Consider the Laplace operator $-\Delta$ on $L^2(\mathbb{R}^3)$ with domain $C_0^{\infty}(\mathbb{R}^3 \backslash \{ 0 \})$, the smooth functions with compact support not near 0. I have had a look at this Solving Poisson Equation with domain $L^2(\mathbb{R}^3 \backslash \{ 0 \})$.
But feel that what was missing here was that since we are away from zero, we can't simply solve $(\Delta^{\ast} \pm i)u =0$, it is better to solve ($\Delta^{\ast} \pm i)u =\delta$ for some $\delta$ function. In so doing, taking the Fourier transform yields that $$(4\pi^2 \left| \xi \right|^2 \pm i) \hat{u}(\xi) = 1,$$ and therefore, $$u = \frac{1}{4\pi} \frac{e^{i\sqrt{\pm i} \ |x|}}{|x|}. $$
From this, I want to show that all self-adjoint extensions of $-\Delta$ have a negative eigenvalue, except possibly one. I'm fairly sure we need to use the above formula that I've obtained for $u$, but am unsure. Also, how would we obtain the Friedrich's extension from this?
You're solving an adjoint equation by looking for all $f \in L^2(\mathbb{R}^3)$ for which $$ \int_{\mathbb{R}^3}\{(\Delta\mp i)\varphi\}\overline{f}dx =0, \;\;\; \varphi \in C_{0}^{\infty}(\mathbb{R}^3\setminus\{0\}). $$ A function $f\in L^2(\mathbb{R^3})$ satisfies the above adjoint equation iff $f \in \mathcal{D}(\Delta^*\pm iI)$ with $(\Delta^*\pm iI)f=0$. Because the Laplacian is nice, weak solutions of this type are classical solutions in $\mathbb{R}^3\setminus\{0\}$. There are no non-trivial solutions $f\in L^2(\mathbb{R}^3)$ of $$ \int_{\mathbb{R}^3}\{(\Delta\mp i)\varphi\}\overline{f}dx = 0,\;\;\;\varphi\in L^2(\mathbb{R}^3). $$ The classical solutions of $(\Delta - i)f_{+} = 0$ in $\mathbb{R}^3\setminus\{0\}$ include radial solutions $$ \frac{e^{\pm \sqrt{i} r}}{r}=\frac{\exp(\pm(1+i)r/\sqrt{2})}{r} $$ However, the only solution in $\mathcal{D}(\Delta^*)$ is the one that is in $L^2(\mathbb{R}^3\setminus\{0\})$, which is $$ f_{+} = \frac{e^{-r/\sqrt{2}}e^{-ir/\sqrt{2}}}{r}. $$ Conjugation gives $f_{-}$ that is a solution of $(\Delta+i)f_{-}=0$: $$ f_{-} = \frac{e^{-r/\sqrt{2}}e^{ir/\sqrt{2}}}{r}. $$ There is a one-parameter family of selfadjoint extensions obtained by adding one element to the native domain of the form $$ h_{\lambda}=\frac{1}{r}e^{-r/\sqrt{2}}\{ e^{ir/\sqrt{2}}e^{i\alpha}-e^{-ir/\sqrt{2}}e^{-i\alpha} \} = C\frac{1}{r}e^{-r/\sqrt{2}}\sin(r/\sqrt{2}+\alpha). $$ To check if there are negative eigenvalues, start by checking if $(-\Delta h_{\lambda},h_{\lambda})$ is negative. For example, \begin{align} (\Delta h_{\lambda},h_{\lambda})&=(\Delta(e^{i\alpha}f_+-e^{-i\alpha}f_-),e^{i\alpha}f_+-e^{-i\alpha}f_+) \\ &= i(e^{i\alpha}f_++e^{-i\alpha}f_-,e^{i\alpha}f_+-e^{-i\alpha}f_-) \\ &= i\|f_+\|^2-i\|f_-\|^2+i(e^{-i\alpha}f_-,e^{i\alpha}f_+)-i(e^{i\alpha}f_+,e^{-i\alpha}f_-) \\ &= 2\Im\{ e^{2i\alpha} (f_+,f_-)\}. \end{align}