I'm reading about the spherical representation and the Riemann sphere, and the projection transformation that takes a point on the sphere to a point on the (extended) complex plane. An exercise is to show that if $Z,Z'$ are the stereographic projections onto the sphere of $z,z'$ lying on the complex plane, then the triangles $NZZ'$ and $Nzz'$ are similar ($N$ is the north pole).
I think that one way would be to calculate the stereographic projections explicitly, using the formula involving coordinates. For such a nice statement, however, I'm sure there must be a way to look at it geometrically. I've tried drawing pictures, but still can't see why the statement must be true. One observation I have is that the longer $NZ$ is, the shorter $Nz$ is. So that means the similarity should be in the order $\triangle NZZ'\sim\triangle Nz'z$. But how can I actually show the similarity?
Best Answer
Triangles are similar if they agree in all their corner angles. And stereographic projection is conformal, i.e. it preserves angles. If you have both of these, you are pretty much done. If you don't have proven conformality yet, I'd tackle that, since it will be useful in other cases as well.
One approach might be using the fact that Möbius transformations are conformal (if you know that), and you can show that a stereographic projection onto the sphere combined with a rotation of the sphere combined with a stereographic projection back to the plane will result in a Möbius transformation. For this reason, it is enough to show conformality in a single point of the sphere, e.g. the one opposite the center of projection where things are very symmetric.