Whats the connection between functions with curl 0 and holomorphic functions

Solution 1:

The relationship here is one that comes out in geometric calculus. There, we consider the 2d plane as having 3 kinds of objects: scalars, vectors, and bivectors, which serve the role of the imaginary of complex analysis.

Here's how this works: we have a product of vectors such that $e_x e_y = -e_y e_x$, like the cross product, but $e_x e_x = e_y e_y = 1$, like the dot product. This "geometric product" produces a bivector, which represents an oriented plane, just as vectors represent oriented lines or directions.

Call $i = e_x e_y$, and let $f = u + iv$, in analogy to complex analysis. The condition $\nabla f= 0$ reproduces the Cauchy-Riemann condition. Explicitly, this is

$$\begin{align*} \nabla f &= (e_x \partial_x + e_y \partial_y )f \\ &= e_x \partial_x u + e_x e_x e_y \partial_x v + e_y \partial_y u + e_y e_x e_y \partial_y v \\ &= e_x \partial_x u + e_y \partial_x v + e_y \partial_y u - e_x \partial_y v \\ &= e_x (\partial_x u - \partial_y v) + e_y (\partial_y u + \partial_x v)\end{align*}$$

This is a well-known result: that $\nabla$ corresponds more to $\partial/\partial \bar z$ than it does with $\partial/\partial z$. What's interesting is that the integrability condition of $\nabla f = 0$ applies in higher dimensions also, even when the notion of complex differentiation is no longer clear.

To find the relationship between $f$ and a vector field $F$, multiply by $e_x$ on the right so $F = f e_x = u e_x - v e_y = U e_x + V e_y$, with $V = -v$. This generates the integrability condition

$$\nabla F = (\partial_x U + \partial_y V) + i (\partial_x V - \partial_y U)$$

The scalar part is the divergence and the bivector part is the curl. Both of these must be zero to satisfy the same integrability condition as the Cauchy-Riemann condition.

Edit: you might find that the vector field has a negative $y$-component compared to the complex function, but this is also a known result when converting between, say, velocity fields in 2d and complex functions. Geometric calculus only helps prove it mathematically.

Solution 2:

An integral $\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,dt$ can be written as the sum of a real line integral and an imaginary line integral. The vector fields showing up in these line integrals are $(u,-v)$ and $(v,u)$ where $f =u+iv$. The Cauchy-Riemann equations then say that these two vector fields have curl zero.