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Complex and Hypercomplex Numbers

  1. Sep 10, 2009 #1
    Complex numbers are said to be algebraically closed, meaning (to my mind) that given any polynomial x = p(z), with complex numbers x and z, the polynomial maps the complex number field back onto itself completely. For any given x, there will be a z.

    It is then stated that this makes any hyper-complex number superfluous for the analysis of arbitrary functions.

    Now it may be true that the hypercomplex number field is closed for any algebraic operation, and raising complex numbers to fixed powers. And most special functions also appear closed on investigation (I'm an engineer, not a mathematician, so hold your fire).

    What prevents me from defining an arbitrary function though, or finding one, that does not map the complex number field back onto itself completely? Some x = f(z), where for some x, there is not a z? finv(x) = z, z is not complex? If I then defined a hypercomplex number i2, and associated operational behavior whereby the field is once again complete, finv(x) = z, z is an element of hypercomplex field, then why is what I have just done fundamentally different than what was done to come up with complex numbers in the first place?

    After all, the original motivation for complex numbers was to define what happened when you inverted certain polynomials p(z) = x.
     
  2. jcsd
  3. Sep 10, 2009 #2

    CRGreathouse

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    Nothing. Here's a perfectly valid function:
    f(z) = sin(z), x > 3
    f(z) = apple, x <= 3
    which maps the complex numbers C to the apple-complex numbers C ∪ {apple}.
     
  4. Sep 10, 2009 #3
    Are complex numbers really closed? I would extend this concept to include complex conjugation. Then one needs another square root of -1 (i.e. quaternions)
    [tex]
    (a+b\mathrm{i})^*=\mathrm{j}(a+b\mathrm{i})\mathrm{j}^{-1}
    [/tex]

    Is that a good idea?
     
  5. Sep 10, 2009 #4

    lurflurf

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    The complex numbers are algebraically closed, they are not closed in other ways. A stupid example is there are no complex numbers such that |z|=i. We can add in any new numbers we want, the issue is what properties such an extended system has. For example the quaternions are noncommutative. The complex numbers are not ordered.
    Gerenuk you can construct the quaternions that way. The conjugation of a system is effected naturally by an element of the next higher system. Look up The Cayley-Dickson Construction. http://math.ucr.edu/home/baez/octonions/node5.html
     
  6. Sep 10, 2009 #5
    OK, I see.

    To just include the conjugation operation algebraically however one can stick with quaternions, right? Wikipedia says one can express quaterion complex conjugation as multiplications only and in fact one can extract each component of the quaternion this way. Basically if one wants to include the complex conjugation, then quaternions are the choice?!
     
    Last edited: Sep 11, 2009
  7. Sep 11, 2009 #6

    CRGreathouse

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    The complex numbers are closed under conjugation as well as under addition, multiplication, powers, etc.
     
  8. Sep 11, 2009 #7
    Please read the post carefully! The aim is to express congujation with normal algebraic operations. You can't do that with complex numbers, but you can with quaternions.
     
  9. Sep 11, 2009 #8

    CRGreathouse

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    I could say the same to you: I contradicted none of your post.
     
  10. Sep 11, 2009 #9
    Sure. You can say the grass is green and not contradict my statements.
    lurflurf had got the point and given me as very useful link.
     
  11. Apr 19, 2010 #10
    This is actually a good observation. Do we really need more than the four rules of arithmetic to express other constructions in hypercomplex algebras, or can we find ways to represent every operation in terms of the four rules?

    See web article on quaternions:

    http://web.archive.org/web/20070928...mplex.com/education/intro_tutorial/nabla.html


    In section on Josiah Willard Gibbs' contribution we find div and curl can be expressed

    div A = 1/2 . ( d->A + A<-d )
    curl A = 1/2 . (d->A - A<-d)

    which is really the limit of the infinitesimal change in the rotation operation.

    A' = qAq^-1

    is a rotation in quaternions, where the vector A' is the rotated result of A being acted upon by the quaternion q that describes the parameters of the rotation. But, if a and b are two vectors that represent the initial and final states of a rotation, then the quaternion q that does the job is given by

    q = (b/a)^(1/2)

    that "square root" of the ratio of vectors tells us that q = (1 + h.d)^(1/2) with "h" a small infinitesimal parameter and "d" the differential operator is the quaternion operator that when multiplied and divided from opposite sides gives the infinitesimal rotation

    A' = (1 + h.d)^(1/2) . A . (1 + h.d)^(-1/2)

    = (1 + 1/2 . h . d + ...) . A . (1 - 1/2 . h . d + ... )

    = A + 1/2 . h . (d.A - A.d) + ...

    so,

    lim{h - > 0} (A' - A)/h = 1/2 (d.A - A.d)

    quaternions non-commute so d.A != A.d, and in both cases the operator d acts on the variable A, which we make emphatic by replacing the dot . with an arrow -> to remember that this is critical in non-abelian algebras so

    lim{h -> 0} (A' - A)/h = 1/2 ( d->A - A<-d )

    this happens to be exactly the "curl" of vector algebra. So, we clearly see the link between the rotation "qAq^-1" of quaternion algebra, and the "curl" operator of vectors. The latter is derived using only the four rules together with the noncommuting property of the hypercomplex algebra.

    Like the case with complex conjugation *, Hamilton introduced six operators "S,V,K,N,T,U" for "Scalar, Vector, Conjugate, Norm, Tensor, Versor" and treated them as "add on operators" in his calculus, which made the quaternion algebra seem alot more complicated. But, these six operations can be represented in terms of the four rules again. See the URL for that article on Nabla.
     
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