came to me because of the Cauchy-Riemann equations.
If $f$ is an holomorphic function ($f:\mathbb R^2\simeq \mathbb C\to \mathbb C$), $P(x,y)=\mathfrak R(f)$ and $Q=\mathfrak I(f)$ then
$$\begin{cases} \displaystyle\frac{\partial P}{\partial x}=\frac{\partial Q}{\partial y} \\ \displaystyle\frac{\partial P}{\partial y}=-\frac{\partial Q}{\partial x}. \end{cases}$$
With those equations, we can say that if $f(\mathbb C)\subset \mathbb R$, then $f$ is constant (we then have both partial derivates of $Q$ which are null).
May be we could find a solution here, because if $f(\mathbb C)\cap \mathbb R=\emptyset$, then $Q$ doesn't change sign. But we lack a lot of informations to conclude...
May be we can use Liouville theorem to prove a partial result: if 
$$\vert f(x)\vert \xrightarrow[\vert z \vert \to \infty]{} +\infty$$
then $f$ is constant.
But again, this is not really satisfying.
 A: The comment by basket shows a way towards a proof. 
An entire function that avoids all reals, cannot have dense image, as its image must lie in one hyperplane. Yet an entire function that does not have a dense image is easily seen to be constant. 
For let $w_0$ be at a positive distance from the image of $f$. Then $z \mapsto (f(z)-w_0)^{-1}$ is entire and bounded, whence constant.
A: This follows from Picard's little theorem, which states that any non-constant entire function has range $\mathbb{C}$ minus at most one point.

EDIT: As quid's answer shows, invoking Picard here is massive overkill. I answered using it because it's a really cool theorem.
A: For the sake of completeness, here is what's happening for holomorphic functions on general domains (open connected subsets) in ${\mathbb C}$:
A domain $\Omega\subset {\mathbb C}$ is said to be long to the class $O_{AB}$ if every bounded holomorphic function on $\Omega$ is constant. Since half-plane is conformal to the unit disk, your question is effectively about bounded holomorphic functions. It turns out that domains in  $O_{AB}$ admit a very neat geometric characterization which should be as well-known as Liouville's and Picard's theorems:
Theorem. A domain belongs to $O_{AB}$ if and only if its complement has zero 1-dimensional Hausdorff measure.  
(For instance, the standard ternary Cantor set has linear measure zero and, hence, every bounded holomorphic function on its complement is constant. On the other, if $C$ is a "fat" totally disconnected subset of ${\mathbb R}$ then its complement admits bounded nonconstant holomorphic functions.) 
A proof of this theorem (not sure who proved it first) can be found in the book "Riemann surfaces" by Ahlfors and Sario.  
