Cauchy-Riemann Equations

Theorem

Let $G\subseteq ℂ$ be an open subset. Let the function $f=u+iv$ be differentiable at a point $z=x+iy\in G$. Then all partial derivatives of $u$ and $v$ exist at $(x,y)$ and the following Cauchy-Riemann equations hold:

$$\frac{\partial u}{\partial x}(x,y)=\frac{\partial v}{\partial y}(x,y)\frac{\partial u}{\partial y}(x,y)=-\frac{\partial v}{\partial x}(x,y)$$

In this case, the derivative of $f$ at $z$ can be represented by the formula

$f^{\prime }\left(z\right)=\frac{\partial u}{\partial x}(x,y)-i\frac{\partial u}{\partial y}(x,y)=\frac{\partial v}{\partial y}(x,y)+i\frac{\partial v}{\partial x}(x,y)$

Proof

The proof can be decomposed into 3 main steps:

• calculate the partial derivative for the real part,
• calculate the partial derivative for the imaginary part,
• due to property of $f$ being complex differentiable both derivatives yield the same complex value. This leads to Cauchy-Riemann equations.

Proof - Step 1

Let $h:=k+i0(k\in ℝ)$. Then

$\begin{array}{ccc}{f}^{\prime }\left(z\right)& =& \lim {\mathrm{\backslash limits}}_{h\to 0}\frac{f(z+h)-f\left(z\right)}{h}\\ & =& \lim {\mathrm{\backslash limits}}_{k\to 0}\frac{u(x+k,y)+iv(x+k,y)-u(x,y)-iv(x,y)}{k}\\ & =& \lim {\mathrm{\backslash limits}}_{k\to 0}\frac{u(x+k,y)-u(x,y)}{k}+i\frac{v(x+k,y)-v(x,y)}{k}\\ & =& \frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\end{array}$

Proof - Step 2

Let $h:=0+il(l\in ℝ)$. Then

$\begin{array}{ccc}{f}^{\prime }\left(z\right)& =& \lim {\mathrm{\backslash limits}}_{h\to 0}\frac{f(z+h)-f\left(z\right)}{h}\\ & =& \lim {\mathrm{\backslash limits}}_{l\to 0}\frac{u(x,y+l)+iv(x,y+l)-u(x,y)-iv(x,y)}{il}\\ & =& \lim {\mathrm{\backslash limits}}_{l\to 0}\frac{1}{i}\frac{u(x,y+l)-u(x,y)}{l}+\frac{v(x,y+l)-v(x,y)}{l}\\ & =& \frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)\end{array}$

Proof - Step 3

Both partial derivatives must provide a same complex value due to the fact that $f$ is complex differentiable: $f^{\prime }\left(z\right)=\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)=\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)$

Proof - Step 4

Equating the real and imaginary parts, we get the Cauchy-Riemann equations. The representation formula follows from the above line and the Cauchy-Riemann equations.

See also

• Complex Analysis

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