Complex differentiation of a function Assume $ f: \mathbb{C} \to \mathbb{C} $ is $ \mathbb{C} $ differentiable. Denote its derivative as $ f'(z) $.
Next, if we denote $ f(z)=u(z)+iv(z) $ where $u,v $ are functions to $ \mathbb{R} $, then the partial derivatives of $$ f\left(x,y\right)=\begin{pmatrix}u\left(x,y\right)\\
v\left(x,y\right)
\end{pmatrix} $$
Satisfies:
$$ \frac{\partial f}{\partial x}\left(z\right)=-i\frac{\partial f}{\partial y}\left(z\right)=f'\left(z\right) $$
Why is this true?
I couldnt find formal proof for this equation, just intuitive explanations (that werent very convincing).
I'll be glad to see a formal proof for this equation.
Thanks in advance.
 A: Since $f'(z)=\lim_{h\to0}\frac{f(z+h)-f(z)}h$, then, in particular,$$f'(z)=\lim_{h\to0,\,h\in\Bbb R}\frac{f(z+h)-f(z)}h=\frac{\partial f}{\partial x}(z).$$On the other hand, if $z=a+bi$, then$$\frac{\partial f}{\partial x}(z)=\frac{\partial u}{\partial x}(a,b)+\frac{\partial v}{\partial x}(a,b)i.\tag1$$But you can use here the Cauchy-Riemann equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. So, $(1)$ tells you that\begin{align}\frac{\partial f}{\partial x}(z)&=\frac{\partial v}{\partial y}(a,b)-\frac{\partial u}{\partial y}(a,b)i\\&=-\left(\frac{\partial u}{\partial y}(a,b)+\frac{\partial v}{\partial y}(a,b)i\right)i\\&=-i\frac{\partial f}{\partial y}(z).\end{align}If you want to use this to deduce the Cauchy-Riemann equations, then note that you also have\begin{align}f'(z)&=\lim_{h\to0,\ h\in\Bbb R}\frac{f(z+hi)-f(z)}{hi}\\&=-i\lim_{h\to0,\ h\in\Bbb R}\frac{f(z+hi)-f(z)}h\\&=-i\frac{\partial f}{\partial y}(z).\end{align}
