The value of i? (complex algebra)?

Question

i = Square Root of -1
S.R. of 16 = S.R. of 4*4 = 2 * S.R. of 4
So...
-i = S.R. of -1 * 1 = S.R. of -1
So...
-i = i
So doesn't...
i = 0
???

Answer 1

You reasoning is faulty.

SR (-1) = SR (i * i) = i

-i = ( SR (-1) ) * -1

i^2 = -1 by definition

(-i)^2 = (-i * -i) = ( (-1 * i) * (-1 * i) ) = (-1 * -1) * (i * i) = (+1) * (-1) = -1

Answer 2

-1 has TWO complex square roots. They are i and -i. Just because they are both square roots of the same number doesn't mean they are equal. By the way, here is a better "proof" of the same thing.
(-1)/1 = 1/(-1)
S.R. {(-1)/1} = S.R. {1/(-1)}
S.R.(-1) / S.R.(1) = S.R.(1) / S.R.(-1)
i/1 = 1/i
i = -i
i = 0

Hope this helps :-)

Answer 3

nope thats wrong..

-i = - S.R. of -1*1 = -1 S.R -1

Answer 4

no

What you don't get is that -1 * 1 is still -1.

You mean sqrt(-1) * -1, which certainly doesn't equal sqrt(-1) * 1.

Answer 5

-i=-Sqrt[-1*1]=-Sqrt[-1]=-i
-i=-i
i=i