# Existence of the absolute value of the limit implies that either $f \$ or $\bar{f} \$ is complex-differentiable

This is an exercise from Remmert.

Let D be domain in $\mathbb{C} \$, and $f : D \rightarrow \mathbb{C} \$ a real-differentiable function. Suppose that for some $c \in D$ the limit

$\lim_{h \to 0} |\frac{f(c + h) - f(c)}{h}|$

exists. Prove that either $f \$ or $\bar{f} \$ is complex-differentiable at c.

Any hint ?

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Since $f$ is real-differentiable, one may write that $f(c+h)-f(c)=Ah+B\bar h+o(|h|)$, where $A$ and $B$ are constants. So, – serge May 5 '12 at 8:31

Since $f$ is real-differentiable, one may write that $f(c+h)-f(c)=Ah+B\bar h+o(|h|)$, where $A$ and $B$ are constants (actually, $A=\partial f/\partial z$, $B=\partial f/\partial\bar z$). So, $|\frac{f(c+h)−f(c)}h|=|A+B\frac{|h|}h+\frac{o(|h|)}h|$. If both $A$ and $B$ are non-zero, this limit cannoy exist: denoting $h/|h|$ by $e^{i\theta}$, one sees that $|A+Be^{i\theta}|$ depends on $\theta$, i.e., on the direction along which $h$ tends to zero.
If $B=0$, the function is holomorphic, if $A=0$, the function is antiholomorphic.