Cauchy–Riemann equations

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be a function. Identify $\mathbb{C}$ with $\mathbb{R}^2$ by $x + i y \leftrightarrow (x, y)$. Let $u = \operatorname{Re} f$ and $v = \operatorname{Im} f$ be the real and imaginary parts of $f$, as real functions of $(x, y)$ on $U$. Let $z_0 = x_0 + i y_0 \in U$. Then $f$ is complex differentiable at $z_0$ if and only if the following two conditions hold. First, $u$ and $v$ are differentiable at $(x_0, y_0)$ as functions of two real variables. Second, they satisfy $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$$ at $(x_0, y_0)$. In this case $f'(z_0) = \frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)$.

Remarks

For $f(z) = z^2$ we have $u = x^2 - y^2$ and $v = 2xy$. Then $u_x = 2x = v_y$ and $u_y = -2y = -v_x$. The real differentiability of $u$ and $v$ is necessary. The function $u(x, y) = \sqrt{|xy|}$, $v = 0$ satisfies the two equations at $(0, 0)$, but $u + iv$ is not complex differentiable at $0$. The two equations together are the one complex equation $\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$.

Explanations

L1 human

A complex derivative must be the same in each direction. If you compare the derivative along the real axis with the derivative along the imaginary axis, you get the two equations. So the equations are a test with partial derivatives for a complex derivative. For example, $\bar z = x - i y$ has $u_x = 1$ and $v_y = -1$, so it fails the test.