1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    2017 Award

    Staff: Mentor

  5. Oct 14, 2017 #4


    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


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted