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What's the thing with $\sqrt{-1} = i$

What is wrong with the following?:

$$ \frac{1}{\text{i}}=\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=\text{i} $$

There have been a few posts here with similar queries as above saying that

$$\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}$$ does not hold.

But if we are simplifying an expression containing two variables $a$ and $b$ for example in

$$\sqrt{\frac{\left (a+b \right )^2}{-\left( a+ b \right )^2}}$$

we do get


and $\sqrt{-1}$ eventually.

Something must be fundamentally wrong here. What is it?

Thanks a lot...

Update: How would we simplify

$$\sqrt{\frac{\left (a+b \right )^2}{-\left( a+ b \right )^2}}$$ then?

We can say

$$\sqrt{\frac{\left (a+b \right )^2}{-\left( a+ b \right )^2}} = \sqrt{\frac{1}{-1}}$$ and then I'm stuck!


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marked as duplicate by lhf, mixedmath, J. M., Asaf Karagila, Mike Spivey Sep 26 '11 at 20:50

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.

up vote 2 down vote accepted

The familiar "exponent rules", are only valid if all quantities involved are real numbers. This subtlety is not taught when the rules are first introduced because, well, these rules are typically introduced long before students see complex numbers. Then when we move to the field $\mathbb{C}$, suddenly these "iron-clad" rules conflict with complex arithmetic!

Indeed, we have to think of $1/i$ as the number to which multiplication by $i$ produces $1$. In other words, we never want to use the identification $i = \sqrt{-1}$ in $\mathbb{C}$ because $\sqrt{a}$ is no longer unambiguous, as @mixedmath stated above.

Hope this helps!

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Downvoter, please add a comment explaining your downvote. – Shaun Ault Sep 26 '11 at 19:30
Thanks for the link to wolfram. – yCalleecharan Sep 27 '11 at 6:30

You are assuming that $\dfrac{\sqrt a}{\sqrt b} = \sqrt{ \dfrac{a}{b}}$ and that is indeed true when $a$ and $b$ are positive real numbers.

But you cannot hope extend this to all real numbers or to complex numbers ($b \not = 0$) without a definition of the $\sqrt{ }$ function, and if this is single valued then the equality may not always hold.

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No, you were right. That does not hold. One does not 'distribute' square roots. And no, even when 'simplifying,' one cannot distribute square roots in general. The bigger idea is that while in real numbers, we can say $\sqrt 4 = 2$, this only makes sense when we can distinguish between the positive and the negative answer. We can't do so in the complex numbers.

In particular, you sequence is simply an obfuscation of the fact that both $i$ and $-i$ square to give $-1$.

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Thanks for your answer. Square roots and imaginary numbers do not mix well apparently. – yCalleecharan Sep 26 '11 at 12:03
See my updated post: Now, how to simplify [;\sqrt{\frac{\left (a+b \right )^2}{-\left( a+ b \right )^2}};] ? – yCalleecharan Sep 26 '11 at 12:07

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