# 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

## 1 Answer

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
• @sequence It should, but it is not. Are you sure of that result? Where did you read it? – Aretino Feb 26 '18 at 22:13
• I think it's not AE in this case, but the segment from 1 to E. Thus it is not the radius of the circle from the origin to the point of intersection of the ray from the stereographically projected point to the axis of ordinates, but the difference of radii of the annulus with inner radius 1. – sequence Mar 1 '18 at 16:19
• Are you suggesting that my answer in incorrect? I don't think so and would then urge you to point me to the error. – Aretino Mar 1 '18 at 17:33
• 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