= Solution
Write $z=x+iy$ and $\overline z=x-iy$. A convenient <spectral parameter for a linear boundary value problem> is any $\lambda\in\mathbb C\setminus\{0\}$. Define the <modified Helmholtz adjoint plane wave>
$$
\boxed{v_\lambda(z,\overline z)=\exp\!\left[\frac{ik}{2}\left(\lambda z-\frac{\overline z}{\lambda}\right)\right]=e^{A(\lambda)x+B(\lambda)y},}
$$
where
$$
A(\lambda)=\frac{ik}{2}(\lambda-\lambda^{-1}),\qquad B(\lambda)=-\frac k2(\lambda+\lambda^{-1}).
$$
Indeed $A^2+B^2=k^2$, so $(\Delta-k^2)v_\lambda=0$. Equivalently, $4\partial_z\partial_{\overline z}v_\lambda=k^2v_\lambda$. The exclusion of zero is only the coordinate singularity of this spectral parametrization; it still supplies a whole complex one-parameter family. The conjugate companion is
$$
\widetilde v_\lambda=\exp\!\left[-\frac{ik}{2}\left(\lambda\overline z-\frac z\lambda\right)\right]=e^{-A(\lambda)x+B(\lambda)y}=\overline{v_{\overline\lambda}}.
$$
It also solves the adjoint equation.
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