# The square root of $1$ [duplicate]

This question already has an answer here:

I know this is wrong but I don't know why. In the set of complex numbers: $\sqrt1 = \sqrt{i^2\cdot i^2} = i\cdot i = -1$ What is wrong with this?

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## marked as duplicate by MJD, Pedro Tamaroff, Zev Chonoles, Git Gud, Maisam HedyellooJul 28 '13 at 19:54

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.

What are you using $\sqrt{}$ to mean? –  Chris Eagle Jul 28 '13 at 19:40
See this post, please. The square root function is not well-defined on the complex numbers. –  amWhy Jul 28 '13 at 19:41
Note that $\sqrt{xy}=\sqrt{x}\sqrt{y}$ is no longer an identity for complex numbers. –  Adriano Jul 28 '13 at 19:43

## 2 Answers

One must take care when defining $\sqrt{}$ for complex numbers; it's not generally true that the square root of a product is the product of the square roots.

Although this is true for non-negative real numbers, it doesn't extend in full generality.

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Last time I've checked $-1$ squared was equal to one. That would be the usual way of checking if the root is calculated correctly.

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$\sqrt x$ is normally defined to be the non-negative number $y$ such that $y^2 = x$. –  MJD Jul 28 '13 at 19:56
Well, that might be one of the reasons why this question is really a bit imprecise. "Wrong" as OP stated it would require more context. –  Bartek Banachewicz Jul 28 '13 at 19:59