Complex differentiability implies real differentiability First, we think of $\mathbb{R}^2$ as related to the complex plane by the following: $(x,y) \leftrightarrow x+iy$.
Show that if $f(x,y)=(u(x,y),v(x,y))$ is complex differentiable at $z_0$, then $f$ is differentiable at $z_0$.  Also, at $z_0$ establish the Cauchy Riemann equations.
I have not taken Complex analysis, but I am familiar with the definition of complex differentiable.
Suppose $f$ is complex differentiable.  Then $\lim\limits_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}=a+bi=\beta$.  Then $\lim\limits_{z \to z_0}|\frac{f(z)-f(z_0)-\beta(z-z_0)}{z-z_0}|=0$.
At the above step, do I just sub in $z=x+yi$ and $z_0=x_0+iy_0$ to show real differentiability?
 A: When our domain is a field ($\mathbb R$ or $\mathbb C$) we have the luxury of defining derivative as the limit of divided differences. Not so in $\mathbb R^2$ (and in other vector spaces), where there is no division of vector by another vector. Thus, we should use the more flexible definition of derivative (equivalent to the usual one in $\mathbb R$ and $\mathbb C$): a function $f:\mathbb R^2\to\mathbb R^2$ is $\mathbb R$-differentiable at  $\vec x\in\mathbb R^2$ if there exists a $2\times 2$ matrix $A$ such that 
$$\lim_{\vec h\to 0}\frac{1}{|\vec h|}\left|f(\vec x+\vec h)-f(\vec x)-A\vec h\right|=0 \tag1$$
Let us also rewrite the divided-difference definition of complex differentiability in a manner similar to (1): a function $f:\mathbb C\to\mathbb C$ is $\mathbb C$-differentiable at  $\vec z\in\mathbb C$ if there exists a complex number $a$ such that 
$$\lim_{h\to 0}\frac{1}{| h|}\left|f(z+ h)-f(z)-ah\right|=0 \tag2$$ 
Thinking of complex numbers as vectors in $\mathbb R^2$, we can represent the linear map $h\mapsto ah$ by the matrix
$$A=\begin{pmatrix} \operatorname{Re}\,a & -\operatorname{Im}\,a \\ \operatorname{Im}\,a & \operatorname{Re}\,a \end{pmatrix} \tag3$$ 
With this interpretation, (2) is clearly a special case of (1). 

It is also instructive to consider the path from (1) to (2). If the matrix $A$ happens to be antisymmetric with constant diagonal, then we can form the complex number $a$ as in (3), which, according to (2), is the complex derivative $f'(z)$. Thus, the following are equivalent: 


*

*$f$ is $\mathbb C$-differentiable at $z$

*$f$ is $\mathbb R$-differentiable at $z$, and its derivative matrix is antisymmetric with constant diagonal


Of course, "antisymmetric with constant diagonal" says precisely that the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$ hold at the point of interest.
