TITLE: Analytic functions with isotopic x-rays
QUESTION [8 upvotes]: Following Arias-De-Reyna, the x-ray of an analytic function $f$ means markings on the complex plane, with one color showing the real locus of $f$ and another color the purely imaginary locus.  
Suppose two analytic (or meromorphic, or analytic with isolated singularities) functions $f$ and $g$ possess isotopic x-rays (so an orientation preserving homeomorphism of the plane moves the x-ray of one precisely onto the x-ray of the other).  What can we conclude about $f$ and $g$?    
Actually I'm not sure that's the best way to frame the question.  Perhaps I should ask for a topological characterization of x-rays, and then ask for the dimension of the space of functions in a given isotopy class, and a method for constructing at least one function with a given x-ray type?  For example, an x-ray of a polynomial locates the polynomials roots and their multiplicities from which we can reconstruct the polynomial; so the space of polynomials in the isotopy class of a given polynomial has dimension equal to the number of distinct roots (right?).
Or perhaps one gets a good theory if one studies jointly the x-rays $f$ and $f'$  (with four colors so you can tell them apart)?  Indeed, how much does the isotopy-type of the x-ray of $f$ determine about that of $f'$?

REPLY [4 votes]: Perhaps it is too late (more than few days passed since the questions were asked)
but I will try
to clarify what happens in the non-compact case. Let us restrict ourselves to the fllowing
special case: suppose that we have a meromorphic function which has at most one critical or
asymptotic value on each of the four quadrants, and no other critical or asymptotic values.
In other words, there is a 4 point set $A=(a_j)$ one point in each quadrant and such that
$$f:S\backslash f^{-1}(A)\to S\backslash A$$
is a covering.
For such functions preimages of the cross can be completely characterized: they are cell decompositions
of the plane, each vertex has degree 4. There can be "infinite faces" which have infinitely many
vertices on the boundary, but the number of vertices on every compact is finite.
Theorem. To each such cell decomposition corresponds a meromorphic function f either in the
unit disc or on the plane.
The free parameters are the 4 ponints $a_j$, plus 2 normalization parameters ($f$ and $f(az+b)$
will have equivalent cell decompositions). The proof of this theorem is an easy application of the Uniformization theorem.
Now the difficult problem arises: how to tell whether this is a disc or the plane from the decomposition. This is called the Type Problem of a simply connected Riemann surface.
There was a lot of research on this problem in 1950-s. Of more recent works I mention
Peter Doyle, On deciding whether a surface is parabolic or hyperbolic, where a probabilistic
interpretation hinted in Thurston's last comment is established. And also a paper of Merenkov and Schramm who disprove a conjecture of Nevanlinna that the type depends on the growth rate of
concentric balls :-)
Of course there is nothing special with taking preimage of a cross from the beginning.
One can start with any cell decomposition of the Riemann sphere, such that $f$ has one critical
or asymptotic value in each face. The equivalence class of the preimage cell decomposition,
together with the critical and asymptotic values themselves and two normalization constants
define your function uniquely.
But it is hard to tell whether it is meromorphic in the disc or in the plane, except for some
special cases.