I cannot see why Ahlfors' statement is true (Extending a conformal map)

In page 234 in Ahlfors’ complex analysis text, the author talks about extending a conformal map. During the proof he states:

We note further that $f'(z) \neq 0$ on $\gamma$· Indeed, $f'(x_0)= 0$ would imply that $f(x_0)$ were a multiple value, in which case the two subarcs of $\gamma$ that meet at $x_0$ would be mapped on arcs that form an angle $\pi/n$ with $n \geq 2$; this is clearly impossible.

Here $\gamma$ is a line segment on the real axis, which is contained in the boundary of the domain $\Omega$ (which is mapped conformally onto $|w|<1$).

I think that his statemet is false, as $\Omega=(-1,1) \times (0,1)$, with $f(z)=z^2$ is a counterexample: Take $x_0=0$, $f(x_0)=0$ is a double value then, and the segment $\gamma=(-1,1)$ breaks into the subarcs $(-1,0),(0,1)$ which are being mapped onto the same arc (the angle between them is $2\pi$).

Is his statement false indeed? If so, do you know what did he mean to say?

Thanks.

Solutions Collecting From Web of "I cannot see why Ahlfors' statement is true (Extending a conformal map)"

You are right: the angle of $\pi$ between two subarcs would become $\pi n$ in the image, not $\pi/n$. This suggests that Ahlfors was really thinking about inverse images. At this point of the proof $f$ is already known to be analytic at $x_0$, and the local structure of an analytic map was established back on page 133. Suppose that $f-f(x_0)$ has the zero of order $n\ge 2$ at $x_0$. Consider a small arc of the circle $|w|=1$ containing $f(x_0)$. Its inverse image under $f$ consists of $2n$ curves emanating from $x_0$ at angles $\pi/n$. Since $\pi/n<\pi$, at least one of these curves must enter $\Omega$, contradicting the maximum principle for $f$.