Prove that if the image of a holomorphic function $g$ is contained in the union of n distinct lines, then $g$ is constant Let $L_1,.., L_n \subset \mathbb R^2$ (or $\mathbb C$) be n distinct lines. Prove that if $g: \mathbb C \to \mathbb C$ is a holomorphic function and $g(\mathbb C) \subset L_1 \cup L_2 \cup ... \cup L_n$, then $g$ is constant.
I would appreciate any hint for this exercise since I don't know what to do. Is there a more general theorem from which I could conclude this? Maybe something related to holomorphic functions where the image is the union of connected sets. 
Another thing I thought of is trying to prove it by the absurd but I got stuck, I mean, suppose $g$ is not constant, then it takes at least two different values $w_1$ and $w_2$, from here how could I arrive to an absurd?
 A: A holomorphic function has the open mapping property whereever the derivative is nonzero, so your function always has the derivative vanish (since a union of lines contains no open set), so it is a constant.
A: As Igor mentioned previously, the open mapping property suffices to solve this problem. However, I got this solution that doesn't rely on such theorem and I think can be useful for others.
First, note that either $u$ or $v$ must take no more than $n$ values (if not, we'd have a grid of $(n+1)\times(n+1)$ values in the image of $g$, which can't be covered by $n$ lines).
Since $g$ is holomorphic, both $u$ and $v$ must be differentiable in $\mathbb{C}$, hence continuous. Therefore, if the image of any of them is a set of finite values, it results constant.
To conclude, recall that if $g=u+iv$ is holomorphic, and either $u$ or $v$ is constant, $g$ must be constant as well.
A: Hint Let $a \notin L_1 \cup L_2 \cup .. \cup L_n$. Show that 
$$h(z)=\frac{1}{g(z)-a}$$
is entire and bounded. 
Note Same way you can prove the more general result: if $g$ is entire you can find some $a \in \mathbb C$ and $r>0$ such that $g(\mathbb C) \cap D_r(a)=\emptyset$ then $g$ is constant.
