Let's consider a two-dimensional electric field $\textbf{E}=\textbf{E}(\mathbf x)$, where $\mathbf x\in \mathbb R^2$, and $\textbf{E}$ is a vector representing the direction and strength of the field at that point.

As a vector field, $\textbf{E}$ is the gradient of a scalar field $\phi(\mathbf x)$, the electric potential at point $\mathbf x$: $$\textbf{E}(\mathbf x)=\nabla\phi(\mathbf x). $$

As E is the gradient of a scalar field, the curl of $E$ should be zero.

Now let's bring in complex numbers. The input and output of $E(\mathbf x)$ are both two-dimensional vectors; therefore the input and output can be written as two complex numbers. So we can define that $E(z)$ as a complex-valued function defined on $\mathbb C$.

Since the curl of $\textbf{E}$ is zero, $\frac{\partial E_x}{\partial y}=\frac{\partial E_y}{\partial x}$. This is very similar to one of the Cauchy Riemann equations.

The Cauchy-Riemann equations are $$ u_x'=v_y'\\ v_x'=-u_y'. $$ More details about Cauchy-Riemann Equations.


I hope that $\textbf{E}(z)$ is analytic, but apparently, that's not necessarily the case. Are there any ways to change this a bit so that $\textbf{E}(z)$ is analytic? Can electric fields be studied using analytic functions?

This is not the type of thing I see in most textbooks, so it is very difficult to explain what I mean. Please make sure you understand what I mean first before leaving a comment or a vote. If I am not clear, please ask me to clarify it.

I don't think this is a duplicate: although there are many posts on PSE about complex numbers, they are often very broad and many of them are just about simple things such as expressing sine waves with complex exponentials.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/197307 $\endgroup$ – Sean E. Lake Apr 25 '19 at 7:58
  • $\begingroup$ @SeanE.Lake That question is very broad. It is about all applications of complex numbers in physics. My question is very specific: it is about electric fields. I cannot see why it is a duplicate. $\endgroup$ – Ma Joad Apr 25 '19 at 8:09
  • $\begingroup$ Well, your question may not duplicate the question, exactly, but the accepted answer could easily be an answer to this one. Short summary: yes. $\endgroup$ – Sean E. Lake Apr 25 '19 at 8:10
  • $\begingroup$ I've changed your $E$'s to $\textbf{E}$. Are you asking whether $E_x(z)$ and $E_y(z)$ separately analytic or not? @MaJoad $\endgroup$ – SRS Apr 25 '19 at 8:15
  • $\begingroup$ $E(z)$ outputs a complex number. I am asking whether $E(z)$ as a whole have something to do with complex analytic function. You may answer anything related to this. $\endgroup$ – Ma Joad Apr 25 '19 at 8:25

$\def\RR{\Bbb R^2} \def\CC{\Bbb C} \def\bE{{\bf E}} \def\D#1#2{{d#1 \over d#2}} \def\PD#1#2{{\partial#1 \over \partial#2}} \def\curl{\mathop {\rm curl}} \def\div{\mathop {\rm div}}$ Warning. This is a re-edited version, where the main change was redefining the relationship between $\bE$ and the holomorphic function $E(z)$. This was necessary because the Cauchy-Riemann equations had been written with the wrong sign. As a consequence some signs had to be changed here and there. I hope not having so introduced other unintentional misprints.

A mathematical clarification. A 2D electric field is a vector field on a 2D real vector space, in other words a mapping $\ \RR\to\RR$. The complex field $\CC$ is a vector space isomorphic to $\RR$ (as a vector field). It has a richer algebraic structure, since in $\CC$ a multiplication is also defined, with properties that make it into a field. But this doesn't cancel its being an $\RR\!$ vector space.

Therefore it's perfectly legitimate to speak of our electric field as a function $E:\ \CC\to\CC$. At the same time coordinates $x$, $y$ in $\RR\!$ are $\Re z$, $\Im z$, with $z\in\CC$. Components $E_x$, $E_y$ could be identified with $\Re E$ and $\Im E$, if not for a sign problem.

You aren't exactly right that $\curl \bE=0$ may be seen as one of the Cauchy-Riemann equations. Actually it says $$\PD{}y\,{\Re E} = -\PD{}x\,{\Im E}$$ so we have to adjust a sign, defining $$E_x = \Re E \quad{\rm but}\quad E_y = -\Im E.$$

What about the second C-R equation? let's try: $$\PD{}x\,{\Re E} = \PD{}y\,{\Im E}.$$ $$\PD{E_x}x = -\PD{E_y} y$$ $$\div \bE = 0.$$ We got it! This is another Maxwell equation $\bE$ obeys in an uncharged homogeneous dielectric.

Finally, about electrostatic potential. It would be nice if we could write $$E = -\D Wz$$ with $W$ some holomorphic function. Observe that $$-\D Wz = -\PD{}x\,(\Re W + i\,\Im W)$$ and (since $W$ is holomorphic) the RHS can also be written $$-\PD{}x\,\Re W + i\,\PD{}y\,\Re W.$$ Then $$E_x = -\PD{}x\,\Re W \qquad E_y = -\PD{}x\,\Re W$$ $$\Re W = \phi.$$

It remains to understand the physical meaning of $\Im W$. Would you like to try? Hint: look for an interpretation of its level curves.

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  • $\begingroup$ Good Answer. Should the expression of $E_y$ start with $\partial/\partial y$? You seem to be making a mistake there. $\endgroup$ – Ma Joad Apr 26 '19 at 7:28
  • $\begingroup$ Also, I don't think $\partial \Re E/ \partial y = \partial \Im E/ \partial x$ is a Cauchy-Reimann equation. Shouldn't it be $\partial \Re E/ \partial x = \partial \Im E/ \partial y$? The latter is certainly the version is most texts. How does that work out? $\endgroup$ – Ma Joad Apr 26 '19 at 7:37
  • $\begingroup$ @MaJoad You're right in both comments. The former was a trivial misprint but the latter is much worse: I totally mistook all signs. I'd already seen that and hoped to correct my post before somebody noticed that, but you've been faster :-) I'm going to entirely rewrite it. $\endgroup$ – Elio Fabri Apr 26 '19 at 8:45
  • $\begingroup$ But I think it is not a big deal, because it is close enough. No matter how hard we try, we cannot change the equation about curl into one of Cauchy-Riemann, so there is no need to struggle with it. We can actually re-define $E$ to swap its real and imaginary part, so we can overcome the problem. $\endgroup$ – Ma Joad Apr 26 '19 at 8:49
  • $\begingroup$ @MaJoad Close but confusing just because it's close. There is more than one solution. Swapping $E$'s components is one, but I prefer to put $E_y=-\Im E$. This saves $\Re W=\phi$. $\endgroup$ – Elio Fabri Apr 26 '19 at 9:44

I am not sure about the question. But I'll give it a try. Since an electrostatic field is irrorational i.e.$\boldsymbol{\nabla}\times\textbf{E}(\textbf{x})=0$, it amounts to $$\frac{\partial E_x}{\partial y}=\frac{\partial E_y}{\partial x}.\tag{1}$$ Please note that $$ \frac{\partial }{\partial z} = \frac{1}{2}\left( \frac{\partial }{\partial x}- i\frac{\partial}{\partial y} \right), \quad \frac{\partial }{\partial z^*} = \frac{1}{2}\left( \frac{\partial }{\partial x} + i\frac{\partial}{\partial y} \right). $$ Now, in Cartesian coordinates the electric field can be decomposed as $$\textbf{E}(\textbf{x})=\textbf{E}(x,y)=\hat{\bf{x}}E_x(x,y)+\hat{\bf{y}}E_y(x,y)$$ when written in terms of $(x,y)$. Writing in terms of a different set of independent variables $z,z^*$, $$E_x(x,y)\to \tilde{E}_x(z,z^*),~{\rm and}~E_y(x,y)\to \tilde{E}_y(z,z^*)\tag{2}$$ where '$\tilde{}$' signifies that the functional dependence, in general, is different. Now the relation $(1)$ becomes $$i\Big(\frac{\partial}{\partial z}-\frac{\partial}{\partial z^*}\Big)\tilde{E}_x=\Big(\frac{\partial}{\partial z}+\frac{\partial}{\partial z^*}\Big)\tilde{E}_y.\tag{3}$$

Now, $f$ is an analytic function of $z$ if $$\frac{\partial f}{\partial z^*}=0.\tag{4}$$ When $\tilde{E}_{x,y}$ are analytic functions, they'll satisfy $\frac{\partial \tilde{E}_x}{\partial z^*}=\frac{\partial \tilde{E}_y}{\partial z^*}=0.$ Therefore, $(3)$ simplifies to $$\frac{\partial}{\partial z}(\tilde{E}_x+i\tilde{E}_y)=0.$$ Therefore, for $E_x$ and $E_y$ to be analytic, the necessary condition is that $\tilde{E}_x+i\tilde{E}_y$ must be independent of $z$.

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  • $\begingroup$ That's not a very useful result. The "necessary condition" at last implies that $E$ is constant everywhere since $E_x$ and $E_y$ are forced to be constant. $\endgroup$ – Ma Joad Apr 26 '19 at 7:55
  • $\begingroup$ @MaJoad No, that's not correct. In this formalism, $z$ and $z^*$ are treated as independent variables, so there can still be dependence on the latter. $\endgroup$ – Emilio Pisanty Apr 27 '19 at 13:44
  • $\begingroup$ That said, @SRS, your final conclusion is badly worded and could be significantly improved. $\endgroup$ – Emilio Pisanty Apr 27 '19 at 13:45

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