Compute $D u$ in the sense of distributions, where $u =(x+iy)^{-1}$ is a function defined in $\mathbb{C}$ and $D=\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}$. (Hint: use the integration by parts formula)
The Solution to the question is $2\pi\delta$
I don't know how to solve it Please help me.
Show that $D\frac1{x+iy}$ is a radial distribution, with $\Phi(z)=\int_0^1 \phi(e^{2i\pi t}z)dt=\varphi(|z|^2)$ we get
$$\langle D\frac1{x+iy},\phi \rangle= \langle D\frac1{x+iy},\Phi \rangle=-\langle \frac1{x+iy}, D\Phi\rangle$$ $$=-\int_0^\infty \frac1{re^{it}} \int_0^{2\pi} D\Phi(re^{it}) dt rdr=-\int_0^\infty \frac1{re^{it}} \int_0^{2\pi} \varphi'(r^2) 2re^{it} dt rdr$$ $$=-2\pi \int_0^\infty \varphi'(r^2)2rdr=2\pi \varphi(0)=2\pi \phi(0)$$