if a complex function $f$ is real-differentiable, then $f$ or $\overline{f}$ are complex-differentiable This is an exercise from Remmert's Theory of Complex functions.
Let $D\subset \mathbb{C}$ be a domain and $f:D\rightarrow \mathbb{C}$ a real-differentiable function. Assume that the following limit exists:
$ \mathrm{lim}_{h\rightarrow 0} \left|  \frac{f(c+h) - f(c)}{h}   \right|.$
Show that either $f$ or $\overline{f}$ is complex-differentiable.
I've tried showing that $\frac{\partial f}{\partial \overline{z}} = 0$ or $\frac{\partial \overline{f}}{\partial z} = 0$ by using the fact that there exist continuous functions $g$ and $h$ such that in $D$ one can write 
$f(z) = f(c) + (z-c)g(z) + (\overline{z} - \overline{c})h(z)$ 
and that  $g(c)= f_{z}(c)$ and $h(c) = f_{\overline{z}}$ and then plugging this into the limit above. Does this approach works and I just can´t see how to do it? Can someone give a hint or a guideline solution to this?
 A: The assumption implies that the derivative matrix $A:=Df(c)$ satisfies $|Av| = |Aw|=:a$ for any two unit vectors $v$ and $w$ (where $a$ depends on $c$, but not on $v$ or $w$). This implies that $A$ is a scalar multiple $A = aB$ of an orthogonal matrix $B$, which means that $\frac{\partial f}{\partial z}(c) = 0$ or $\frac{\partial f}{\partial \bar{z}}(c) = 0$, depending on whether $A$ reverses or preserves orientation. In the first case, $\bar{f}$ is complex differentiable, in the second $f$ is.
A: This is not really different from what Lukas Geyer wrote, but is a little less linear-algebraic. 
The fact that $f$ is real differentiable at $c$ can be expressed as $f(c+h)-f(c) = \alpha h+\beta \bar h+o(|h|)$ 
for some $\alpha,\beta\in \mathbb C$. Writing $|\alpha h+\beta \bar h| = |h| |\alpha+\beta (\bar h/h)|$, 
observe that  a certain argument of $h$ yields  $|\alpha+\beta (\bar h/h)| = |\alpha|+|\beta|$, while 
another value of the argument achieves $|\alpha+\beta (\bar h/h)| = ||\alpha|-|\beta||$. Therefore, 
$|\alpha|+|\beta| = ||\alpha|-|\beta||$, which means that either $\alpha$ or $\beta$ is zero. 
