Derivative of a logarithm and Dirac delta function I'm reading Polyakov's book, Gauge fields and strings. There is this formula (9.247) which I do not really understand how to get.
The formula states that in two dimensions, taking $z$ as my holomorphic and $\bar{z}$ as my antiholomorphic variable, the following relation holds
$$\partial_{\bar{z}}\frac{1}{z-w}=-\pi \delta(z-w)$$
How does one find this?

Multiplying the LHS for an holomorphic test function $f(z)$, which then satisfies $\partial_\bar{z}f(z) = 0$, and integrating over the unit disc $D$, we get 
$$\begin{align}
\iint_D f(z)\color{blue}{\partial_\bar{z}\left(\frac{1}{z-w}\right)}\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i}&=\iint_D \partial_\bar{z}\left(\frac{f(z)}{z-w}\right)\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i}=\int_D \frac{f(z)}{z-w}\frac{\mathrm dz}{-2i}=-\pi f(w)\\
&=\iint_D \color{blue}{-\pi\delta(z-w)}f(z)\frac{\mathrm dz\wedge \mathrm d\bar z}{-2i}
\end{align}
$$where in the second and third equalities we used respectively Stokes' theorem and Cauchy's equality. 

Thus
  $$
\partial_\bar{z}\left(\frac{1}{z-w}\right)=-\pi\delta(z-w)
$$

