# Intersection of a Plane with the Riemann Sphere

While reading Fundamentals of Complex Analysis by Saff and Snider, I came across an example (see page 47, edition 3) where it is shown that "all lines and circles in the $z$-plane correspond under stenographic projection to circles on the Riemann sphere". It is also mentioned that this argument can be reversed. My problem is that I don't really get how I can use it in a situation such as:

How do I determine whether the intersection of a plane, say $3x+3y+5z=5$, with the Riemann sphere correspond to a line or a circle in the complex plane?

• Does the circle in the sphere pass through the north pole? If it does, it's a straight line in the plane, otherwise a circle. – Daniel Fischer Sep 2 '14 at 20:25
• Ah, how good to hear… Thanks a lot Daniel! – lon126don Sep 2 '14 at 20:30
• @DanielFischer Just to be clear, that would mean the intersection above generates a line, right? And, say $3x+2y+7z=6$, creates a circle, or am I wrong? – lon126don Sep 2 '14 at 20:43
• Right. The north pole $(0,0,1)$ satisfies $3x+3y+5z = 5$, but not $3x+2y+7z = 6$. – Daniel Fischer Sep 2 '14 at 20:45
• Of course, the intersection of a plane with the sphere can also be empty or consist of a single point, these cases are to be ruled out. – Daniel Fischer Sep 2 '14 at 20:46

Let the plane be

$$A\xi + B \eta + C \zeta = D$$

and let a point on the sphere be $(\xi, \eta, \zeta)$ and a point in the complex plane $(x,y,0)$.

The relation between points on the sphere and points on the plane is given by stereographic projection and the following equations hold:

$$\xi = {x\over x^2 + y^2 + 1}, \eta = {y \over x^2 + y^2 + 1}, \zeta = {x^2 + y^2 \over x^2 + y^2 + 1}$$

Substituting into the plane equation you get

$$A{x\over x^2 + y^2 + 1} + B {y\over x^2 + y^2 + 1} + C {x^2 + y^2\over x^2 + y^2 + 1} = D$$

or, equivalently,

$$Ax + By + C(x^2 + y^2) =D(x^2 + y^2) + D$$

or, equivalently,

$$Ax + By + (C-D)(x^2 + y^2) = D$$

This is the equation of a circle if $C-D \neq 0$ so you see that points $(x,y,0)$ satisfying $Ax + By + (C-D)(x^2 + y^2) = D$ are points on the plane that correspond to a circle on the sphere.

If $C=D$ then

$$A\xi + B\eta = C(1-\zeta)$$

or, equivalently,

$$A\xi + B\eta + C(\zeta-1)=0$$

which you can compare to the general form of a plane equation

$$a(x-x_0)+b(y-y_0) + c(z-z_0) = 0$$

so you see that if $C=D$ then $(x_0,y_0,z_0) = (0,0,1)$ which means it is a plane through $(0,0,1)$.

To summarize:

If the plane is through $(0,0,1)$ then the resulting set on the complex plane turns out to be a line, namely, $Ax + Bx = D$.

On the other hand, if the plane does not intersect $(0,0,1)$ the resulting set on the complex plane is a circle.