# B Imaginary Number

1. Oct 13, 2017

### Tian En

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. Oct 13, 2017

### NFuller

My interpretation is that both of these equations are correct since a square root yields both a positive and a negative answer.

3. Oct 13, 2017

### Staff: Mentor

4. Oct 14, 2017

### 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, $\sqrt 4 = \pm 2$.

5. Oct 14, 2017

### FactChecker

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.

6. Oct 14, 2017

### 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.

7. Oct 16, 2017

Thank you.