# Circle of $\varepsilon$-radius on Riemann sphere related to exterior of $1/\varepsilon$-circle

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.

• $\delta=2\beta-\pi/2$. – Aretino Feb 25 '18 at 12:45

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 $AE=\sqrt{4/\epsilon^2-1}$.
• AE should be $1/\varepsilon$ – sequence Feb 26 '18 at 21:41
• I don't think that it is incorrect, but the idea is to prove that the length of the segment from $1$ to $E$ is $1/\varepsilon$. – sequence Mar 2 '18 at 22:51