Solving a transcendental equation consisting of a quadratic part and a part involving inverse Lambert W functions Question statement
I would like to solve the following equation in the two variables $x$ and $y$:
\begin{gather}
0 = x^2 - a y^2 + i b [x y - W^{-1}(x)W^{-1}(y)] ,
\end{gather}
where $a$ and $b$ are purely real parameters, and $W^{-1}(x)$ is the inverse Lambert W function of $x$. Domain constraints exist due to physicality: for all physical solutions $y$ is purely imaginary, and for some physical solution $x$ is purely imaginary.
Answer attempt
Put the above equation in the alternative form thus:
\begin{gather}
0 = x^2 - a y^2 + i b x y (1 - e^x e^y) .
\end{gather}
Then, by inspection (separate terms into pairs), we have at least the solutions
\begin{gather}
\begin{aligned}
\text{(i)}&\text{ } x = \sqrt{a} y \text{ and } y = \frac{2 i n \pi}{1 + \sqrt{a}} \text{, where } n \in \mathbb{Z} \text{,}\\
\text{(ii)}&\text{ } x = - \sqrt{a} y \text{ and } y = \frac{2 i n \pi}{1 - \sqrt{a}} \text{, where } n \in \mathbb{Z} \text{,}\\
\text{(iii)}&\text{ } x = - i b y \text{ and } y = \frac{\operatorname{Log}(- a / b^2)}{1 - i b} \text{, and}\\
\text{(iv)}&\text{ } x = - i \frac{a}{b} y \text{ and } y = \frac{\operatorname{Log}(- a / b^2)}{1 - i (a / b)} .
\end{aligned}
\end{gather}
The method of this answer attempt is quite obviously not general; hence I post. I might ask more specifically: Can we prove that (i) to (iv) are the only solutions? and if not, is there a more elegant method that derives all solutions?
Thanks
I extend sincere thanks for any help.
 A: Using the basic idea behind the method of dominant balance I can get a good approximation for the solutions mentioned in my comment.
Let $a>0$, $b>0$, and $y=(2n+1)\pi i$.  By expanding the parentheses the equation becomes
$$
a\pi^2 + 4 a \pi^2 n + 4a\pi^2 n^2 - b\pi x + x^2 - b\pi xe^x - 2b\pi nx - 2b\pi nxe^x = 0.
\tag{1}
$$
Numerically it appears that as $n \to +\infty$ the root $x$ also tends to $+\infty$ so let's go ahead and assume this.  The two largest terms in the equation are then $4a\pi^2 n^2$ and $-2b\pi nxe^x$, so as $n \to \infty$ the equation is approximately the same as
$$
4a\pi^2 n^2 - 2b\pi nxe^x = 0
$$
or, upon dividing through by $2\pi n$,
$$
2a\pi n - bxe^x = 0.
$$
This has an explicit solution
$$
x = W(2\pi n a/b),
\tag{2}
$$
which appears to give a good approximation for the solutions of $(1)$ as $n \to +\infty$.  This approximation appears to get better when the ratio $a/b$ is large.
If you'd like you can get a more elementary leading order approximation by using the known asymptotic
$$
W(q) \approx \log q - \log \log q + \frac{\log \log q}{\log q} + \cdots
$$
as $q \to \infty$.
Numerics.
Let's fix $a=b=1$.  In the plots below $n$ is on the horizontal axis.
The blue dots in this plot are the numerical solutions of $(1)$ for $x$ when $n=1,2,\ldots,40$ and the black line is the asymptotic solution $(2)$ we obtained.

This second plot shows the absolute error between the numerical solutions to $(1)$ and the asymptotic solution $(2)$ for $n=10,11,\ldots,5\times 10^4$.  The $n$-axis is plotted with a logarithmic scale for clarity.

