Cauchy-Riemann Equations

Theorem

Let $G\subseteq \mathbb {C}$ 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 $\left(x,y\right)$ and the following Cauchy-Riemann equations hold:

{\dfrac {\partial u}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)\quad \quad {\dfrac {\partial u}{\partial y}}\left(x,y\right)=-{\dfrac {\partial v}{\partial x}}\left(x,y\right)

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

f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)

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 $h:=k+i0\left(k\in \mathbb {R} \right)$ . Then

${\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\&=&\lim \limits _{k\to 0}{\dfrac {u\left(x+k,y\right)+iv\left(x+k,y\right)-u\left(x,y\right)-iv\left(x,y\right)}{k}}\\&=&\lim \limits _{k\to 0}{\dfrac {u\left(x+k,y\right)-u\left(x,y\right)}{k}}+i{\dfrac {v\left(x+k,y\right)-v\left(x,y\right)}{k}}\\&=&{\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)\end{array}}$

Proof - Step 2

Let $h:=0+il\left(l\in \mathbb {R} \right)$ . Then

${\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\&=&\lim \limits _{l\to 0}{\dfrac {u\left(x,y+l\right)+iv\left(x,y+l\right)-u\left(x,y\right)-iv\left(x,y\right)}{il}}\\&=&\lim \limits _{l\to 0}{\dfrac {1}{i}}{\dfrac {u\left(x,y+l\right)-u\left(x,y\right)}{l}}+{\dfrac {v\left(x,y+l\right)-v\left(x,y\right)}{l}}\\&=&{\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)\end{array}}$

Proof - Step 3

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

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.

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