Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Imaginary Numbers

  1. Jul 1, 2008 #1

    Just reading through a good physics book about games and came across some stuff on complex numbers. The book states that the following statements are true.


    I have a problem with this one. Surely doesn't this mean the same as:




    How is minus one mulitplied by k equal to minus one when k is the square root of minus one?

    It also states that:


    Which again I don't agree with. Surely i*j=-1 not k? Also why is the order important when they are all supposed to hold the same scalar value?

    I should point out that this is for use with Quaternions but this part of the book makes no reference to that it's just describing complex numbers. Am I going mad or am I correct in my assumptions that these statements aren't true?

    Many thanks :wink:
  2. jcsd
  3. Jul 1, 2008 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    These are quarternions, and there's no getting around that. Those are the rules that define quarternions.

    That sentence would seem to back that up (unless you meant to put some punctuation to make 'it's just describing complex numbers' stand alone).
  4. Jul 1, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Then how does the book define j and k at this point?
  5. Jul 1, 2008 #4

    D H

    Staff: Mentor

    These aren't complex numbers. They are, as Matt has noted (and as you noted later in your OP) part of the definition of the quaternions. The unit quaternions i, j, and k are three distinct things. As an analogy, think of the unit vectors in Euclidean three space. The unit vectors i, j, and k are three distinct things.

    -1 has many roots in the quaternions. An infinite number of roots, in fact. So it is best to stop thinking of quaternions as complex numbers. They are a different beast.

    This is another part of the definition of the quaternions. It is a postulate of the quaternions. You can't argue with postulates. This postulate right away indicates that quaternion multiplication is not commutative. Writing ijk = -1 without parentheses implies that quaternion multiplication is associative, which it is.

    Mathematicians are not bound by the real world. They can (and do) make up many different number systems that are different from the integers, rationals, reals, and complex numbers with which you are familiar. The quaternions are but one example of a somewhat esoteric number system.

    Some of the number systems devised by mathematicians are quite useful. The quaternions are one of those. They were developed before and motivated the development of vectors. Maxwell's equation were written in quaternion form before they were written in vector form. The unit vectors i, j, and k are labeled as such precisely because of the unit quaternions are called i, j, and k. Quaternions fell by the wayside for some time after the development of the 3-vector dot and cross product, but have made quite a comeback as of late because they are extremely useful for describing rotations in 3-space.
  6. Jul 1, 2008 #5
    That's lovely, thanks. So I see that this is a special thing and such straightforward attempts at rationalising its rules and statements is folly as they obey special conditions which have to be set to make the concept work in the first place.

    I think from what you guys are saying (I'm having a little trouble following all of it) it is best to see i, j and k as three kind of vectors which simply cannot be multiplied and added together in a scalar way.
  7. Jul 1, 2008 #6
    Well, you seem to have the idea. Another good example is modular addition (I think I'm using that term correctly.) For example, 10pm + 4hours = 2am. Obviously that disobeys the rules of addition, but it is how we have defined the system that makes it work.
  8. Jul 1, 2008 #7

    D H

    Staff: Mentor

    You are trying to apply the rules that describe the complex numbers to something else. If quaternions vex you, try octonions, or sedenions, or ... The reals are an algebra. The complex numbers, quaternions, octonions, etc are algebras over [itex]\mathcal R^2[/itex], [itex]\mathcal R^4[/itex], [itex]\mathcal R^8[/itex], etc. Think of the quaternions and other algebras over [itex]\mathcal R^{2^n}[/itex] as extensions of the complex numbers.

    You can visualize quaternions this way. Thinking of them as comprising a scalar real part and a 3-vector imaginary part works. Let [itex]Q_1[/itex] and [itex]Q_2[/itex] be quaternions. Writing them in the form

    [tex]Q_1 = (q_{1s}\, , \; \mathbf q_{1v})[/tex]
    [tex]Q_2 = (q_{2s}\, , \; \mathbf q_{2v})[/tex]

    Quaternion addition is simple:

    [tex]Q_1 + Q_2 = (q_{1s}+q_{2s}\, , \; \mathbf q_{1v}+\mathbf q_{2v})[/tex]

    Quaternion multiplication can be written using the vector dot and cross products:

    [tex]Q_1 \cdot Q_2 =
    (q_{1s}q_{2s} - \mathbf q_{1v}\cdot\mathbf q_{2v} \, , \;
    q_{1s}\mathbf q_{2v}+q_{2s}\mathbf q_{1v} + \mathbf q_{1v}\times\mathbf q_{2v})[/tex]

    This was done historically in the reverse manner. Quaternion multiplication was defined first. The 3-vector dot product and cross product were initially defined based on quaternion multiplication.

    There are even things like the quaternion exponential. Back to complex numbers: The complex exponential of a pure imaginary number is a unit complex number and is related to rotations in two-space via Euler's equation. Back to quaternions: The quaternion exponential of a pure imaginary quaternion is a unit quaternion and is similarly related to rotations in three-space.
  9. Jul 1, 2008 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The complex numbers, and quarternions are algebras over R, not R^n (unless you make n=1).

    The octonions are not an algebra, as they are not associative, although that is a strict interpretation of the term algebra, I admit. For the OP an algebra is just a vector space over a field with an associative product.
  10. Jul 1, 2008 #9


    User Avatar
    Science Advisor
    Homework Helper


    Hi Adder_Noir! :smile:

    Think of i, j and k as three perpendicular 180º rotations, and 1 + θi as being a rotation through an angle 2θ, for very small θ. :smile:
  11. Jul 1, 2008 #10

    D H

    Staff: Mentor

    Re: rotations

    No need for small angle approximations with quaternions! Let

    [tex]Q = (\cos \frac{\theta}2\,,\, \sin \frac{\theta}2 \hat{\mathbf u})[/tex]

    The quaternion product

    [tex](0\,,\,\mathbf x') = Q\cdot (0\,,\,\mathbf x)\cdot Q^{\ast}[/tex]

    represents the rotation of a vector [itex]\mathbf x[/itex] about the rotation axis [itex]\hat{\mathbf u}[/tex] by an angle [itex]\theta[/itex], large or small.
  12. Jul 1, 2008 #11


    User Avatar
    Science Advisor
    Homework Helper

    keeping it simple …

    Yes there is …

    :biggrin: he's an Adder, not a Multiplier! :biggrin:
  13. Jul 1, 2008 #12

    D H

    Staff: Mentor

    There is a place for small angle approximations in quaternions, but no need for them. My expression in post #10 is completely general because the unit quaternions are a chart on SO(3).
  14. Jul 1, 2008 #13
    Wow some great stuff there. I'l have to work through it all when I've had my tea. Many thanks to all who posted :wink:
  15. Jul 2, 2008 #14
    "I should point out that this is for use with Quaternions but this part of the book makes no reference to that it's just describing complex numbers."

    Somtimes quaternions are called hyper-complex numbers.
  16. Jul 3, 2008 #15
    Thanks pal :wink:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Imaginary Numbers
  1. Imaginary Numbers (Replies: 4)