- #1

Philip

- 19

- 3

$$(\sqrt{x^{2} + y^{2}} -a)^{2} + z^{2} = b^{2}$$

When solving for the z-axis in the torus equation, we get complex solutions, from the empty intersection:

$$z = - \sqrt{b^{2} - a^{2}}$$

$$z = \sqrt{b^{2} - a^{2}}$$

I was told by someone that these solutions have nothing to do with the initial torus. What's the reasoning here?

It would seem that the real set of points on the torus would still be an important part of the complex solutions. Even though z makes no intercepts initially, it will if translating the torus along x or y, by the value of plus or minus 'a' . Doing so will cancel the complex component, bringing two solutions into the real plane.Another question I had was regarding the solutions of z, and whether they can simplify/reduce any more than,

$$z = - \sqrt{(b-a)(b+a)}$$

$$z = \sqrt{(b-a)(b+a)}$$

More specifically, can we get the expected complex conjugates out of this expression?