since i^2=-1 does -i = sqrt 1?

Question

is neg. i equal to the square root of 1?

Answer 1

Not quite. I see where you're going with this, though.

You know that i = sqrt(-1)
But that does not mean that -i = sqrt(-(-1).

To see this, consider 4 = sqrt(16).
You would not say that -4 = sqrt(-16).
In fact, you know that -4 = sqrt(16). Technically, it would be that sqrt(16) = +/-4.

Since i is imaginary, it follows its own rules. One could say that sqrt(-1) = +/-i. I am not so sure about that, but we do have something that gives us -i. Consider the following:
i = sqrt(-1)
i^2 = sqrt(-1) * sqrt(-1) = -1
i^3 = i^2 * sqrt(-1) = -1 * i = -i
i^4 = i^3 * sqrt(-1) = -i * i = -i^2 = -(-1) = 1
i^5 = i^4 * sqrt(-1) = 1 * i = i
And so on.

So, i does have some nifty properties, but not that one.

Answer 2

No.
Starting with i^2=-1 we can say that
-i^2=1
Or we could say that
i=sqrt(-1) which is the definition of i.

Sqrt 1 is, of course, +/- 1

In fact, (-i)^2 = (-1)^2 x i^2
= 1 x i^2
= i^2
-1

Answer 3

Hi,
No, “i” is by definition √(-1). So, using the definition -i = -√(-1), but if you then use the definition on that, you just have –i again.
Suppose that we have -√(-36). Then we could write that as this:
-√(36)√(-1)
=-6i

Hope this helps a little.
FE