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Homework Help: Proving i=sqr(-1)

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Our teacher said we can try to prove that the complex numbers will only work so that they are an extension to real numbers if i=sqr(-1).

    2. Relevant equations
    We know that |i|=1, and that for any complex numbers |a||b|=|ab|, and of course that the complex numbers are commutative, associative, and distributive.

    3. The attempt at a solution

    I didn't have much ideas of how to prove this, I figured I need to try proving i^2=-1, so denoting u for the real number 1 unit, I wrote |u+i||u-i|=|u^2-i^2|, knowing that |u|=|i|=1 we can write sqr(2)*sqr(2)=|u^2-i^2|, and so 2=|1-i^2|.

    And from here, since |i|=1 and so |i*i|=1, the only way this can be is if i^2 = -1 and so i=sqr(-1). Is this an entirely precise and complete proof or did I miss something?
  2. jcsd
  3. Oct 1, 2011 #2
    Just putting this out there, [itex]i^2 = 1[/itex] doesn't necessarily mean [itex]i = \sqrt{-1}[/itex]. It could be [itex]-\sqrt{-1}[/itex]...
  4. Oct 1, 2011 #3
    i = (0, 1)

    To multiply complex numbers: (a, b)*(a', b')= (aa'-bb', ab'+a'b)

    keep in mind (a, b) = a+bi where a and b are real numbers.

    Can you now see why [itex] i^2= -1 [/itex] ?

    This is also why one should take care to understand what the definition of i is. It can be... over-simplifying, to say [itex] i= \sqrt{-1} [/itex]
  5. Oct 2, 2011 #4
    @ArcanaNoir: in that multiplication rule you already used that i^2=-1.

    Okay so then how do you prove whether i is + or -sqr(-1)?
  6. Oct 2, 2011 #5
    I didn't "use" i^2= -1

    I showed how you can show it by computing (0,1)*(0,1)

    As for i, i= (0,1), it's no more possible to show whether you use (0,1)*(0,1) or -(0,1)*-(0,1) than it is in real numbers.
  7. Oct 2, 2011 #6


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    Defining "i" as the number whose square is -1 is ambiguous because there are two such numbers. one of them is i and the other is -i. We cannot say "the positive root" as we can in the real numbers because the complex numbers are not an "ordered" field. That is why, as ArcanaNoir says, it is better to define complex numbers as pairs of numbers with the defined multiplication.

    That way, we define i to be the pair (0, 1), not (0, -1), but, other than that, there is no way to distinguish between "i" and "-i".
  8. Oct 2, 2011 #7
    Oh, I see it now. Thanks!
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