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

A: I am confused by the above answer, and also do not see where Stokes' theorem is being applied. The following is hopefully more clear. 
Work on a sufficiently small disk $\Delta$ centred at $w$, and let $f$ be a holomorphic function on $\Delta$. The distributional derivative of $1/(z-w)$ with respect to $\overline{z}$ is the distribution $g$ such that \begin{eqnarray*}
\int_{\Delta} f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} &=& \int_{\Delta} g(z) \frac{1}{z-w} \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}}. 
\end{eqnarray*}
Note that since $f$ is holomorphic, $\dfrac{\partial f}{\partial \overline{z}} \equiv 0$, and therefore \begin{eqnarray*}
\frac{\partial}{\partial \overline{z}} \left( \frac{f(z)}{z-w} \right) &=& \frac{1}{(z-w)^2} \left( (z-w) \frac{\partial f}{\partial \overline{z}} - f(z) \frac{\partial}{\partial \overline{z}} (z-w) \right)  \\
&=& - \frac{f(z)}{(z-w)^2} \frac{\partial}{\partial\overline{z}} (z-w) \\
&=& f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right).
\end{eqnarray*}
Therefore, \begin{eqnarray*}
\int_{\Delta} f(z) \frac{\partial}{\partial \overline{z}} \left( \frac{1}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} &=& \int_{\Delta} \frac{\partial}{\partial\overline{z}} \left( \frac{f(z)}{z-w} \right) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}} \\
&=& \frac{1}{-2\sqrt{-1}} \int_{\Delta} \frac{f(z)}{z-w} dz \\
&=& -\pi f(w), 
\end{eqnarray*}
by the Cauchy integral formula. Observe, however, that $f(w)$ can be expressed as the distribution $\delta(z-w)$ acting on $f$. Hence, \begin{eqnarray*}
-\pi f(w) &=& \int_{\Delta} -\pi \delta(z-w) f(z) \frac{dz \wedge d\overline{z}}{-2\sqrt{-1}}. 
\end{eqnarray*}
From this, we see that $g = -\pi \delta(z-w)$, as required. 
Note also, that $dz \wedge d\overline{z} = -2\sqrt{-1} dx \wedge dy$, hence the reason for the $-2\sqrt{-1}$ term hanging around. 
