The differentiability of $f(z)$ if $ \lim\limits_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$? 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?
 A: You are right that $\bar f$ satisfies the condition $\lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$. This is however the only exception as the following theorem shows.
Theorem. Suppose that $f$ satisfies the condition 
$$\lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$$
in some region $\Omega$. Assume additionally that $f$ is sufficiently smooth (e.g. $f \in C^1({\mathbb R}^2, {\mathbb R}^2)$) and $k\neq 0$ (at all points $z_0\in \Omega$). Then either $f$ is holomorphic in $\Omega$ or $\bar f$ is holomorphic in $\Omega$.
(The additional assumption can be relaxed.)
Proof: Write $f(z) = u(z) + i v(z)$ and $z = x+ i y$. Consider the Jacobian matrix $J_f$ of 
$$J_f = \begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{pmatrix}.
$$
Note that for every $(a,b) \in {\mathbb R}^2 \setminus \{0\}$,
$$k = \lim_{t \to 0} \frac {|f(z_0 + (a+bi)t)-f(z_0)|}{|a+bi|\cdot |t|}=\frac{\left|J_f 
\begin{pmatrix}
a\\ b\end{pmatrix}\right|}{|(a,b)|}.$$
Therefore, for every $v\in {\mathbb R}^2$, $|\left(\frac{1}{k} J_f\right) v| = |v|$. Thus $\frac{1}{k} J_f$ is an orthogonal operator.
There are two possibilities: $\frac{1}{k} J_f$ is either a rotation or reflection matrix. So if $\det(J_f) > 0$ then $J_f$ is a scaled rotation matrix:
$$J_f = \begin{pmatrix}
\alpha & \beta\\
-\beta & \alpha
\end{pmatrix}.
$$
Thus $f$ satisfies Cauchy–Riemann equations. Otherwise, $J_f$ is a scaled reflection matrix:
$$J_f = \begin{pmatrix}
\alpha & \beta\\
\beta & -\alpha
\end{pmatrix},
$$
and $\bar f$ satisfies Cauchy–Riemann equations.
Note also that since $k\neq 0$ either $f$ satisfies Cauchy–Riemann equations at all points $z\in \Omega$, or $\bar f$ satisfies Cauchy–Riemann equations at all points $z\in \Omega$. It also follows that $f'(z)\neq 0$.
