24 - Stereographic Projection

Stereographic Projection

Let

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

be the unit sphere of radius $1$ centred at the origin in ${\mathbb{R}}^3$

Let $N=(0,0,1)$ of $\mathbb{S}$ (the north pole)

Define bijective map $S:\mathbb{C}\to \mathbb{S}\setminus \{N\}$ as follow
Join $z$ to $N$ by a straight line and let $S(z)$ be the point where the line meets sphere $\mathbb{S}$
then $S$ is called the stereographic projection

Viewing the Complex Plane inside ${\mathbb{R}}^3$

By viewing the complex plane $\mathbb{C}$ as the copy of ${\mathbb{R}}^2$ inside ${\mathbb{R}}^3$ by plane

$$\{(x,y,0)\in \mathbb{R}:x,y\in \mathbb{R}\}$$

then $z=x+iy$ corresponds to point $(x,y,0)$

Explicit Formula for the Stenographic Projection lemma

$$S(z)=\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)$$

Proof

General Point on the line joining $z$ and $N$ is

$$t(0,0,1)+(1-t)(x,y,0)$$

There is a unique value of $t$ with $t\ne 1$ such that the point lies on sphere, that is

$$t=\frac{(x^2+y^2-1)}{(x^2+y^2+1)}$$

This can also be rewritten as

$$S(z)=\frac{1}{1+|z{|}^2}(2\mathrm{Re}(z)+2\mathrm{Im}(z),|z{|}^2-1)$$

Distance Metric of the Stenographic Projection

Let $z,w\in \mathbb{C}$

$$d(z,w):=||S(z)-S(w)||=\frac{2|z-w|}{\sqrt{1+|z{|}^2}\sqrt{1+|w{|}^2}}$$

Proof

Since $||S(z)||=||S(w)||=1$ then

$$||S(z)-S(w)|{|}^2=2-2⟨S(z),S(w)⟩$$

where $⟨\cdot ,\cdot ⟩$ refers to the usual inner product in ${\mathbb{R}}^3$

So using the formulae and computation then

$$⟨S(z),S(w)⟩=1-\frac{2|z-w{|}^2}{(1+|z{|}^2)(1+|w{|}^2)}$$

Thus

$$||S(z)-S(w)|{|}^2=\frac{4|z-w{|}^2}{(1+|z|{)}^2(1+|w{|}^2)}$$

Extension of the Distance Metric to Infinity lemma

As $|z|\to \infty$ then $S(z)\to N$ hence

$${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$$

then for $z\in \mathbb{C}$

$$d(z,\infty )=\frac{2}{\sqrt{1+|z{|}^2}}$$

Proof

By definition we have

$$d(z,\infty )=||S(z)-S(\infty )||=||S(z)-N||$$