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B Imaginary Number

  1. Oct 13, 2017 #1
    I ran into such problem. Not sure if some one can help.

    $$\sqrt{-i^2}=\sqrt{-1\times i^2}=\sqrt{-1\times -1}=\sqrt{1}=1$$

    I also have

    $$\sqrt{-i^2}=\sqrt{-1}\times \sqrt{i^2}=\sqrt{-1}\times i=i\times i=-1$$

    Can anyone explain to me the inconsistencies?
     
  2. jcsd
  3. Oct 13, 2017 #2
    My interpretation is that both of these equations are correct since a square root yields both a positive and a negative answer.
     
  4. Oct 13, 2017 #3

    fresh_42

    Staff: Mentor

  5. Oct 14, 2017 #4

    Mark44

    Staff: Mentor

    I disagree. The first equation is find, because the radical on the left is essentially ##\sqrt 1##, which is 1.
    The second equation is not correct, because the property that ##\sqrt a \sqrt b = \sqrt {ab}## is applicable only if both a and b are nonnegative. This is pointed out in the Insights article that @fresh_42 cited.

    In addition, the real square root of a nonnegative number represents a single number, so it's not correct to say that, for example, ##\sqrt 4 = \pm 2##.
     
  6. Oct 14, 2017 #5

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Either answer is incomplete. The best answer is ±1. Once complex numbers are allowed, you need to be aware that the square root function always has two possible answers.
    Using the '√' notation together with its assumption of a positive answer is treacherous in a context where complex numbers have been introduced. The '√' radical notation implies that the positive square root of a positive number will be used. Use '±√' if you want both to be considered. But once complex numbers are involved, those conventions do not apply.
     
  7. Oct 14, 2017 #6

    Mark44

    Staff: Mentor

    You make a good point, but the first equation starts off with ##\sqrt{-i^2}##. Since -i2 = -(-1) = 1, we are taking the square root of a positive number, and complex numbers are not involved.
     
  8. Oct 16, 2017 #7
    Thank you.
     
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