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I'm confused with a proof in "Complex Analysis" by Ahlfors.(P.73 2.3 comformal mapping) I need some help for the last part of the proof.

"Let $f:\Omega\to \mathbb C$ and fix $z_0\in \Omega$. Suppose that $\displaystyle \lim_{t\to t_0} \frac{|f(z(t))-f(z(t_0))|}{|z(t)-z(t_0)|}=k$ for every regular arc $z(t)=x(t)+iy(t)$ that maps into $\Omega$ and $z(t_0)=z_0$ for some $t_0$.(k is a constant and independent of a choice of z) If $\displaystyle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ both exist and both are continuous at $z=z_0$, $f$ has a derivative at $z=z_0$."

Let $w(t)=f(z(t))$.

By using the continuity of the partial derivatives, we can write $\displaystyle {w'(t_0)}= \frac {\partial f} {\partial x}x'(t_0)+\frac {\partial f} {\partial y}y'(t_0)$.

This can be rewritten as following: $\displaystyle {w'(t_0)}=\frac 1 2 \Big ( \frac {\partial f} {\partial x}+i\frac {\partial f} {\partial y}\Big ) \bar {z'(t_0)}+\frac 1 2 \Big ( \frac {\partial f} {\partial x}-i\frac {\partial f} {\partial y}\Big ) {z'(t_0)}$.

Since $z$ is regular we can divide both side by $z'$.

$\displaystyle \frac {w'(t_0)}{z'(t_0)}=\frac 1 2 \Big ( \frac {\partial f} {\partial x}+i\frac {\partial f} {\partial y}\Big ) \frac{\bar {z'(t_0)}}{ {z'(t_0)}}+\frac 1 2 \Big ( \frac {\partial f} {\partial x}-i\frac {\partial f} {\partial y}\Big ) $.

Thus the modulus of the left side is a constant and it means that the modulus of the right side must be a constant too. Since the right side represents a circle, either the radius is 0 or the center is at the origin. In the former case, $\displaystyle \frac {\partial f} {\partial x}+i\frac {\partial f} {\partial y}=0$ and in the latter case, $\displaystyle \frac {\partial f} {\partial x}-i\frac {\partial f} {\partial y}$. The former case implies Cauchy-Riemann for $f$ and the latter case implies Cauchy-Riemann for $\bar f$.

The proof in the book ends here, but the existence of the derivative of $\bar f$ does not always imply the existence of the derivative of $f$. However, I cannot figure out how to end this proof. Could anyone help on this?

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