---
tag: 0T71SSYZ39
kind: statement
role: theorem
name: Cauchy–Riemann equations
aliases: [cauchy-riemann-conditions]
msc: [30A99]
uses: [0D1C9Y1EXA, 02ZE6JHJ7G]
sources:
  - url: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations
    locator: Complex differentiability
  - url: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations
    locator: lead section
same_as:
  wikidata: Q622741
  wikipedia: 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|02ZE6JHJ7G]] 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|0D1C9Y1EXA]] 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.