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± sqrt(b^2 - a^2) = ± b ± ai ?

  1. Jan 8, 2015 #1
    I came across this strange relationship when deriving the degree-4 equation for a torus. First thing that comes to mind is the 'Freshman's Dream'. Apparently, it was pure coincidence that they are equal. But, I don't believe in coincidences when it comes to a math expression. There is something about this one.

    I found it when thinking about the 1D intercepts of a torus, along x,y, and z. Defined implicitly,

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

    The major diameter 'a' lays flat in the xy plane. This means the X and Y axes intercept as four points along a line, and have the four solutions x = ± a ± b , and y = ± a ± b . These two can be seen the implicit function, by canceling out two dimensions, and observing the result:

    (sqrt(x^2) - a)^2 -b^2 = 0
    x = ± b ± a

    (sqrt(y^2) - a)^2 -b^2 = 0
    y = ± b ± a

    ( - a)^2 +z^2 -b^2 = 0
    z = ± b ± ai

    The X and Y axes are fairly straightforward. The Z axis is what really interested me. I was thinking of how to express the Z-cut which threads right through the hole, making no intercepts at all. Graphing the plane equation XZ or YZ will give us two circles in a row. The Z axis will go right through the middle of the two circles, with no contact.

    This reminds me of a parabola that sits above the x-axis in the +y region, which makes complex solutions when we solve for x. Borrowing a little intuition there, I felt that perhaps the Z-cut can be expressed as four complex solutions of ± b ± ai , where the 'complex intercept region' is the hole of the torus, and the 'real intercept region' is the ring around it. If we move the torus by +ai (relative to Z), we bring two of the four solutions into the real plane, making z = ± b and ± b + 2ai . The two real solutions ± b are the Z-axis skewering the ring top to bottom, The two complex solutions ± b + 2ai are now twice as far away in the complex plane, as the other 2 potential reals, on the other side of the empty hole. Using Z-intercepts to represent a torus will only give us half of all possible real solutions, unlike X or Y, which go through the side.

    When I combined these 1D intercepts into a single equation, there were some extra repeated terms that needed to be removed, by use of a consolidation compliment , 2(a^4 + b^4). But, just the 1D intercepts and compliment weren't enough, since the oblique regions are not defined as a circular ring, but come out as blobs. Comparing the fully expanded degree-4 to what I ended up with needed the oblique compliment , 2((xy)^2 + (xz)^2 + (yz)^2). These terms define all possible oblique angles of a 1D intercept into a circular ring.

    The final result, which graphs a perfect torus using complex conjugates for the Z-intercept:

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

    expands into

    (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-ai)(z+b-ai)(z-b+ai)(z+b+ai) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(a^4 + b^4) = 0

    setting a = 3 , b = 1

    (x-1-3)(x+1-3)(x-1+3)(x+1+3) + (y-1-3)(y+1-3)(y-1+3)(y+1+3) + (z-1-3i)(z+1-3i)(z-1+3i)(z+1+3i) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(3^4 + 1^4) = 0

    which will graph into a perfect torus:


    But, the question still remains, in the case of

    z = ± sqrt(b^2 - a^2) = ± b ± ai

    The z-intercepts can also be written this way, and also graphs the identical torus.

    Perhaps what's really going on is that this is a product of solutions,

    z = (-sqrt(b^2 - a^2)(+sqrt(b^2 - a^2) = (z - b - ai)(z + b - ai)(z - b + ai)(z + b + ai)

    The above method of complex conjugate intercepts also works for defining holes of a 4D toroidal shape as well. So far, I tested it with two other degree-4 equations in 4D, and it functions perfectly. However, in 4D there are two more toroidal shapes that are degree-8, with TWO such holes. One of them makes all complex intercepts in 1D. I have not yet derived their oblique angle terms, since a degree-8 surface is significantly more complex. They do make the proper, albeit blobby, intercepts of two degree-4 tori in 3D, though.

    Let me know what you all think.

    Last edited: Jan 8, 2015
  2. jcsd
  3. Jan 8, 2015 #2


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    Staff: Mentor

    The second line does not follow from the first one, it is also wrong in terms of the geometrical interpretation (where x,y,z are real).
  4. Jan 13, 2015 #3
    Ah, probably what you were looking for was,

    full torus equation
    (sqrt(x^2 + y^2) - a)^2 +z^2 -b^2 = 0

    set x, y to 0, making equation for z-intercepts

    (-a)^2 + z^2 - b^2 = 0
    (-a)^2 + z^2 = b^2
    z^2 = b^2 - (-a)^2
    z = ± sqrt(b^2 - (-a)^2)

    (apologies if this flow is wrong, if so, how should it be?)

    which is also equal to the four solutions, as graphed in above pic,

    z = - b - ai , b - ai , - b + ai , b + ai

    Understanding that x,y,z are in the real plane, my question goes to the intercept that Z is making with the torus, as being entirely empty. Solving for z-intercepts, Z makes no real solution, but a complex one, as I tried to show by the product of four solutions in the complex plane.
  5. Jan 13, 2015 #4


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    It is not, and repeating this completely unjustified claim won't make it right.
  6. Jan 13, 2015 #5
    Well, that's the thing. When I put those proposed four solutions for z in the torus equation, as graphed in above image, it came out correct. The imaginary numbers define the hole, and the empty space. I understand how it seems wrong. I'm not sure how to properly prove any of it to you. Other than to show you what this function makes when graphed, using the complex numbers in this particular way. If it is wrong, then it seems like the end result would also show this conclusion, as well. That is what's confusing me right now.
  7. Jan 14, 2015 #6


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    Consider b=1,a=1 for example
    z = ± sqrt(b^2 - (-a)^2)=0 has the only solution z=0.

    None of those four options is a solution: z = - 1 - i , 1 - i , - 1 + i , 1 + i

    How do you even "put complex z into a graph"? Putting them into the original equation (sqrt(x^2 + y^2) - a)^2 +z^2 -b^2 = 0 does not give a correct result.
  8. Jan 14, 2015 #7
    It's the other equation I wrote that used complex numbers, not the standard implicit form of

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

    I used the function

    (x-1-3)(x+1-3)(x-1+3)(x+1+3) + (y-1-3)(y+1-3)(y-1+3)(y+1+3) + (z-1-3i)(z+1-3i)(z-1+3i)(z+1+3i) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(3^4 + 1^4) = 0

    which is equal (discovered by expanding the equation and/or plotting the function) to

    (sqrt(x^2 + y^2) - 3)^2 +z^2 -1^2 = 0

    You can copy-paste these two into calcplot3D, to see what I mean: http://web.monroecc.edu/calcNSF/

    I tried to manually build a degree-4 torus equation out of using 1D intercepts along x, y, and z. axes. The problem was how to define an axis that makes no intercepts, that threads right through the hole. Axes X and Y have four real solutions, but Z is different. Intuition lead me to using complex numbers to define the empty void, with potential real intercepts spaced in a higher dimension of the real plane ( from the z-perspective). Potential real intercepts is the key phrase, since the hole is not infinite. Moving the torus far enough will make the Z-axis intercept some part of the ring, making a maximum of two real solutions, plus two complex. That strange torus equation ended up making the exact, identical shape, with a=3 and b=1.

    It's very weird, I know. You're not the only one who tells me Z won't make complex numbers for a solution. It was a hypothesis and experiment that lead me to this strange relation. There doesn't seem to be a direct way to get the complex numbers, through any known algebraic means. That's what made me so curious about it!
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