Solving Poisson Equation with domain $L^2(\mathbb{R}^3 \backslash \{ 0 \})$. I want to solve the following equation $$\Delta u \pm i u =0,$$ where $u \in C_c^{\infty}(\mathbb{R}^3\backslash \{ 0 \})$ is a smooth function, compactly supported away from the origin. Note that this is an attempt to determine the deficiency indices of the operator $-\Delta $. I can use the Fourier transform, this gives me $$(\left|\xi \right|^2 \pm i)\hat{u} =0.$$ It is possible to treat $u$ as a $\delta$-function? 
I'm unsure. 
 A: If you define $Lf=-\Delta f$ on $C_c^{\infty}(\mathbb{R}^3\setminus\{0\})$, then $L$ is symmetric on its domain $\mathcal{D}(L)$, and the domain is dense in $L^2(\mathbb{R}^3)$. The deficiency spaces are associated with the adjoint operator $L^*$. So you're looking for functions $f \in L^2$ such that
$$
       \int_{\mathbb{R}^3} \{(L\pm iI)g\}\overline{f} dx = 0,\;\;\; g\in\mathcal{D}(L).
$$
Any such $f$ is going to be an $L^2$ weak solution of $(-\Delta\mp iI)g=0$ in $\mathbb{R}^3\setminus\{0\}$. And, it would admit solutions of the form you stated, assuming such functions are in $L^2$.
There are radial solutions of your equations, which are the solutions of the following for any given complex $\lambda$:
$$
       Lf(r) = -\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right)= \lambda f.
$$
There is a simplication when substituting $f=\frac{1}{r}g$. The new equation is
$$
     -\frac{1}{r^2}\frac{d}{dr}\left(r^2\left(\frac{1}{r}\frac{dg}{dr}-\frac{1}{r^2}g\right)\right)=\frac{\lambda}{r}g \\
      -\frac{d}{dr}\left(r\frac{dg}{dr}-g\right)=\lambda rg \\
    -r\frac{d^2g}{dr^2}-\frac{dg}{dr}+\frac{dg}{dr}=\lambda rg \\
     \frac{d^2g}{dr^2}+\lambda g = 0.
$$
The solutions are
$$
           f(r)=\frac{g}{r} = A\frac{\sin(\sqrt{\lambda}r)}{r}+B\frac{\cos(\sqrt{\lambda}r)}{r} \\
     = A'\frac{e^{i\sqrt{\lambda}r}}{r}+B'\frac{e^{-i\sqrt{\lambda}r}}{r}.
$$
The second forms are better because exponential decay puts the solution in $L^2_{r^2}[0,\infty)$. Boundary functionals are the real and imaginary parts of
$$
          \Phi(f) = (Lf,f_{i})-(f,Lf_{i}) .
$$
