Is there any intuition or meaning regarding Cauchy-Riemann equations?

Question

Holomorphic functions are one of the most beautiful objects in mathematics. However, Cauchy-Riemann equations are a little bit of a mystery for me. What does that mean intuitively when a function satisfyes those equations?

Answer 1

A matrix of the form $A=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ is a pure scaling and rotation (it does not scale different directions differently, nor does it have any "shear"). This is exactly the type of operation that is represented by multiplication by a complex number. Indeed you can identify $A \begin{bmatrix} x \\ y \end{bmatrix}$ with $(a+bi)(x+yi)$, using the usual identification $\begin{bmatrix} x \\ y \end{bmatrix} \leftrightarrow x+yi$.

What the Cauchy-Riemann equations say is that the Jacobian of the functions $u(x,y),v(x,y)$ is of this form (the diagonal elements are equal, the off-diagonal elements are exactly opposite) which means that it can be represented as multiplication by some complex number. This is needed in order to have the derivative of your complex function at a point be another complex number rather than a 2x2 matrix, as it would be for a generic differentiable $\mathbb{R}^2$-valued function of a $\mathbb{R}^2$ variable.

Answer 2

They're oddly absent from most treatments of complex analysis, but I like the Wirtinger derivatives, which also make Cauchy--Riemann transparent.

The idea is that any complex function $f(x+iy)$ of two real variables $x, y$ can be thought of instead as a function $f(z, \overline{z})$ where $z = x+iy$ and $\overline{z} = x-iy$. For example, the "modulus squared" function is $$f(x+iy) = |x+iy|^2 = x^2 + y^2 = z\overline{z}.$$

This substitution makes it natural to differentiate with respect to $z$ or $\overline{z}$ instead of $x$ or $y$. Formally, $$\begin{align*}\frac{\partial}{\partial z} &= \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right) \\ \frac{\partial}{\partial \overline{z}} &= \frac{1}{2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right).\end{align*}$$

As a sanity check, $\frac{\partial}{\partial z} z = 1$, $\frac{\partial}{\partial z} \overline{z} = 0$, $\frac{\partial}{\partial \overline{z}} z = 0$, $\frac{\partial}{\partial \overline{z}} \overline{z} = 1$. The usual product rule works for both of these derivatives, so for instance $$\frac{\partial}{\partial \overline{z}} z\overline{z} = \left(\frac{\partial}{\partial \overline{z}} z\right)\overline{z} + z \left(\frac{\partial}{\partial \overline{z}} \overline{z}\right) = z.$$

More generally, as far as polynomials in $z$ and $\overline{z}$ are concerned, we may treat $z$ and $\overline{z}$ as independent variables exactly as if we were applying $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ to a bivariate polynomial in independent variables $x$ and $y$.

The Cauchy--Riemann equations say precisely that $\frac{\partial}{\partial \overline{z}} f = 0$. The proof is easy: $$\begin{align*} 2\frac{\partial}{\partial \overline{z}} (u+iv) &= u_x + i u_y + iv_x - v_y \\ &= (u_x - v_y) + i(u_y + v_x) = 0+0i \\ &\Leftrightarrow u_x = v_y \text{ and } u_y = -v_x. \end{align*}$$

This relates very nicely to holomorphic functions as limits of polynomials. Namely, given a bivariate polynomial $\sum_{i, j=0}^N c_{i, j} z^i \overline{z}^j$, the derivative with respect to $\overline{z}$ is 0 if and only if $c_{0, j} = 0$ for all $j \geq 1$, i.e. if and only if it was a polynomial in $z$ after all.

In this sense, the Cauchy--Riemann equations are just saying we're restricting to limits of polynomials in $z$ and not polynomials in $z$ and $\overline{z}$.

Answer 3

A function $f:\Bbb C \to \Bbb C$ may be seen as a map $f:\Bbb R^2\to \Bbb R^2$. When differentiable, you can look at the total derivative $Df(x,y):\Bbb R^2\to \Bbb R^2$, which is a $\Bbb R$-linear map. But you started with a complex function, so $\Bbb R$-linearity won't cut it. You want to set back $\Bbb R^2 = \Bbb C$, $z=x+iy$ and say that $Df(z):\Bbb C \to \Bbb C$ is a $\Bbb C$-linear map. When can you say this? Precisely when the Cauchy-Riemann equations are satisfied. In which case $Df(z)$, as a linear map between $1$-dimensional (complex) vector spaces, is multiplication by a (complex) scalar. What is this scalar? $f'(z)$.

Bottom line: the Cauchy-Riemann equations measure the deviation from $Df(z)$ being $\Bbb C$-linear.

Answer 4

Preliminary fact: any complex number $w$ can be represented as $$w= a+ib \sim \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \qquad a,b \in \mathbb{R}$$ This representation respects the algebraic rules of $ \mathbb{C}$. E.g., check that $i^2=-1$: $$ 1 \sim \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad i \sim \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \quad \Rightarrow \quad i^2 \sim \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \sim -1 $$

The Cauchy-Riemann equations are the requirement that, given the complex function $f(z)$, the derivative $df/dz$ is a complex number itself: given $f=f_R(x,y)+i f_I(x,y)$ and $z=x+iy$, the Jacobian matrix must represent a complex number, namely $$ \begin{bmatrix} \partial_x f_R & \partial_x f_I \\ \partial_y f_R & \partial_y f_I \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \qquad \Leftrightarrow \qquad \partial_x f_R=\partial_y f_I \qquad \partial_x f_I=-\partial_y f_R \, . $$

Answer 5

There are two intuition I can give:

1. Analytic function will map infinitesimal square to square/ preserves orthogonality.

Explanation: Suppose $p$ is a point on the complex plane, then $\epsilon$ be an infinitesimal complex vector parallel to the $x$ axis and $i\epsilon$ one to $y$. Under the complex mapping $f$, they will be mapped to vectors $ \epsilon \partial_x f$ and $ \epsilon \partial_y f$, if squares go to squares it must mean that rotating $\epsilon \partial_x f$ by ninty degree gives $\epsilon \partial_y f$. We have from that:

$$ i \partial_x f = \partial_y f $$

Letting $f=u+iv$, we get by equating components:

$$ -\partial_x v= \partial_y u$$

$$ \partial_x u = \partial_y v$$

More details of this approach is discussed in Tristan Needham's Visual Complex Analysis book.

2. We can think of this in terms of complex valued differential forms. We know that the integral of a one form between two point is path independent if it's exterior derivative is vanishing. Consider the complex one form $fdz$, if we take the exterior derivative and impose the previously mentioned condition:

$$ d(fdz)= df \wedge dz = (\partial_x f dx+ \partial_y f dy) \wedge (dx +idy)=0$$

We work out second last equality:

$$i \partial_x f dx \wedge dy - \partial_y f dx \wedge dy = 0 $$

We have,

$$i \partial_x f = \partial_y f$$

Expanding we get CR back. The upshot of this approach is it is directly clear what the connection with Cauchy's theorem is. Secondly, we can also use the idea that exterior derivative is a local integration to get more geometry out of it.

Bonus: The forms approach also explains $ \overline{z} dz$ gives the area when integrate over a contour $\partial C$

$$ \int_{ \partial C} \overline{z} dz = \int_C d(\overline{z} dz) = \int_C 2i dx \wedge dy= 2i A$$

Answer 6

For a function of a real variable to have a derivative, the limit of the difference quotient from the left or from the right exists, and the two are equal.

For a function $f$ of a complex variable to have a derivative, the limits of the difference quotient from any direction are equal. This is a much stronger requirement. Necessary, and sufficient to just take the directions along the $x-$ and $y-$axes.

The limits are partial derivatives of the real and imaginary parts of $f$.