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The problem is that \(\sqrt{1}=\pm1\).

no, sqrt(1) = 1

check your calculator

If you use the second way, then \(\sqrt{-1}\cdot\sqrt{-1}=|-1|\cdot|-1|=1\cdot1=1\).

Although there are problems when you do what I just did.

\[\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}\]

but sqrt(-2) = sqrt(-1 * 2) = sqrt(-1) * sqrt(2)

oh, at least one of a,b must be positive, ok

i sqrt 2 is fine

convert to i wherever possible instead of messing with sqrt signs

sorry i can only give one medal

micah, then you have no choice but to convert two negative signs to positive in a product

but if we use king georges idea, thatn sqrt(-1) = + - i

because (-i)^2 = -1, (+i)^2 = -1

there are 2 roots of -1, i and -i

right, but what does sqrt(-1) equal to, is it multivalued or not?

in context

scroll above, and see king georges response

, king george, sqrt( x^2) = |x| , so i think you are wrong there

ok so in the real numbers this is not valid

Ugh....i^2 = -1 and x^2 = -1 has 2 solutions
Can you get abiguity, yes - so be careful

Oops. I totally was wrong there :(

, what is principal branch of the complex square root

You don't want to know....

estudier, i think i can handle it :)
please send me a link

http://en.wikipedia.org/wiki/Branch_point