differentiable as an R2 function v.s. as a complex function Let $f: \mathbb C \to \mathbb C$ be a complex function. Denote its real and imaginary part by $u: \mathbb C \to \mathbb R$ and $v: \mathbb C \to \mathbb R$ respectively.
Consider the function $\widetilde f : \mathbb R^2 \to \mathbb R^2$ defined by $\widetilde f (x,y) = (u(x+iy),v(x+iy))$.
I am aware that $f$ is differentiable (in the complex sense, i.e. $\lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$ exists) iff $\widetilde f$ is continuously differentiable and $u$ and $v$ satisfy Cauchy-Riemann equation.
I think if "continuously differentiable" above is replaced by "differentiable (in the real sense, i.e. $\lim_{\mathbf h \to 0} \frac{\lvert \widetilde f(\mathbf x+ \mathbf h)-\widetilde f(\mathbf x) - D \widetilde f \mathbf h\rvert}{\lvert \mathbf h \rvert} = 0$)", the "if" part will not hold.
Can anyone give a counterexample for this? (Or if it holds, can anyone give a proof?)
 A: You can sort this thing one point in the domain at a time. 
Fact: 
$f\colon U \to \mathbb{C}$ is complex differentiable at $z$ 
if and only if 
$f$ is differentiable at $z$ ( as a function from $U$ to $\mathbb{R}^2$ ) and 
the partial derivatives at $z$ ( which exist since $f$ is differentiable at $z$) satisfy the Cauchy-Riemann equations. 
Note that differentiable at $z$ implies continuous at $z$.
Therefore:
$f\colon U \to \mathbb{C}$ is complex differentiable on  $U$
if and only if 
$f$ is differentiable on $U$ ( as a function from $U$ to $\mathbb{R}^2$ ) and 
the partial derivatives  ( which exist since $f$ is differentiable on $U$) satisfy the Cauchy-Riemann equations. 
The first equivalence is elementary and the second follows from the first and definitions.
From the theory of complex functions it follows that a function $f$  is complex differentiable at $U$ if and only if it is complex analytic, that is for every point $z_0$ in $U$ the values of $f$ on an open  disk centered at  $z_0$ and contained in $U$ are given by a power series in $(z-z_0)$.
