Theorem

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

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

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

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

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

Proof

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

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

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

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

Hence:

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

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.