Let $f$ be an entire function, show that the following conditions imply that $f$ is constant. Let $f$ be an entire function, show that the following conditions imply that $f$ is constant.
(i)  $f(\mathbb C)\cap\{x \in \mathbb{R} | x < 0 \}= \emptyset$
(ii) $f(\mathbb C)\cap\{x \in \mathbb{R} | 0 < x < 1 \}= \emptyset$
(iii) $f(\mathbb C)\cap\{z \in \mathbb{C} | |z - 1| < \frac{1}{4} \} = \emptyset$
I know how to prove the (iii) using Liouville's Theorem. I know that the $f(\mathbb C)$ is dense in $\mathbb{C}$ if $f$ is a nonconstant entire function.
Then the intersection of the range of $f$ with any open set in $\mathbb{C}$ can not be empty. But for (i) and (ii) I don't know because those two are not open sets in $\mathbb{C}$. Show I consider that sets as open sets in the relative topology? 
 A: Throwing little Picard or the Riemann mapping theorem at (i) seems overkill to me. We just need to recall that the map
$$ g(z)= \frac{\sqrt z - 1}{\sqrt z + 1} $$
is biholomorphic from $ U = \mathbb C \setminus (-\infty,0] $ onto the open unit disc. So if $ f $ is entire and $ f(\mathbb C)\subset U, $ then $ g\circ f $ is a bounded entire function. Hence it is constant by Liouville. It follows that $f$ is constant.  
A: For the first, we can use the Riemann mapping theorem. The complex plane minus a line (here we take $x\leq -1$) is conformaly mapped to the unit disk. Let $g$ be that mapping. Now, take $h=g\circ f$. Observe that $h$ is a bounded entire function hence it is constant, which in turn implies that $f$ is constant.
For the second one, notice the set $\mathbb{C}- [a,b]   $ can be conformaly mapped to the unit disk minus a point. For instance when $a=0,b=1$ take $g= \dfrac{2}{1 + \sqrt{1 - \frac{1}{z}   }} -1$. Now, arguing as before we can conclude that $f$ is constant.
A: They all follow form the little Picard theorem.
