conformal but not real differentiable, still holomorphic? can one prove that a conformal map (one which preserves angles (in sense as well as in size) between arcs with non-zero derivatives at a point z and for which the limit of the absolute value of the difference quotient exists at z) has a complex derivative at z without assuming that it is differentiable at z as a map between IR^2? If so, how? If not can you give me a counterexample?
Thanks
 A: As @Sam pointed out, it is not immediately clear how to define the angle between images of curves. One way is to simply assume 
(V) for every differentiable curve with nonzero velocity vector its image under $f$ is also differentiable and has nonzero velocity vector 
One may suspect that assumption (V) is equivalent to $\mathbb R$-differentiability of $f$, but it is not. For example, the function $f(re^{i\theta})=r\exp (\theta+i\sin 2\theta)$ satisfies (V) but is not $\mathbb R$-differentiable at $0$. (Aside: there was a similar question A sufficient condition for differentiability of a function of several variables in terms of differentiability along paths. but it had the extra assumption that the directional derivative is linear.) So the questions makes sense with the addition of (V), but is somehow less appealing in this form since (V) does not look a very natural thing.
So I'm inclined to forget the angles and to consider only the second condition in the question: $$(*)\qquad \lim_{z\to a}\frac{|f(z)-f(a)|}{|z-a|}$$
exists at every point $a$. If such $f$ is a homeomorphism (at least locally), then it satisfies the definition of quasiconformal map with the coefficient of quasiconformality $K=1$. It is a well known, but nontrivial fact, that $1$-quasiconformal maps are conformal. The linked article by Heinonen attributes it to Menshov (1937) but does not give a precise reference and I can't track it down right now. Nowadays one simply appeals to the equivalence of the metric and analytic definitions of quasiconformality, which is established, e.g., in Lectures on Quasiconformal Mappings by Väisälä.
By the way, 1-quasiconformality is substantially weaker than (*): it only requires that $|f(z)-f(a)|$ is close to $|f(\zeta)-f(a)|$ whenever $|z-a|=|\zeta-a|$. There is no requirement for $|f(z)-f(a)|$ to be close to $|z-a|$.
