How to define a delta function on complex plane?

Question

I understand that it makes perfect sense to define a 2-dimensional delta function on the complex plane by $$\int dz\wedge d\bar{z}\delta(z)\delta(\bar{z})=1.$$ However, is there any chance to define a 1-dimensional holomorphic delta function $\delta(z)$, which equals to 1 under certain kind of integration, other than $\frac{1}{2\pi i}\frac{1}{z}$?

Answer 1

There's two kinds of delta functions for contour integrals.

One will look like the standard sort of delta function: it has the property that $\int_Cf(z)\delta(z-z_0)dz = f(z_0)$ as long as the contour $C$ passes through the point $z_0$, and is zero otherwise.

The other is $1/(2\pi iz)$, the "delta function" with the property that $\int_C f(z) \delta(z-z_0)dz = f(z_0)$ as long as the contour $C$ is closed and $z_0$ is on the interior. But this doesn't look like the standard sort of delta function at all. What gives?

To illustrate, consider the following 2D vector field. $$ \mathbf{A}(x,y) = \frac{-y\hat{\mathbf{x}} +x\hat{\mathbf{y}}}{2\pi(x^2+y^2)} $$ You'll find that the line integral $\oint \mathbf{A}(x,y)\cdot d\boldsymbol\ell = 1$ as long as the path surrounds the origin and zero otherwise. So $\mathbf{A}(x,y)$ is acting like a delta function in the way $1/(2\pi i z)$ is. And now you might remember Stokes' theorem, and notice that $\mathrm{curl} \,\mathbf{A} = \delta(x)\delta(y)$.

So what's the analog of curl in the complex plane? Well, if we consider a holomorphic complex valued function $f(x+iy)$, then the derivative of the function is $$ \frac{df}{dz} = \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y} $$ and the Cauchy-Riemann equations require $$ \frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} = 0. $$ This last operation, the "conjugate derivative", is the curl analog. You'll find that $1/(2\pi i z)$ has zero conjugate derivative everywhere except the origin, where it's undefined. And it's undefined in just the right way so that $\int_Cdz/(2\pi iz) =1$ for all contours $C$ that surround the origin. And that's how it looks like a "standard" delta function.

Answer 2

I want to add a third useful kind of complex delta distribution to the two mentioned by eyeballfrog. It is a 'holomorphic' version of the first $\delta$ mentioned by them, which integrates to 1 if the function passes through 0. 'holomorphic' means that it is obtained as a limit of holomorphic functions and in consequence the Cauchy integral theorem holds. The price we pay for this is that $\delta$ can't vanish everywhere away from the origin. Here is the definition: $$\delta(z) = \lim_{\lambda \rightarrow \infty} \sqrt{\lambda \over \pi} e^{-\lambda z^2}$$ This is divergent in the quadrants $|\Re z|\leq |\Im z|$ and converges to zero everywhere else. But if we have an integration contour with both endpoints in the convergent sector, we can deform the contour to pass through the origin and get $\int_C dz f(z)\delta(z) = f(0)$ if for instance the contour starts with $-|\Im z| > \Re z < 0$ and ends with $|\Im z| < \Re z > 0$ and $f$ is holomorphic.

I encountered this in Berry's paper "Faster than Fourier", and it's expanded in "Delta function expansions, complex delta functions and the steepest descent method" by Lindell.