I've recently read in a book on complex variables that the points which are located on the spherical sector within the radius of a circle of radius $\varepsilon$ on the Riemann sphere are projected stereographically to the points lying in the complex plane beyond the circle of radius $1/\varepsilon$. So I decided to prove this, and my proof follows next, but I believe there should be a simpler, more straightforward proof based on trigonometry. Please let me know if you can find one.
So we have the segment $NC$ of length $\varepsilon$ and we need to find the length of the segment $AE$. We know the following:
- The angle $\delta$.
- The length of the segment $AN$ is $1$.
- $h=1$
We can calculate the angle $\beta$ from the given data. We can also find $\varepsilon = 2\cos(\beta)=\frac{2}{\sec\beta}$.
$\delta = \frac{\pi}{2}-\theta=\pi - 2\beta$
So we can find $AE = \frac{\sin\beta}{\sin\delta}=\frac{\sin\beta}{\sin(\pi-2\beta)}=\frac{\sec\beta}2$.
Thus we have the needed relation.
Let $H$ be the midpoint of $NC$. Triangles $HNA$ and $ANE$ are similar, thus: $$ NH:NA=NA:NE, \quad\hbox{that is:}\quad {\epsilon\over2}:1=1:NE, $$ whence $NE=2/\epsilon$ and $AE=\sqrt{4/\epsilon^2-1}$.