In the appendix B of a physics paper arXiv: 1902.01434, it says $$ \partial_z\frac{1}{\bar{z}}=\partial_{\bar{z}}\frac{1}{z}=2\pi\delta(z)\delta(\bar{z}), $$ same as 2-dimensional delta function (complex plane) and A puzzle with derivative of delta-functions. However, from the definition of Wirtinger derivatives, one can also get $$ \partial_z\frac{1}{\bar{z}}=0, $$ such as What is $\partial_z \frac{1}{\bar{z}}$?. So, my question is, which is the right way to do the calculation? For example, we know $\partial_{z}\bar{z}$ is not differentiable, but we can still have $$ \partial_z\bar{z}=\partial_z \frac{1}{\frac{1}{\bar{z}}}=-2\pi\bar{z}^2\delta(z)\delta(\bar{z}), $$ what is wrong here? What about $\partial_z\frac{\bar{z}-a}{\bar{z}-b}$?
I'm really confused here, thank you for any help.
What measure are they using in the complex plane when they get the factor $2\pi$?
Changing from $(z, \bar{z})$ to $(x,y)$ and using well-known divergence and rotation of two vector fields I get: $$ \partial_z \frac{1}{\bar{z}} = \partial_z \frac{z}{|z|^2} = \frac12 (\partial_x - i \partial_y) \frac{x+iy}{x^2+y^2} \\ = \frac12 \left[ \left(\partial_x \frac{x}{x^2+y^2} + \partial_y \frac{y}{x^2+y^2} \right) + i \left(\partial_x \frac{y}{x^2+y^2} - \partial_y \frac{x}{x^2+y^2} \right) \right] \\ = \frac12 \left[ 2\pi\,\delta(x,y) + i\,0 \right] = \pi\,\delta(x,y) = \pi\,\delta(x)\,\delta(y) . $$
This is with respect to the area measure $dx \wedge dy = \frac{i}{2} dz \wedge d\bar{z}.$
How do they get a factor $2\pi$?