Question related to Liouville's Theorem Prove or disprove: Let $D$ be an unbounded domain in $\mathbb{C}$. If $f$ is a bounded, analytic function on $D$, then must $f$ be constant on $D$?
This is clearly related to Liouville's Theorem, yet I am not sure how to approach this question using an arbitrary unbounded domain rather than all of $\mathbb{C}$.
 A: Look at $e^z$ on $\{ z : \Re(z)<0\}$, for example. You'll find this function is bounded, analytic and nonconstant.
A: I would like to try to show exactly why boundedness is essential for the function $f$. It's not really a disproof, but there have already been counter-examples given, and there's no reason to be redundant. Let's look at a proof of Liouville's Theorem (which can be found anywhere really, but this one comes from Brown and Church's book on elementary complex analysis), obviously with m own emphasis.
Theorem: If a function $f$ is entire and bounded on $\mathbb C$, then $f(z)$ is constant in the plane.
Proof: Since $f$ is entire, we can apply Cauchy's inequality with any $z_0$ we like and $R$ as the radius of the circle used as the contour in the inequality. By the deformation of paths principle, we can change the shape of this circle in to some contour so that it lies entirely in your unbounded region. In particular, since $n=1$ in the inequality, we know that
$$|f'(z_0)| \leq \frac{M_R} {R}$$
where $M_R$ is the maximal value attained on the contour.
$\color{red}{\text {However, since $f$ is bounded over all of $\mathbb C$, there is a constant $M$ which bounds this value for any $R$,}}$
 i.e. $R$ can be taken arbitrarily large, and $\frac{M_R}{R} \leq  \frac {M}{R}$, so replace $M_R$ with $M$ and observe that $M$ is now totally independent of $R$. But since $R$ can be arbitrarily large, we must have that $|f'(z_0)|=0$ everywhere. This implies that $f$ is constant, and we're done.
So the real trick here was to extend Cauchy's inequality, which is defined over a circle, to the whole complex plane. For this to be permissible, we can't have $f$ blow up as we go far away from the origin. As a last note, it should be clear that boundedness over $\mathbb C$ is not strictly necessary, but the boundedness of $f$ over your domain is absolutely essential.
