stereographic projection $\sigma:\mathbb{R}\rightarrow S^1$

Question

The line which connects a Point $x\in\mathbb{R}\subset\mathbb{C}$ with $i$ intersects the unit circle in one Point. This Point shall be called $\sigma(x)$. Now I Need to find a closed form for $\sigma(x)$. There are hints and a solution but I don't understand neither. Hope somebody can explain why the hints are true and how one can derive the solution from them.

Let $\sigma(x)=\xi+i\eta $

$\xi$ is the real part and $\eta$ is the immaginary part.

The hints are $\xi:x=(1-\eta):1$

and

$\xi^2+\eta^2=1$

I understand the second hint because $\sigma(x)$ must be on the unit circle but not the first.

So in the picture below why is $a:b=c:d$. Is there a proof that uses congruencies?

And finally with those hints in mind how can I derive the solution that

$\sigma(x)=\frac{2x+i(x^2-1)}{x^2+1}$

?

Answer 1

The triangle defined by $i, \sigma(x)$ and $i\eta$ is similar to the triangle defined by $i, 0$ and $x$. Thus the ratio of short legs in those two triangles, which is $\frac ab$, must be equal to the ratio of the long legs in those triangles, which is $\frac cd$.

The first hint tells you that $\xi=x(1-\eta)$. Inserting that into the second hint, we get $$ x^2(1-\eta)^2+\eta^2=1 $$ This is a quadratic equation in the unknown $\eta$ that you can easily solve (remember that $x$ isn't an unknown). Even more easily since you know that $\eta=1$ is one solution.

Answer 2

You're almost there. The reason for the proportion is that in your lower drawing, $a$ is to $a+b$ as $c$ is to $d$ (by similar triangles). (I"m assuming the thing labelling the distance from the origin to $x$ is a "d".)

In general, it's probably unwise to do almost-parallel labelling like this; you might expect, given the names, to get the proportion $$ \frac{a}{b} = \frac{c}{d} $$ but that's not right, because $b$ is part of one leg of the triangle, but $d$ is all of the other leg.