# Dirac delta distribution in the complex plane

In the complex plane, one can write $$\delta(x) \delta(y) = \dfrac{1}{\pi} \partial_z \dfrac{1}{\bar{z}}.$$

How to prove this relation has been answered before and identifies

$$2\pi \delta(x) \delta(y) = \partial_x \left(\dfrac{x}{x^2 + y^2} \right) + \partial_y \left(\dfrac{y}{x^2 + y^2} \right).$$

However, I have not been able to show that this relation indeed holds. How can I prove that it is indeed true?

Let me write $$z = (x,y)$$ and $$|z|^2 = x^2+y^2$$. Then, your second relation can be written $$\nabla\cdot(z/|z|^2) = 2\pi\,\delta(z)$$ where $$\delta(z) = \delta(x)\,\delta(y)$$ is the Dirac delta in $$\Bbb R^2$$. This follows from the well-known fact that $$-\ln |z|$$ is a fundamental solution of the Poisson equation, since then $$\nabla\cdot(z/|z|^2) = \nabla\cdot(\nabla(\ln|z|)) = \Delta \ln |z| = 2\pi\,\delta(z).$$

There are a lot of derivations of the Poisson fundamental solution. One can remark that taking the derivative in the classical sense $$\Delta \ln |z| = 0$$ for $$z\neq 0$$, use a scaling argument to deduce that $$\Delta \ln |z| = C\,\delta(z)$$. Another that I prefer is to use the Fourier transform: since (see e.g. here) $$\mathcal{F}(\ln(z)) = C\,\delta(z) - \mathrm{pf}\frac{1}{2\pi\,|z|^2}$$, one deduces that $$\mathcal{F}(\Delta\ln(z)) = -C\,4\pi^2 |z|^2\,\delta(z) + 2\pi = 2\pi$$ because $$|z|^2\,\delta(z) = 0$$, and so $$\Delta\ln(z) = 2\pi\,\delta(z)$$.

The strategy is to show that the delta function at the origin in the plane can be approximated by a family of smooth density functions that depend on a parameter $$\epsilon \to 0$$ that ultimately concentrate their mass at the origin.

First make use of the approximation $$\frac{ 1}{\overline z} = \frac{ z}{ z\overline z} \sim f(z,\overline z)=\frac{ z}{ z\overline z+\epsilon}$$ and then we use the Wirtinger calculus and quotient rule to compute

$$\partial_z f(z,\overline z)=\partial_z \frac{ z}{z\overline z+ \epsilon} =\frac{ (\epsilon + z\overline z)-z\overline z }{(|z|^2+ \epsilon)^2} =\frac{\epsilon}{(|z|^2+ \epsilon)^2} :=\mu(z,\overline z)$$.

Next check that this mass density expression always has the same total mass in the plane; that is, show that $$I=\int \int \mu \ dx \ dy$$ is always the same constant regardless of the choice of $$\epsilon$$. (This is easy to verify by making the scaled change of variables $$x=\sqrt{\epsilon} x', y= \sqrt{\epsilon} y'$$ for which $$dx dy = \epsilon dx' dy'$$.)

Next you can evaluate this normalization constant by setting $$\epsilon =1$$ and then using polar coordinates:

$$I= \int_0^{2\pi} \int_{r=0}^{\infty} \frac {r \ dr \ d\theta}{ (r^2+ 1)^2} = \pi$$

Lastly, (handwaving somewhat) you can confirm by graphing the density functions that the mass density distributions concentrate their mass in zones whose radii ultimately tend to zero.

Footnote. The function $$f$$ and the measure $$\mu$$ are connected to the Laplacian of perturbations of the log function as follows: $$\mu=\frac{\partial^2 \ln(\epsilon+ z \overline z)}{\partial z \partial\overline z}$$ and $$f= \frac{ \partial}{\partial \overline z} \ln(\epsilon+ z \overline z)$$.

The usual real-variable Laplacian can be written in Wirtinger notation as $$u_{xx}+u_{yy}= 4 u_{z\overline z}$$.

Another strategy starts by applying Stokes' theorem to Cauchy's integral formula : $$f(\zeta) = \oint\frac{\mathrm{d}z}{2\pi i} \frac{f(z)}{z-\zeta} = \int\frac{\mathrm{d}\bar{z}\wedge\mathrm{d}z}{2\pi i} \frac{\partial}{\partial\bar{z}}\frac{f(z)}{z-\zeta}$$ With Cartesian coordinates, i.e. $$z = x + iy$$, one has $$\mathrm{d}\bar{z}\wedge\mathrm{d}z = 2i \,\mathrm{d}x\wedge\mathrm{d}y$$, hence the following representation of the 2D Dirac delta on the (Cartesian) complex plane : $$\delta^{(2)}(x,y) = \delta(x)\delta(y) = \partial_{\bar{z}}\frac{1}{\pi z}$$ and, by taking the complex conjugate of this relation, $$\delta(x)\delta(y) = \partial_z\frac{1}{\pi\bar{z}}.$$ As for your second identity, it can be proven by interpreting $$\partial_z$$ as a Wirtinger derivative, as follows : $$\begin{array}{rcl} 2\pi\delta(x)\delta(y) &=& \displaystyle 2\partial_z\frac{1}{\bar{z}} \quad\mathrm{with}\quad \frac{1}{\bar{z}} = \frac{z}{|z|^2} \\ &=& \displaystyle (\partial_x - i\partial_y) \frac{x+iy}{x^2+y^2} \\ &=& \displaystyle \partial_x\frac{x}{x^2+y^2} + \partial_y\frac{y}{x^2+y^2} + i\partial_x\frac{y}{x^2+y^2} - i\partial_y\frac{x}{x^2+y^2} \\ &=& \displaystyle \partial_x\frac{x}{x^2+y^2} + \partial_y\frac{y}{x^2+y^2} \end{array}$$ because the last two terms compensate each other.

• Thank you. For your second proof, I don't understand the following: you say the last two terms compensate each other, which I agree with. However, applying the derivatives on the first two terms also gives me zero... why don't these vanish as well? Commented Mar 20, 2023 at 13:47