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$i^2$ why is it $-1$ when you can show it is $1$?

Consider the set of complex numbers. Does the equation $ \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1}$ holds? Why?

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marked as duplicate by lhf, Thomas Andrews, Cameron Buie, Micah, Henry T. Horton Dec 13 '12 at 20:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

How do you define $\sqrt{-1}$? – Nameless Dec 13 '12 at 19:52
If you define $\sqrt{x}$ as a single valued function on complex numbers, it cannot be defined so that $\sqrt{xy}=\sqrt{x}\sqrt{y}$, only so that $\sqrt{xy} = \pm \sqrt{x}\sqrt{y}$ – Thomas Andrews Dec 13 '12 at 19:57
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No. The question is not 'Why does this equation fail to hold?' - it is 'Why should this be true?'. We can prove that $\sqrt{xy}=\sqrt{x}\sqrt{y}$ when at most one of $x,y$ is negative, but not when both are.

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