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: $\mu \in \mathbb{R}$.