Certain condition on an entire function implies the function is constant ? Let $f=u+iv$ be an entire function such that $v \ge 2u+1$ , then is it true that $f$ is constant ? 
 A: Put $f(z)=u+iv$, $g(z)=\exp((2+i)f(z))$. Then we have $|g(z)|=\exp(2u-v)\leq \exp(-1)$, and we finish easily.
A: It's true. Let $g$ be a bijective holomorphic map which send $\{x+iy \in \mathbb C: y \geq 2x+1\}$ onto the unit disc. Then $g\circ f$ is entire and bounded, thus is constant. Thus $f$ is also constant. 
(question: Is there such a holomorphic map $g$?)
Remark: The map $h(z) = e^{-i\theta} (z-i)$, where $\theta = \tan^{-1} 2$ is a linear map which send $\{x+iy \in \mathbb C: y \geq 2x+1\}$ onto the upper half plane. 
A: Just to throw another (almost equivalent) method in, $0$ is not in the image of the function and there is $r>0$ such that the disc $D_r=\{z\ :\ |z|<r\}$ is not either.
For example, take $r=1/4$: if $|z|<1/4$, then, $z=x+iy$ with $|x|,|y|<1/4$, so
$3/4\geq|y-2v|$, hence it is impossible that $y\geq 2x+1$ when $x+iy=z\in D_{1/4}$.
Now, take $g=1/f$. This is a holomorphic entire function. We know that $|f|>1/4$, therefore $|g|<4$, so by Liouville's theorem, $g$ is constant and so is $f$.
