When one solves the wave equation
$ ( \partial_t^2 - v^2 \nabla^2) \mathbf{E}(\mathbf{x},t) = 0 $
in $\mathbb{R}^3 \times \mathbb{R} $ using the Fourier transform method, the general solution is found to be
$ \mathbf{E}(\mathbf{x},t) = \int_{\mathbb{R}^3} \mathbf{E}_+(\mathbf{k}) e^{i ( \mathbf{k} \cdot \mathbf{x} + v |\mathbf{k}| t ) } + \mathbf{E}_-(\mathbf{k}) e^{i ( \mathbf{k} \cdot \mathbf{x} - v |\mathbf{k}| t ) } d^3\mathbf{k} $
For an overview of how to get this result see for example this answer. Note that here I am using the typical physics conventions for electromagnetic waves.
Now, an outgoing monochromatic plane wave is defined as
$ \mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 e^{i ( \mathbf{n} \cdot \mathbf{x} - v |\mathbf{n}| t ) } $
for $\mathbf{n} \in \mathbb{R}^3$. This indeed satisfies the wave equation and can be obtained from the above solution by putting $\mathbf{E}_+(\mathbf{k}) = 0$ and $\mathbf{E}_-(\mathbf{k}) = \delta(\mathbf{k} - \mathbf{n})$.
If one allows $\mathbf{n} = \mathbf{n_R} + i \mathbf{n_I}$ to be imaginary, then the now so-called 'inhomogeneous' outgoing monochromatic plane wave
$ \mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 e^{- \mathbf{n_I} \cdot \mathbf{x} } e^{i ( \mathbf{n_R} \cdot \mathbf{x} - v |\mathbf{n}| t ) } $
also satisfies the wave equation. However, now it is not obvious to me what $\mathbf{E}_+(\mathbf{k})$ and $\mathbf{E}_-(\mathbf{k})$ should be to obtain this solution from the general solution above. It wouldn't seem correct to put $\mathbf{E}_-(\mathbf{k}) = \delta[\mathbf{k} - (\mathbf{n_R} + i \mathbf{n_I})]$ since the integration is over real $ \mathbf{k}$-space. Does the above general solution somehow miss these 'inhomogeneous' plane wave solutions? How can this situation be reconciled or explained?