Proving Complex Integral Identity: \ln|z|^2

[ismaili]

Homework Statement

Prove the following identity,
$$\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}\ln|z|^2 = 2\pi \delta^2(z,\bar{z})$$
where the delta function is defined such that
$$\int dz d\bar{z} \detla^2(z,\bar{z}) = 1$$

Homework Equations

The Attempt at a Solution

While $$z$$ is not zero, the identity is easily seen to be hold, because,
$$\ln|z|^2 = \ln z + \ln\bar{z}$$
So both sides are zero.
To include the point $$z=0$$, I tried to integrate both sides,
$$\int dzd\bar{z}...$$
The right hand side is obviously $$2\pi$$.
But I don't know how to deal with the left hand side?

Anyone got any ideas?
Thanks!

 

[mighty2000]

Thank you for bringing this interesting identity to our attention. I would like to provide a proof for this identity using some basic concepts from complex analysis and the properties of the delta function.

Firstly, we can rewrite the given identity as:

\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}\ln|z|^2
= 2\pi \delta^2(z,\bar{z}) = 2\pi \delta(z)\delta(\bar{z})

where the second equality follows from the definition of the two-dimensional delta function.

Now, let's consider the function f(z,\bar{z}) = \ln|z|^2. This function is analytic everywhere except at z = 0, where it has a singularity. We can therefore use the Cauchy-Riemann equations to express the derivatives of f in terms of its partial derivatives:

\frac{\partial}{\partial z} f = \frac{1}{2}\left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right)
\frac{\partial}{\partial\bar{z}} f = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)

where z = x+iy.

Substituting these expressions into the original identity, we get:

\frac{1}{4}\left(\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} - i\frac{\partial^2 f}{\partial x\partial y} - i\frac{\partial^2 f}{\partial y\partial x}\right) = 2\pi \delta(z)\delta(\bar{z})

Now, we can use the fact that \frac{\partial}{\partial x}\delta(x) = -\delta'(x) and \frac{\partial}{\partial y}\delta(y) = -\delta'(y) to simplify the above expression as:

\frac{1}{4}\left(-\delta'(x) + \delta'(y) - i\delta''(x) -i\delta''(y)\right) = 2\pi \delta(z)\delta(\bar{z})

Using the definition