Stereographic projection is conformal

Statement

Setup

Further information: Stereographic projection

Consider ${\mathbb{R}}^3$, three-dimensional Euclidean space, with coordinates $x,y,z$. Denote:

$S^2:=\{(x,y,z)\in {\mathbb{R}}^3\mid x^2+y^2+z^2=1\}$

Identify $\mathbb{C}$ with the $xy$-plane under the map:

$(x,y,0)\mapsto x+iy$

Let $N=(0,0,1)$ denote the north pole in $S^2$, and define the stereographic projection as a bijective map:

$S^2\setminus \{N\}\to \mathbb{C}$

which sends a point $P\in S^2\setminus \{N\}$ to the unique point in $\mathbb{C}$ that is collinear with $N$ and $P$.

Actual statement

Pick any point $P\in S^2\setminus \{N\}$. Then there is a natural induced map from the tangent space to $P$ in $S^2$ to the tangent space to its image, in $\mathbb{C}$. This map is conformal, i.e. it preserves angles.

In other words, if we make two smooth curves in $S^2$ that intersect at $P$, the angle of intersection between those curves at $P$ equals the angle of intersection of their images under stereographic projection.

Alternative formulation

Consider $S^2\setminus \{N\}$ and $\mathbb{C}$ as Riemannian manifolds, with the former getting the induced structure from its embedding in ${\mathbb{R}}^3$.

Then, the stereographic projection is a conformal map of Riemannian manifolds.

Alternatively, the Riemannian metric on $\mathbb{C}$ obtained using that on $S^2\setminus \{N\}$, is conformally equivalent to the standard metric.