Let $f$ be entire and suppose that $\text{Im} f\ge0$. Show that $\text{Im} f$ is constant. How do you show that if $f$ is entire and if $\text{Im} f\ge0$, then $\text{Im} f$ is constant? Is this even true in the first place?
 A: Based on @kobe's suggestion:
Let $g(z) = e^{if}$. Write $f = u + iv$, then $g(z) = e^{ui - v} = e^{-v}e^{iu}$. If $v\geq 0$ then $|g| = e^{-v} \leq 1$, so $g$ is bounded. Also $g$ is entire, so $g$ is constant by Liouville's theorem. Then $f$ is constant.
A: You could also notice that the upper half plane is equivalent via a linear fractional to the disc.  That is, the map
$$
w \mapsto \frac{w-i}{w+i},
$$
takes the upper half plane to the unit disc (to see this, notice that it takes $i$ to 0, and it takes the real line to the unit circle).  So the function
$$
z \mapsto \frac{f(z)-i}{f(z)+i}
$$
is a bounded entire function.
The moral of the story is that the upper half plane is the same as a disc, as far as analytic functions are concerned.  Many domains are equivalent to the disc like this; any simply connected domain in $\mathbb{C}$ that is not the entire complex plane in fact by the Riemann Mapping Theorem.  This is why lots of results are stated for the disc.
In fact, an entire function must hit everything except possibly one point, that's Picards Little Theorem.  So it definitely cannot miss an entire half plane.  Though of course, you probably don't want to use such a big hammer here.
A: If $f=u+iv$ let $g:=f+i$, hence $g=u+i(v+1)$ with $Img=v+1 \ge 1$. Thus $g$ is entire and $g$ has no zeroes. Therefore $1/g$ is entire.
From $|g|^2=u^2+(v+1)^2 \ge (v+1)^2 \ge 1$ we see that $1/g$ is bounded. By Liouville, $1/g$ is constant, thus $f$ is constant.
