# Stereographic projection | Complex Analysis

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## The Riemann Sphere

Extended Complex Numbers
The extended complex numbers consist of the complex numbers $$\Bbb{C}$$ together with ∞. The extended complex numbers is denoted by $$\Bbb{C_\infty}$$.
Geometrically, the set of extended complex numbers is referred to as the Riemann sphere (or extended complex plane).

Definition: The Riemann Sphere denoted $$\Bbb{C_\infty}=\Bbb{C}∪$${∞} is the topological space adjoining the single point to $$\Bbb{C}$$.

## Stereographic projection

Let $$\Bbb{C}$$ be the complex plane. At first, we visualize that this plane is finite but very large. Through the origin construct the line perpendicular to $$\Bbb{C}$$. Let this be γ axis of three dimensional euclidian space in which a point has co-ordinates (α, β, γ).
Two cases
(i) Considering a sphere with center at (0, 0, 0) and radius 1. In this case the sphere will intersect the plane along the unit circle.
(ii) Considering a sphere with center at (0, 0, 1/2) and radius 1/2. In this case the plane touches sphere through origin.
Both these cases will have the upper most point (0, 0, 1).

Consider the radius of sphere of radius 1 and center at (0, 0, 0). i.e.
S ={(α, β, γ) ∈ $$\Bbb{R}: α^2+ β^2+ γ^2=1$$}. The plane γ = 0 coincide with the complex plane and at this situation α and β are coordinates correspond to x axis and y axis respectively. Let Q(x, y, 0) be any point of the plane. Through the point N = N(0, 0, 1) draw a straight line NQ, intersecting the sphere at a point (α, β, γ). Then (α, β, γ) is called a stereographic projection or image of (x, y, 0) on the surface of the sphere.

Now we will express α, β, γ in terms of x and y.
The line in $$\Bbb{R}^3$$ passing through (0, 0, 1) and (x, y, 0) (which contains the point (α, β, γ)) is given by
$$\frac{α-0}{x-0} = \frac{β-0}{y-0} = \frac{γ-0}{0-1}$$
This implies α=x(1-γ), β=y(1-γ)
Thus, $$α^2+ β^2+ γ^2=1$$ gives
$$γ= \frac{x^2+y^2-1}{x^2+y^2+1}$$
Hence, $$\alpha = \frac{2x}{x^2+y^2+1}$$ and $$\beta = \frac{2y}{x^2+y^2+1}$$
So the point (α, β, γ) is given by
$$( \frac{2Re z}{ \mid {z}^2 \mid+1 } ,\frac{2Im z}{ \mid {z}^2 \mid +1} , \frac{ \mid {z}^2 \mid -1}{ \mid {z}^2 \mid +1} )$$, where z=x+iy
Hence for every (x, y) in the finite plane there exist a point
$$( \frac{2Re z}{ \mid {z}^2 \mid+1 } ,\frac{2Im z}{ \mid {z}^2 \mid +1} , \frac{ \mid {z}^2 \mid -1}{ \mid {z}^2 \mid +1} )$$ on the surface of the sphere S.

##### Prob.(1): Find the point corresponding to 1 +i √3 in the Riemannian sphere.

Solution.
Let (x, y) = (1,√3) in the plane. Let the corresponding point on Riemannian sphere be (α, β, γ). Then
$$( \frac{2Re z}{ \mid {z}^2 \mid+1 } ,\frac{2Im z}{ \mid {z}^2 \mid +1} , \frac{ \mid {z}^2 \mid -1}{ \mid {z}^2 \mid +1} )$$, where z=x+iy
⇒α =2/5, β=2√3/5, γ=3/5
Hence the point is (2/5, 2√3/5, 3/5)

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