Are solutions of the following equation countable : $ \frac{a\exp(ix)}x + \frac{b \exp(iy)}y = c $? I would like to prove that:
For given non zero complex numbers $a,b$ and $c$, the set of positive real numbers $x>0$, $y>0$ satisfying the equation:
$$
 \frac{a\exp(ix)}x + \frac{b \exp(iy)}y = c
$$
is countable, where $i$ is the imaginary (complex) unit.
 A: Hint:
The two spirals are making an infinity of turns around two distinct centers. If neither of them passes through the center of the other, the number of intersections is finite.
If one of the loops goes through a center, it meets the spiral an infinity of times, but in orderly fashion: you can number the intersections by increasing parameter, with an increment close to $\pi$. (Very close to that center, the other spiral is virtually a striaght line.)
A: This is a very partial answer attempting to cover all the cases, with a proven case and other cases at the state of conjectures without proofs.
First, have a look at the figures at the bottom ; it suffices to know at present that the common points to the two spirals are in correspondence with solutions. The first figure has a finite number of intersection points, the second one, where one of the spiral passes through the center of the other one, has a - denumerable - infinite number of intersection points. The third one illustrates the case of a common center with, again a (denumerable) infinite number of intersection points.
Let us consider the curve (S) parameterized in this way:
$$\gamma: t \to \dfrac{1}{t}e^{it} \ \ \text{for} \ \ t>0.$$
It is an hyperbolic spiral * ((https://en.wikipedia.org/wiki/Hyperbolic_spiral) with equation $r\theta=1$), different from other classical families of spirals, Archimedean, logarithmic, and Cornu spiral.
Two extreme cases:


*

*If $t \to 0$, $\gamma(t) \to +\infty$, i.e., (S) possesses an asymptote which is a straight line parallel to the real axis. More precisely, by considering $\dfrac{1}{t}e^{it}=\dfrac{1}{t}(1+it-\dfrac{1}{2}t^2+\cdots)$, this asymptote has equation  $y=1$.

*If $t \to \infty$, $\gamma(t) \to 0$ ; the curve spirals around a limit point, the origin that will be called the "center" of the spiral.
The equation given in the question can be transformed into:
$$\tag{1}\gamma(x)=c'+b'\gamma(y) \ \ \text{with} \ \ b'=-b/a, \ \ c'=c/a.$$
We have thus to consider the intersection of two spirals, the one in the LHS which is $(S)$, the other in the RHS, let us name it $(S')$, which is an enlarged, rotated and translated version of $(S)$, the enlargement factor being $|b'|$, the rotation angle being arg($b'$), and the translation given by $c'$.
Thus, two cases can occur :
Case 1 : (illustrated by Figure 3) : if $c'=0$ [the spirals' centers coincide] : in this case, let $b'=re^{i\theta}$ (thus $r$ and $\theta$ are fixed quantities ; $r$ is assumed $\neq 1$). (1) can be written under the form:
$$\tag{2}\gamma(x)=b'\gamma(y) \ \iff \ \dfrac{1}{x}e^{ix}=re^{i\theta} \dfrac{1}{y}e^{iy}$$
Equating modulus and argument on both sides gives 
$$\tag{3} \ \iff \begin{cases}y=rx \\ x=\theta+y+k2\pi \end{cases}$$
where each $k \in \mathbb N$ is associated to a specific turn around the origin.
Now, consider the resulting equation:
$$\tag{4}x=\theta+rx+k2\pi \ \iff x=\dfrac{1}{1-r}(\theta+k2\pi)$$
(take $k \in \mathbb N$ if $r<1$ and $k \in -\mathbb N$ if $r>1$) giving a unique solution for each turn (because once $x$ is known, $y=rx$ is known). Thus there are a denumerable infinite number of solutions.
Case 2: $c' \neq 0$. There are two subcases:
Subcase 2.1: one of the spirals passes through the center of the other (illustrated by Figure 2).
Conjecture: there is an (denumerable) infinite - number of intersection points.
Subcase 2.2: no spiral passes through the center of the other (illustrated by figure 1). 
Conjecture : there is a finite number of intersection points.

Remark about subcase 2.1: the fact that for example spiral (S') passes through the origin can be transcribed into the constraint 
$$\exists y \in \mathbb R, \ c'-b'\dfrac{1}{y}e^{iy}=0  \ \iff \ \ d'y=e^{iy}  \ \iff \ \ yre^{i \theta}=e^{iy}$$
(by setting $d':=\tfrac{c'}{b'}:=re^{i \theta}$). Thus, we must have simultaneously $yr=1$ and $\theta = y + k 2 \pi$. Therefore, the constraint is to have $d'$ such that $|d'|=1/(arg(d')+k 2 \pi)$ (which is a family of spirals).

(Figure 1: Random case : at most a finite number of intersection points. Here $b'=1-i, \ \ c'=0.1+0.2i$.)

(Figure 2: Case where one of the spirals passes through the center of the other.)

(Figure 3: case where the spirals have a common center : the common points have been computed using formula (4).)


*

*[I'm indebted to Yves Daoust for pointing to me the name 'hyperbolic spiral" I didn't know]

