Difference sets of holomorphic injections Let $D$ be a bound domain in $\mathbb C$ and let $f$ and $g$ be injective holomorphic functions on $D$. Is it possible that the set $\{{w}:f(w) - g(w) = z\}$ is infinite for all but perhaps one complex numbers $z$?
 A: This is not a complete answer, but only some thoughts that I think could maybe help solve the problem. I will expand this post every time I have some new idea/result.
Let $D\subset\mathbb{C}$ be a bounded domain (i.e. open and connected), $A\subseteq\mathbb{C}$ any infinite subset of $\mathbb{C}$. Assume we have $f,g:D\rightarrow\mathbb{C}$ two holomorphic functions such that for every $a\in A$ we have that $E_a:=\{z\in\mathbb{C}|f(z)-g(z)=a\}$ is infinite. Then since $D$ is bounded, and thus precompact, whe have that $E_a$ has an accumulation point $z_a$ in $\bar{D}$ (the closure of $D$). If $z_a\in D$, then by the identity theorem we have that $f=g$, which is a contradiction. Thus $E_a$ must have an accumulation point on $\partial D$ for every $a\in A$.
I think that this can be useful in the search of a contradiction to the question, or in restricting the research for an example of two such functions. Personally, I propend to think that no two such functions exist.
A: This is not an answer, but too long to put as a comment.  I have an idea for creating an example of some such $f$ and $g$.
Consider the functions $f_1(x)=e^x(\cos(x)+10x)$ and $g_1(x)=e^x(\sin(x)+10x)$.  On $(0,\infty)$, these functions are injective (strictly increasing) and the set $\{x>0:f_1(x)-g_1(x)=r\}$ is infinite for all $r\in\mathbb{R}$.  Moreover, if we pre-compose with $\phi(x)=\tan\left(\frac{\pi}{2}x\right)$, then we can view $f_1$ and $g_1$ as having domain $(0,1)$, in which case the set $\{x\in(0,1):f_1(x)-g_1(x)=r\}$ is infinite for all $r\in\mathbb{R}$.
I don't know enough about the behavior of the trig functions to know whether this would extend nicely to the disk or not, but it is an idea.
