How to show the following formula in the sense of distribution? How to show the following formula in the sense of distribution in $\mathbb{R}^2$?
\begin{equation}(\partial_x +i\partial_y)\frac{1}{x+iy}=2\pi\delta_{(0,0)}
\end{equation}
 A: Let ${\psi} \left(x , y\right)$ be a test function and let
${\varphi} \left(r , {\theta}\right) = {\psi} \left(r \cos  {\theta} , r \sin  {\theta}\right)$.
We compute
\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \frac{\partial  {\varphi}}{\partial  r}&=&\displaystyle  \cos  {\theta} \frac{\partial  {\psi}}{\partial  x}+\sin  {\theta} \frac{\partial  {\psi}}{\partial  y}\\
\displaystyle  \frac{\partial  {\varphi}}{\partial  {\theta}}&=&\displaystyle -r \sin  {\theta} \frac{\partial  {\psi}}{\partial  x}+r \cos  {\theta} \frac{\partial  {\psi}}{\partial  y}
\end{array}
\end{equation}
Hence
\begin{equation}
\frac{1}{r} \frac{\partial  {\varphi}}{\partial  r}+\frac{i}{{r}^{2}} \frac{\partial  {\varphi}}{\partial  {\theta}} = \frac{1}{x+i y} \left(\frac{\partial  {\psi}}{\partial  x}+i \frac{\partial  {\psi}}{\partial  y}\right)
\end{equation}
Hence
\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle\left\langle \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle &=&\displaystyle -\int_{{\mathbb{R}}^{3}}^{}\frac{1}{x+i y} \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left({\psi}\right) d x d y\\
&=&-\displaystyle  \int_{{\left[0 , \infty \right[}\times{\left[0 , 2 {\pi}\right]}}^{}\left(\frac{\partial  {\varphi}}{\partial  r}+\frac{i}{r} \frac{\partial  {\varphi}}{\partial  {\theta}}\right) d r d {\theta}
\end{array}
\end{equation}
Using that $\displaystyle  \int_{0}^{2 {\pi}}\frac{\partial  {\varphi}}{\partial  {\theta}} d {\theta} = 0$, it remains
\begin{equation}
\left\langle \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle  =-\int_{0}^{2 {\pi}}\int_{0}^{\infty }\frac{\partial  {\varphi}}{\partial  r} d r d {\theta} = \int_{0}^{2 {\pi}}{\psi} \left(0 , 0\right) d {\theta} = 2 {\pi} {\psi} \left(0 , 0\right)
\end{equation}
