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Regarding the complex solutions of a torus

  1. May 17, 2015 #1
    The equation for a torus defined implicitly is,

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

    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?
     
  2. jcsd
  3. May 17, 2015 #2
    What you've said doesn't make sense. Consider the torus centered at ##h, k##; translating the torus doesn't change the solutions for ##z## obtained at the central axis of the torus ##(x, y)=(h, k)##.


    They simplify precisely when ##a=b## or ##a=-b##, in which case the torus becomes a horned torus.

    What complex conjugates? You never mentioned anything of the sort. Could you elaborate?
     
  4. May 18, 2015 #3
    Thanks for your reply! Assuming a > b, I'm considering the ring torus. Placed at origin, the z-axis will sit in the center of the hole. Translating the torus by the value of 'a' , along x or y, will make z intersect the ring, making two points.

    More precisely,

    translate by 'a' along x (where this could be +a , -a ,along x or y) :

    $$(\sqrt{(x-a)^{2} + y^{2}} -a)^{2} + z^{2} = b^{2}$$
    set x=0 , y=0 , solve for z:
    $$(\sqrt{(0-a)^{2} + 0^{2}} -a)^{2} + z^{2} = b^{2}$$
    $$(\sqrt{(-a)^{2}} -a)^{2} + z^{2} = b^{2}$$
    $$(\sqrt{a^{2}} -a)^{2} + z^{2} = b^{2}$$
    $$(a -a)^{2} + z^{2} = b^{2}$$
    $$z^{2} = b^{2}$$
    $$z = \sqrt{b^{2}} , z = -\sqrt{b^{2}}$$

    where z has two real solutions,
    $$z = b , z = -b$$

    Coincidentally, from a naive 1D point of view in the z-axis, adjusting the major diameter 'a' gradually to zero is visually indistinguishable from translating by 'a' . They both would look the same!

    About the complex conjugates, here's what I've been thinking. When a > b , and setting a=0 will make real solutions of b , then wouldn't it mean 'a' is the imaginary part? It does describe the size of the ring torus hole, which is an empty set of points.

    Similarly, when we graph a parabola that sits above the x-axis, in the +y half, the quadratic equation has two complex solutions. X makes a completely empty intersection. The set of points of the parabola sit "outside" the x-axis, and a complex conjugate solution is the algebraic way of defining it.

    Considering that, is it reasonable to assume that the set of points on the torus, that lie "outside" the z-axis, can also be described as complex conjugate solutions?


    Going back to $$z = - \sqrt{b^2 - a^2} , z = \sqrt{b^2 - a^2}$$

    when a > b for a ring torus, we get square root of a negative number, which yields an imaginary number. If a = 0 , we get square root of a positive number, yielding a real number. So, if both are working together, since both would not equal zero, wouldn't this be a complex conjugate, with real and imaginary?
     
  5. May 18, 2015 #4
    That is correct--but what does it have to do with the torus?? I still don't understand.

    Translating a quadratic on the ##y##-axis has a very simple effect indeed, easily described via the quadratic formula.
    The equation for a torus however is quartic. It is only when you hold two of ##x, y, z## constant that the equation reduces to a quadratic--in that case, we are restricted to a minuscule portion of the full solution set,
     
  6. May 18, 2015 #5
    Yes, that's a good point. It does reduce to a product of two quadratic equations for a circle. I guess what I'm trying to understand, is by considering this 2D solution of a torus:

    https://www.desmos.com/calculator/qmk3bia8ia

    If solving for x, can we say these circles lie in the complex plane, much like a parabola that makes no intersection? If so, what would those solutions look like, other than ##x = -\sqrt{b^2-a^2} , \sqrt{b^2-a^2}## ? Is it possible to get a more obvious-looking complex conjugate, like ##(x-1-i2)(x-1+i2)## ?
     
  7. May 19, 2015 #6
    Well, after a LOT of playing around, experimenting, and searching for some kind of meaning here, I finally found it. The circles are in the complex plane, and the roots of z, as stated above, are the simplest possible expression, other than something like ##z = -\sqrt{(b-a)(b+a)} , z = \sqrt{(b-a)(b+a)}## . I needed to see where the points on the circle were, that correspond to what the imaginary value will be. It's a product of four of them, as the points that lie on the y-axis.
     
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