Elegant proofs of $\bar{\partial}z^{-1} = 2\pi \delta_0$ For a function $f(x,y)$ on $\mathbb{R}^2,$ defined possibly outside the origin, write
$$\int_\epsilon ' f \,dx\,dy : = \int_{\mathbb{R}^2\setminus D_\epsilon}f \, dx\,dy,$$
(the integral on the complement to the $\epsilon$-disk) and
$$\int' f \,dx \,dy : = \lim_{\epsilon\to 0} \int'_\epsilon f \,dx\,dy,$$ when defined.
We view $\mathbb{R}^2$ as the complex line with coordinate $z = x+iy$. Then I claim that the assignment $$\phi:f(x,y) \mapsto \int_{\mathbb{R}^2}' f\cdot z^{-1} \,dx\,dy$$ makes sense as a functional on compactly supported, smooth functions (indeed, if $f$ is rotationally symmetric then $\phi(f) = 0$ even before taking the limit, and if $f(0) = 0$ then $f\cdot z^{-1}$ is bounded, hence integrable, at $0$; now in the space of smooth compactly supported functions, any function can be written as a rotationally symmetric function plus a function that vanishes at the origin). We can thus formally define a distribution $z^{-1}\in C^{-\infty}(\mathbb{R}^2)$ as $$z^{-1}: = \frac{\phi}{dx\,dy}.$$
Now write $\bar{\partial} = \partial_x + i \partial_y$ .Like any vector field, this acts on distributions, and so we have a distribution $\bar{\partial} z^{-1}.$ Since $z^{-1}$ is holomorphic where it is smooth, we must have $\bar{\partial} z^{-1}$ be a distribution supported at the origin. In fact, it is known to mathematical physicists that the result is a delta function at the origin: $$\bar{\partial} z^{-1} = 2\pi \delta_0,$$ and this fact is useful in conformal field theory.
I would like to see a proof of this result (the physics sources I have seen do not prove this). In fact, I know how to give one using a direct calculation with polar coordinates: but this seems ad hoc and is not very satisfying to me.
I suspect there might be "nicer" proofs using one or more of the following three techniques, and I am hoping that more analytically literate MO users can provide them.

*

*If one can show that $\bar{\partial} z^{-1}$ is determined by its values on holomorphic (near the origin) functions, the Cauchy residue formula would imply that $\bar{\partial} z^{-1} = 2\pi \delta_0.$

*I suspect there should be a proof using the stationary phase approximation. 

*This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators.

 A: The identification of $\partial_{\bar{z}}z^{-1}$ with a delta function follows directly from the Cauchy–Pompeiu formula
$$f(\zeta) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) \,dz}{z-\zeta} - \frac{1}{\pi}\iint_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dx\wedge dy}{z-\zeta},\qquad\qquad(\ast)$$
for $f$ a complex-valued $C^1$ function on the closure of the disk $D$ in $\mathbb{C}$. (Note that $f$ need not be holomorphic, as in the usual Cauchy formula.) If the support of $f$ is within $D$ the boundary integral $\int_{\partial D}$ does not contribute, while the area integral $\iint_D$ states that$^\ast$
$$\frac{\partial }{\partial \bar{z}} \frac{1}{z-\zeta}=\pi\delta^{(2)}(z-\zeta).$$
Historical note: Wikipedia gives a 1905 paper as the source of the formula ($\ast$), but I could not locate it there. I did find it in a 1913 paper by Pompeiu (equation 3):

The first term $h(z)$ is the boundary integral. The capital $S$ denotes the area integral, the function $\varphi$ is the derivative $\partial f/\partial \bar{\zeta}$.

$^\ast$ The OP has a factor $2\pi$ instead of $\pi$, because of a different definition of the Wirtinger derivative. Here I follow the definition $\partial/\partial\bar{z}=\tfrac{1}{2}(\partial/\partial x+i\partial/\partial y)$, while in the OP there is no coefficient $\tfrac{1}{2}$.
A: You can conclude it from the well known fact that $\frac{1}{2\pi}\log|z|$ is a fundamental solution to the Laplace equation. The argument presented below is taken from my lecture notes:
Harmonic Analysis, see Proposition 6.18 on pg. 89. Please, refer to the notes for more details.


A: Following Gelfand-Graev, Grothdieck, and Schwartz: The right-most pole of $u_s=|z|^{2s}=z^s\cdot \overline{z}^s$ (or,
properly, the distribution given by integration-against that
function) is at $s=-1$, when the function ceases to be $L^1_{\mathrm{loc}}$.
As usual,
$$
u_s(f)
\;=\;
u_s(f-f(0)\cdot e^{-z\overline{z}})+f(0)\cdot u_s(e^{-z\overline{z}})
$$
The value $u_s(f-f(0)\cdot e^{-z\overline{z}})$ is computable by integration
against $|z|^{2s}$, since (by design) that difference is a Schwartz
function vanishing at $0$. This vanishing does not imply
divisibility by $z$, nor by $\overline{z}$, nor by $x$, nor by $y$, but still
does imply, by Taylor-Maclaurin expansion with remainder, that
$$
f(z) - f(0)\cdot e^{-z\overline{z}} \;=\; O(|z|)
\hskip30pt \hbox{(as $|z|\to 0$)}
$$
In particular, the residue of $u_s$ at $s=-1$ is a multiple
of $\delta$ (because $f(0)=\delta(f)$), and the multiple can be
determined by integrating against $e^{-z\overline{z}}$:
$$
\int_{\mathbb C} e^{-z\overline{z}}\,(z\overline{z})^s\;dx\,dy
\;=\;
2\pi \int_0^\infty e^{-r^2}\,r^{2s}\;r\,dr
\;=\;
2\pi \int_0^\infty e^{-r^2}\,r^{2s+2}\;{dr\over r}
\;=\;
\pi \int_0^\infty e^{-r}\,r^{s+1}\;{dr\over r}
\;=\;
\pi\cdot \Gamma(s+1)
$$
The residue at $s=-1$ is $\pi$.
Then, with interchange of evaluation and differentiation justified by
the Schwartz-Grothendieck vector-valued extension of Cauchy-Goursat
theory,
$$
\overline{\partial}\,{1\over z}
\;=\;
\overline{\partial}\Big(z^s \overline{z}^{s+1}\Big|_{s=-1}\Big)
\;=\;
\Big(\overline{\partial}(z^s\overline{z}^{s+1})\Big)\Big|_{s=-1}
\;=\;
\Big((s+1)z^s\overline{z}^s\Big)\Big|_{s=-1}
$$ $$
\;=\;
\mathrm{Res}_{s=-1}(z^s\overline{z}^s)
\;=\;
\pi\cdot \delta
$$
