Re:Learning to Love Complex Numbers
by DanielRibeiro on Tuesday May 27, 2014 @02:13AM
(#7803162)
It is very sound. You can think of i as being a variable 'x', and therefore, you can see Complex numbers are polynomials over real numbers (denoted as R[x]). But you can simplify the polynomials by using the fact that x2 = 1 for this construction (since x = i on this situation).
This is the exact construction of Complex numbers as an algebraic extension over the field of Real numbers[1]
What is more amazing: there is no such way to do the same on the complex numbers. This is because the Complex numbers form an Algebraically closed field[2]
Very sound.
One formal model of complex numbers is R[i]/(i^2 + 1), which means all real polynomials in the variable i (R[i]) after "simplifying" (/) by taking i^2 + 1 = 0 (equivalently i^2 = -1)... which is exactly the process you are describing.
Re:Learning to Love Complex Numbers
by acjohnson55 on Thursday May 29, 2014 @05:23PM
(#7818671)
i's irreducible simply because it's a unit-sized basis vector for how much of a quantity is in the "i direction". It's just something to hang a coefficient on so that it doesn't mix with the real part. We could just as well invent a similar thing for the reals, and call it g. Then your result is 6i + 8g. Or we could use tuples. It doesn't matter, as long as you do proper bookkeeping of the real and imaginary parts.
Example: (i + 3) + (i + 3) == i^2 + 6i + 9 == 6i + 8 is about as logical to me as (x + 3) + (x + 3) == x^2 + 6x + 9. Sure, if we knew what the heck x was this operation would have been easier, but since we don't, we just use FOIL to work around it. In the former, somehow we know what i^2 is but not i, so we can reduce a little further but not completely.
Of course, I don't know how sound this gutfeel impression is.