Let $f:\Omega \to \mathbb C$, where $\Omega$ is a region in $\mathbb C$, and $z_0 \in \Omega$.
Suppose that $\displaystyle \lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$ for some constant k regardless of the direction from which $z$ is approaching to $z_0$.
Does it mean that $f(z)$ is analytic?
I don't think this is true because $\displaystyle \lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=\displaystyle \lim_{z \to z_0} \frac {|\bar f(z)-\bar f(z_0)|}{|z-z_0|}$ and $f, \bar f$ cannot be analytic simultaneously unless they are constants. So this basically would basically mean that all analytic functions are constants, which is clearly wrong.
However, Ahlfors writes (At least, I interpreted so) in his textbook Complex Analysis that this property implies the differentiability of $f$ with "additional regularity assumptions". (P.73-74 3rd edition)
Conversely, it is clear that both kinds of conformality (preserving the angle and the size) together imply the existence of $f'(z_o)$. It is less obvious that each kind will separately imply the same result, at least under additional regularity assumptions
I guess I interpreted wrong way but I don't know what's wrong with my argument. Could anyone help me with this?