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I have a question

  1. Jun 19, 2010 #1
    Question:

    Is the following mathematical argument algebraically valid:

    sqrt(-1) x sqrt(-1) = [sqrt(-1)]2 = -1

    I know sqrt(-1) is a complex number, I just want to know if the argument above is valid.

    Thanks.
     
  2. jcsd
  3. Jun 19, 2010 #2

    jbunniii

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    Yes, it's valid.

    There is one subtlety: there are two numbers whose square is -1, namely [itex]i[/itex] and [itex]-i[/itex].

    Your argument works fine if you choose sqrt(-1) = [itex]i[/itex]:

    [tex]\sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1[/tex]

    and equally well if you choose sqrt(-1) = [itex]-i[/itex]:

    [tex]\sqrt{-1} \times \sqrt{-1} = (-i) \times (-i) = (-i)^2 = -1[/tex]

    But what happens if I choose sqrt(-1) = [itex]i[/itex] for one of the square roots, and sqrt(-1) = [itex]-i[/itex] for the other one? Then I get

    [tex]\sqrt{-1} \times \sqrt{-1} = (i) \times (-i) = -(i^2) = 1[/tex]

    Uh oh!
     
  4. Jun 19, 2010 #3
    LOL @ "uh oh!"

    Thanks. I actually just searched through the threads and see that this question has been asked several times in different ways. But it make sense now. Thanks again.
     
  5. Jun 19, 2010 #4
    That so I was told, is the reason for i. Then the matter becomes [tex]\sqrt(-1)x\sqrt(-1) = i^2 = -1.[/tex]
     
    Last edited: Jun 19, 2010
  6. Jun 20, 2010 #5
    The point is that you have to define sqrt() to be a function, i.e. return only one value. I think this is called principal value. Then your calculations works. You just have to remember that [itex]sqrt(x^2)\neq x[/itex] in general.

    But I'm not completely sure about this Riemann surface business and what it's for :)
     
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