How to find a function whose limit points identify $\mathbb{C}$ with a cylinder? I was considering the set of mobius transformations $$ \frac{a+bx}{c+dx} $$ 
A property they have is that if you consider any line $x(t) = e^{i\theta} t $ then:
$$ \lim_{t \rightarrow \infty} \frac{a + bx(t)}{c+dx(t)} =  \frac{a+be^{i\theta}t}{c+de^{i\theta t}} = \frac{b}{d}  $$
Where curiously $\frac{b}{d}$ is COMPLETELY independent of $e^{i\theta}$. 
In some sense the "limit" points of the mobius transformation, identify every direction with the same complex value, and you might even go as to say the "limit points" identify $\mathbb{C}$ as a sphere. 
All this obvious, but now suppose we turn the question around a bit,
How do we find a complex function whose set of limit points identify something other than a sphere.
The Question:
I wanted to start with something simple like a cylinder.
In this case we are trying to find a complex function $f$ such that if $\frac{\pi}{4} < s < \frac{3\pi}{4} $
Then:
$$  \lim_{t \rightarrow \infty} f(e^{is}t) = f(e^{-is}t) $$ 
However we don't want over identification, that is for distinct $\frac{\pi}{4} < s < \frac{3\pi}{4} $ it should be that $\lim_{t \rightarrow \infty} f(e^{is}t)$ is distinct
Moreover it also must be the case that any $s \in (-\frac{\pi}{4}, \frac{\pi}{4}) \cup (\frac{3\pi}{4}, \frac{5\pi}{4}) $ it should be that $  \lim_{t \rightarrow \infty} f(e^{is}t)  $ takes on unique values. 
A strategy:
So, if we forget that we are working in $\mathbb{C}$ we can just consider the unit disk in $\mathbb{R}^2$ and from here find a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which behaves in the expected way on the unit disk [identifying it with a cylinder]. Now the next step would be to find another function $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ whose limit points in each direction are the coordinates the unit circle corresponding to that direction.
Then $f(g)$ is a function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that behaves how we want and we just try our luck in checking if $f(g)$ obeys the Cauchy-Riemann Equations to give rise to a complex function.
 A: This is impossible by the classification of singularities of holomorphic maps. Suppose that such an $f$ exists, and look at $f$ restricted to a punctured neighborhood $U$ of $\infty$ on the Riemann sphere. There, $f$ is a holomorphic, $\mathbb{C}$-valued function defined on a punctured neighborhood of a point. By pre-composing with a Möbius transformation, we may as well move $\infty$ to $0$.
The classification of singularities of holomorphic functions then says that $f$ has a removable singularity, a pole, or has an essential singularity at $0$. The conditions imposed on $f$ in for angles in $(-\pi/4, \pi/4)$ and $(3\pi/4, 5\pi/4)$ rule out poles and the distinctness requirements on other angles rule out a removable singularity. On the other hand, essential singularities attain every complex value, with at most one exception, infinitely many times in any neighborhood of a singularity. The simple structure requested of $f$ at the singularity is incompatible with this. This produces the desired contradiction.
