I finally figured out how the solution of z is working, in the torus equation, and how the imaginary number solutions locate the circles. That was one tough concept to nail down. I kept thinking it should be something else than what you get algebraically, since I was using the strange equation, with complex conjugate roots for z. It all makes so much more sense. Considering a

parabola, with the lowest point set on y=5 , solving for x yields a complex root, (x - i√5)(x + i√5). The imaginary value is the square root of the height above x. Adjusting the parabola right or left, will make a complex conjugate root.

So, what happens with a circle, and we solve for x? Well, since a circle wraps back around, there are

two points, as an upper and lower bound. Considering a

circle that doesn't intersect x , the imaginary value is a product of both of those upper/lower points, as (x - 2i√2)(x + 2i√2) , which is a simplified form of (x - i√(4*2))(x + i√(4*2)) .

This is exactly equal to (z - √((b-a)(b+a))(z + √((b-a)(b+a)) , which is the solution for z. It really was right in front of me, but I couldn't visually understand the connection. What I needed to see was how it worked with a parabola, then find a working circle equation, expressed as roots of x:

x^2 + (y-a)^2 = b^2

(x-sqrt(-a^2+2 a y+b^2-y^2))(x+sqrt(-a^2+2 a y+b^2-y^2)) = 0

Solving for 1D roots of the tiger equation, which yields the correct complex conjugates of two circles above x:

(sqrt(x^2+0^2)-a)^2 + (sqrt(y^2+0^2)-b)^2 = c^2

solve for x,

x = ±(a+sqrt(-b^2-2 b y+c^2-y^2))

x = ±(sqrt(-b^2-2 b y+c^2-y^2)-a)

x = ±(a+sqrt(-b^2+2 b y+c^2-y^2))

x = ±(sqrt(-b^2+2 b y+c^2-y^2)-a)

express as product of roots for just the upper half of the four,

(x-(a+sqrt(-b^2-2 b y+c^2-y^2)))(x+(a+sqrt(-b^2-2 b y+c^2-y^2)))(x-(sqrt(-b^2-2 b y+c^2-y^2)-a))(x+(sqrt(-b^2-2 b y+c^2-y^2)-a))

a=3 , b=3 , c=1 , y=0 , solve for x-intercepts,

(x-(3+sqrt(-3^2-2*3*0+1^2-0^2)))(x+(3+sqrt(-3^2-2*3*0+1^2-0^2)))(x-(sqrt(-3^2-2*3*0+1^2-0^2)-3))(x+(sqrt(-3^2-2*3*0+1^2-0^2)-3))

(x-3 -2i√2)(x-3 +2i√2)(x+3 -2i√2)(x+3 +2i√2)

x = ±3 ±2i√2

Both circles have an upper/lower bound of 4 and 2 , which become i√8 , and canceling this will put the lower bound of both circles on x = ±3.

Which leads right back to my experimental equation for the torus:

(x-b-a)(x+b-a)(x-b+a)(x+b+a) + (y-b-a)(y+b-a)(y-b+a)(y+b+a) + (z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(a^4 + b^4) = 0

Surprisingly, what I hadn't tried yet, was setting x and y to zero, and seeing what it simplifies to. When I did, it made more sense why (z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) had worked.

(0-b-a)(0+b-a)(0-b+a)(0+b+a) + (0-b-a)(0+b-a)(0-b+a)(0+b+a) + (z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) + 2((0)^2 + (0*z)^2 + (0*z)^2) - 2(a^4 + b^4) = 0

2(a^2-b^2)^2 + (z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) - 2(a^4 + b^4) = 0

(z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) + 2(a^2-b^2)^2 - 2(a^4 + b^4) = 0

(z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) - 4a^2b^2 = 0

Where,

(z-b-ia)(z+b-ia)(z-b+ia)(z+b+ia) - 4a^2b^2 == (z^2+a^2-b^2)(z^2+a^2-b^2)

Which means, we can actually express the solutions for z as a product of four complex conjugates, with additional term of -4a^2b^2 , like a constant. The algebraic solution is z^2 + a^2 - b^2 , but this is only a degree-2 equation, and so needs to be squared to include the missing degree-4 terms.

If setting y=0, z=0, we get

(x-b-a)(x+b-a)(x-b+a)(x+b+a) + (0-b-a)(0+b-a)(0-b+a)(0+b+a) + (0-b-ia)(0+b-ia)(0-b+ia)(0+b+ia) + 2((x*0)^2 + (x*0)^2 + (0)^2) - 2(a^4 + b^4) = 0

(x-b-a)(x+b-a)(x-b+a)(x+b+a) + (a^2-b^2)^2 + (a^2+b^2)^2 + 2((x*0)^2 + (x*0)^2 + (0)^2) - 2(a^4 + b^4) = 0

(x-b-a)(x+b-a)(x-b+a)(x+b+a) + (a^2-b^2)^2 + (a^2+b^2)^2 - 2(a^4 + b^4) = 0

(x-b-a)(x+b-a)(x-b+a)(x+b+a) = 0

The remaining terms from setting y=0, z=0 cancel out completely along with the consolidate compliment - 2(a^4 + b^4) , yielding four real roots of x.