Given the Sturm-Liouville type (time independent Schroedinger) equation


\begin{equation}
\frac{d^2 y}{d x^2} - \left(\mu + V(x)\right) y = \lambda \, y,\quad x \in \mathbb{R}
\end{equation}
where $V(x)$ is symmetric and exponentially decreasing as $|x| \to \infty$, choose the two linearly independent solutions $y\_\pm$ such that $y\_\pm(x)$ behaves as $e^{\pm\sqrt{\mu+\lambda}\,x}$ as $x \to \infty$.


Bounded solutions to this equation are characterised by standard Sturm-Liouville theory, yielding an ordered sequence of *real* eigenvalues $\lambda\_0 > \lambda\_1 > \ldots$ and corresponding eigenfunctions $\phi\_i(x)$. The Sturm-Picone Comparison Theorem states that $\phi\_n(x)$ has exactly $n$ zeroes; in extension, the oscillation properties of the nonbounded solutions $y\_\pm$ are also completely determined for *real* values of $\lambda$.


**Is there anything known for zeroes of $y\_\pm(x)$ for *complex* $\lambda$?** In other words, fix $x\_\* \in \mathbb{R}$ and consider $f(\lambda) := y\_+(x\_\*)$ as a function of $\lambda \in \mathbb{C}$. Can we say anything about the zeroes of $f(\lambda)$?


N.b. Numerical investigations for some concrete $V(x)$ suggest that all zeroes of the corresponding $f(\lambda)$ lie on the real line.


**edit**: I should add: $\mu \in \mathbb{R}$.



Zeroes of Sturm-Liouville solutions as a function of the (complex) eigenvalue
If $\mu$ is real, and $\sqrt{\mu+\lambda}$ is not pure imaginary, the zeros are real, because
they are eigenvalues of one of the problems:
$y(-\infty)=0,\; y(x^\*)=0$ or $y(+\infty)=0,\; y(x^\*)=0$, and these problems are self-adjoint.


