# 柯西－黎曼方程

(1a)     ${\displaystyle {\partial u \over \partial x}={\partial v \over \partial y}}$

(1b)    ${\displaystyle {\partial u \over \partial y}=-{\partial v \over \partial x}.}$

## 注釋和其他表述

### 共形映射

(2)    ${\displaystyle {i{\partial f \over \partial x}}={\partial f \over \partial y}.}$

${\displaystyle {\begin{pmatrix}a&-b\\b&\;\;a\end{pmatrix}},}$

### 複共軛的獨立性

(3)    ${\displaystyle {\frac {\partial f}{\partial {\bar {z}}}}=0}$

${\displaystyle {\frac {\partial }{\partial {\bar {z}}}}={\frac {1}{2}}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right).}$

### 複可微性

${\displaystyle f(z)=u(z)+iv(z)}$

${\displaystyle \lim _{\underset {h\in \mathbb {C} }{h\to 0}}{\frac {f(z_{0}+h)-f(z_{0})}{h}}=f'(z_{0})}$

${\displaystyle \lim _{\underset {h\in \mathbb {R} }{h\to 0}}{\frac {f(z_{0}+h)-f(z_{0})}{h}}={\frac {\partial f}{\partial x}}(z_{0}).}$

${\displaystyle \lim _{\underset {ih\in i\mathbb {R} }{h\to 0}}{\frac {f(z_{0}+ih)-f(z_{0})}{ih}}=\lim _{\underset {ih\in i\mathbb {R} }{h\to 0}}-i{\frac {f(z_{0}+ih)-f(z_{0})}{h}}=-i{\frac {\partial f}{\partial y}}(z_{0}).}$

f沿着兩個軸的導數相同也即

${\displaystyle {\frac {\partial f}{\partial x}}(z_{0})=-i{\frac {\partial f}{\partial y}}(z_{0}),}$

### 物理解釋

${\displaystyle {\bar {f}}={\begin{bmatrix}u\\-v\end{bmatrix}}}$

${\displaystyle {\frac {\partial (-v)}{\partial x}}-{\frac {\partial u}{\partial y}}=0.}$

${\displaystyle {\frac {\partial u}{\partial x}}+{\frac {\partial (-v)}{\partial y}}=0.}$

### 其它解釋

${\displaystyle {\frac {\partial u}{\partial s}}={\frac {\partial u}{\partial n}},\quad {\frac {\partial u}{\partial n}}=-{\frac {\partial u}{\partial s}}}$

${\displaystyle {\partial u \over \partial r}={1 \over r}{\partial v \over \partial \theta },\quad {\partial v \over \partial r}=-{1 \over r}{\partial u \over \partial \theta }.}$

${\displaystyle {\partial f \over \partial r}={1 \over ir}{\partial f \over \partial \theta }.}$

## 非齊次方程

${\displaystyle {\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}=\alpha (x,y)}$
${\displaystyle {\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}=\beta (x,y)}$

${\displaystyle {\frac {\partial f}{\partial {\bar {z}}}}=\phi (z,{\bar {z}})}$

${\displaystyle f(\zeta ,{\bar {\zeta }})={\frac {1}{2\pi i}}\iint _{D}\phi (z,{\bar {z}}){\frac {dz\wedge d{\bar {z}}}{z-\zeta }}}$

## 推廣

### Goursat定理及其推廣

f = u+iv為複函數，作為函數f : R2R2可微。則柯西積分定理（柯西－古爾薩定理）斷言f在開域Ω上解析若且唯若它在該域上滿足柯西-黎曼方程（Rudin 1966，Theorem 11.2）。特別是，f不需假定為連續可微（Dieudonné 1969，§9.10, Ex. 1）。

f在整個域Ω上滿足柯西-黎曼方程是要點。可以構造在一點滿足柯西-黎曼方程的連續函數，但它不在該點解析（譬如，f(z) = z5/|z|4）。只滿足柯西-黎曼方程也是不夠的，（需額外滿足連續性），下面的例子表明了這一點：（Looman 1923，p.107）

${\displaystyle f(z)={\begin{cases}\exp(-z^{-4})&\mathrm {if\ } z\not =0\\0&\mathrm {if\ } z=0\end{cases}}}$

• f(z)在開域Ω⊂C上局部可積，並以弱形式滿足柯西-黎曼方程，則f和Ω上的一個解析函數幾乎處處相等。

### 多變量的情況

${\displaystyle {\bar {\partial }}}$

${\displaystyle {\partial f \over \partial {\bar {z}}}=0}$,

${\displaystyle {\partial f \over \partial {\bar {z}}}={1 \over 2}\left({\partial f \over \partial x}-{1 \over i}{\partial f \over \partial y}\right).}$