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I used to have a vague feeling that the residue theorem is a close analogy to 2D electrostatics in which the residues themselves play a role of point charges. However, the equations don't seem to add up. If we start from 2D electrostatics given by $$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = \frac{\rho}{\epsilon_0},$$ where the charge density $\rho = \sum_i q_i \delta(\vec{r}-\vec{r}_i)$ consists of point charges $q_i$ located at positions $\vec{r}_i$, and integrate over the area bounded by some curve $\mathcal{C}$, we find (using Green's theorem) $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \frac{1}{\epsilon_0} \sum_i q_i.$$ Now, I would like to interpret the RHS as a sum of residua $2\pi i\sum_i \text{Res}\, f(z_i)$ of some analytic function $f(z_i)$ so that I would have the correspondence $$q_i = 2\pi i\epsilon_0 \text{Res}\, f(z_i).$$ For this to hold, the LHS would have to satisfy $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \int_\mathcal{C} f(z)\, dz,$$ however, it is painfully obvious that the differential form $$E_x\, dy - E_y\, dx = -\frac{1}{2}(E_y+iE_x)dz + \frac{1}{2}(-E_y + iE_x)dz^*$$ can never be brought to the form $f(z)dz$ for an analytic $f(z)$.

So, it would appear that there really isn't any direct analogy between 2D Gauss law and the residue theorem? Or am I missing something?

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2 Answers 2

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential.

In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified with, say, the real part of a holomorphic function. In more detail, let us write (with $z=x+iy$) $$ f(z) = \phi(x,y) + i\ \psi(x,y)\ , $$ where we choose to identify the real part with the electrostatic potential. You can check that Cauchy-Riemann conditions imply that $\mathbf{E}\cdot \nabla \psi=0$. This implies that the $\psi=$ constant lines are the electrostatic field lines.

Adding a point charge, implies that it is harmonic everywhere except at the location of the charge. The relevant function is $f(z) = \lambda \log (z-z_0)$, where $z_0$ is the location of the charge and $\lambda$ is proportional to the charge. The connection with the residue theorem follows since $f'(z)$ has a simple pole at $z_0$ with residue $\lambda$.

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Not a point charge but a homogeneous linear distribution, which is perpendicular to the complex plane. –  auxsvr Apr 22 at 11:20
    
The question is on 2D electrostatics. Yes, one can view a two-dimensional point charge as a line charge distribution in three dimensions (or a planar charge distribution in four dimensions and so on) –  suresh Apr 22 at 12:07
    
How do you find the Coulomb potential in dimension 2? –  auxsvr Apr 22 at 12:57
    
@auxsvr Good point. It is taken to be the solution to Laplace's equation. For "point" charges in 2d, it is $\log(x)$ and $d>2$, it is $1/r^{d-2}$. –  suresh Apr 22 at 13:50

The analogy follows with the right definitions. The "flux" of the "vector" $E(z)$ through a contour $\Gamma$ is $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$.

I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential $\Omega$: $E(z) = -(\mathrm{d}_z \Omega(z))^*$.

So it is the conjugate of $E(z)$, not $E(z)$ itself, that is holomorphic, being given by $E(z)^* = -\mathrm{d}_z\Omega(z)$. Work out the direction of the vectors from the Cauchy-Riemann relations and you'll see that you need a conjugate to make $-\nabla \phi$ equal to the real part of the derivative $-(\mathrm{d}_z\Omega(z))^*$.

Hence $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$ is the real part of an everyday contour integral and the flux is thus the real part of $2\,\pi\,i$ times the sum of residues at the poles of $E(z)^*$, whence the 2D Gauss law follows.

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