# 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?

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i dont think this is true (as stated in the title). if $f:\mathbb{R}^2\to\mathbb{R}^2$ is differntiable, it has a derivative $$df=\left( \begin{array}{cc} a&b\\ c&d\\ \end{array} \right)$$ if this is complex differentiable, then we need $$df=\left( \begin{array}{cc} a&-b\\ b&a\\ \end{array} \right)$$ –  yoyo Mar 29 '11 at 0:53
I'm aware of that. I'm sorry for the misleading title. But the exercise itself should clarify what I'm talking about. –  Chu Mar 29 '11 at 3:57
If you tweak a function like f(x,y)=($x^2$-$y^2$, 4xy) ; this is real-differentiable, since the partials exist and are continuous, but it is not complex-differentiable; the left part is the real part of $z^2$, but the right part is not the imaginary part of $z^2$, and using conjugation will not change that. –  gary May 24 '11 at 1:26

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.
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.