5
$\begingroup$

I have trouble understanding some basic concepts in Complex Analysis:

For $z=x+\mathrm{i}y$, we define: $$\partial \equiv \frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right)$$

The following is stated as obvious: $$\partial \frac{1}{z^{*}}=\pi \delta(x) \delta(y)$$

In order to prove this equality I was told to integrate the left and right hand sides over a small square centered at the origin. However I do not recover the desired result.

My main obstacles is to understand why: $$\int_{-\epsilon}^\epsilon\int_{-\epsilon}^\epsilon\pi \delta(x) \delta(y)\mathrm{d}x\mathrm{d}y\neq\pi\text{ ?}$$ $$\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right)\frac{1}{x-\mathrm{i}y}\neq-\frac{1}{(x-\mathrm{i}y)^2}+\frac{1}{(x-\mathrm{i}y)^2}=0 \text{ ?}$$

Edit: I suspect it has something to do with the fact that $\frac{1}{z^*}$ does not have a series expansion....

$\endgroup$

1 Answer 1

6
$\begingroup$

The key here is to understand that $\partial_z\frac{1}{z^*}=0$ everywhere but at the origin. This is how a delta-singularity can arise in this context. Once one is aware of this, integrating this function on a square shouldn't be too difficult, and can be done without use of multivariable tools like Stokes' theorem. To wit, using the fundamental theorem of calculus and Fubini's theorem we obtain:

$$\int_{[-a,a]\times[-a,a]}\partial_z\frac{1}{z^*}dxdy=\frac{1}{2}\int_{-a}^{a}\int_{-a}^{a}(\partial_x-i\partial_y)\frac{1}{x-iy}dxdy\\\begin{align}&=\frac{1}{2}\int_{-a}^ady\frac{1}{x-iy}\Bigg|_{(-a,y)}^{(a,y)}-\frac{i}{2}\int_{-a}^adx\frac{1}{x-iy}\Bigg|_{(x,-a)}^{(x,a)}\\&=\frac{1}{2}\int_{-a}^a{dy}\frac{2a}{a^2+y^2}-\frac{i}{2}\int_{-a}^{a}dx\frac{2ia}{a^2+x^2}\\&=2a\int_{-a}^a\frac{dy}{a^2+y^2}\\&=\pi\end{align}$$

which shows there is some kind of mass at the origin since the integral is non-zero for arbitrary sizes of the square. Of course, a full proof of the fact that this is a delta-function requires applying this to a test function, but for illustration purposes, this calculation is generally enough.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .