Cauchy-Riemann Equations

Theorem
Let $$G \subseteq \mathbb{C}$$ be an open subset. Let the function $$ f = u + i v $$ 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:

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

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

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

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

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

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

Hence:

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

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.