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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.