Why $\frac{|1-z|}{1-|z|}\le K$ corresponds to the region defined by the Stolz angle? In his presentation of Abel's theorem, Ahlfors mentions that for a fixed positive number $K$, the region defined by \begin{equation}
\frac{|1-z|}{1-|z|}\le K
\end{equation} corresponds to the region inside the unit circle and in a certain angle with vertex $1$, symmetric around the x-axis.
This should not be too difficult, but I cannot actually see how that inequality is related to that geometric picture.
Can someone give a hint?
Thanks!

It's not an equality. The Stolz angle with opening $\alpha > 0$ and radius $r$ is
$$S(\alpha,r) = \{1 - \rho e^{i\varphi} : 0 < \rho < r,\; \lvert\varphi\rvert < \alpha\},$$
a circular sector that for $\alpha < \pi/2$ and small enough $r$ (depending on $\alpha$) is contained in the unit disk. Its boundary consists of two straight line segments, and one circular arc.
For $0 < K < \infty$, the region
$$R(K) = \left\lbrace z \in \mathbb{D} : \frac{\lvert 1-z\rvert}{1-\lvert z\rvert} < K \right\rbrace$$
is bounded by a curve that contains no straight line segment. But as $z$ approaches $1$ on the boundary of $R(K)$, the angle between between the real axis and $1-z$ approaches a limit $< \pi/2$ (namely $\arccos K^{-1}$).
The equivalence of the two conditions is to be understood in the sense that if we consider $R(K,r) = \{z \in R(K) : \lvert 1-z\rvert < r\}$, then for every $K \in (0,\infty)$, there is an $\alpha < \pi/2$ with $R(K,r) \subset S(\alpha,r)$, and for every angle $\alpha \in (0,\pi/2)$, there is a $K < \infty$ with $S(\alpha,r) \subset R(K,r)$ (where $r$ is always supposed small enough that $\overline{S(\alpha,r)}\setminus \{1\}$ is contained in the unit disk). Demonstrations at wolfram.
