2-dimensional delta function (complex plane)

Question

I have a task to show that

$$\partial_{\bar{z}} \frac{1}{z - \zeta} = \pi \delta^{(2)}(z - \zeta) $$

But I thought, that delta-function is determined by $\int f(\zeta) \delta(z-\zeta) d\zeta = f(z)$, so in such point of view function $\frac{1}{z-\zeta}$ should be a delta function.

Where am I wrong? And how to proof this relation, because function $\frac{1}{z-\zeta}$ is holomorphic, so its $\partial_{\bar{z}}$ equals zero.

Answer 1

Note that $C(z) = \frac{1}{z-\zeta}$ is not holomorphic at $z=\zeta$, so you can't quite say that $\partial_{\bar z} C = 0$.

Also, you want the integral to give $f(z)$, not just for holomorphic functions (otherwise you can't say that $\partial_{\bar z} C = \delta$.

The easiest route is probably through Cauchy's integral formula for $C^1$ functions (whose proof is more or less the same as the one for holomorphic functions, even though it's missing from many textbooks):

If $\Omega$ is a domain in $\mathbb{C}$ with say $C^1$-boundary, and $f \in C^1(\bar \Omega)$, then $$ f(z) = \frac1{2\pi i}\int_{\partial\Omega} \frac{f(\zeta)}{\zeta-z}\,d\zeta - \frac{1}{2\pi i} \iint_{\Omega} \frac{\partial_{\bar \zeta}f(\zeta)}{\zeta-z}\,d\bar\zeta \wedge d\zeta. $$

Now, if $\phi$ is any test function, take $\Omega$ as a large disc containing the support of $\phi$. Then $$ \phi(z) = -\frac{1}{2\pi i} \iint_{\Omega} \frac{\partial_{\bar \zeta}f(\zeta)}{\zeta-z}\,d\bar\zeta \wedge d\zeta = -\frac{1}{\pi} \iint_{\Omega} \frac{\partial_{\bar \zeta}f(\zeta)}{\zeta-z}\,dA, $$ where $dA$ is the usual area form (or Lebesgue measure if you prefer) on $\mathbb{C}$. This last equality is just a reformulation of the distributional derivative, i.e. $$ \frac1\pi \langle \partial_{\bar z} \frac{1}{\zeta-z}, \phi \rangle = \phi(z) = \langle \phi, \delta_z \rangle $$ so $\partial_{\bar z} \frac{1}{\zeta-z} = \pi\delta_z$ as requested.