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$$i=\sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{1}{i}=\frac{-i^2}{i}=-i$$

I'm sure there is a mistake above but I can't figure out where. What's exactly wrong in the above situation & why?

Git Gud
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Omur
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    why $\sqrt{-1}=i$ and not $-i$? – Ofir Schnabel Nov 05 '15 at 10:51
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    See http://math.stackexchange.com/questions/49169/why-sqrt-1-times-1-neq-sqrt-12. – lhf Nov 05 '15 at 10:52
  • otherwise $2i=0$ which is impossible – Surb Nov 05 '15 at 10:52
  • The main problem is using $\sqrt{x}$ on negative numbers and expecting it to behave nicely. Note that what you've really found here is that $(i)^2=(-i)^2=-1$ – Arthur Nov 05 '15 at 10:52
  • @ Arthur The answer you are getting is same as i,-i are conjugates and square of conjugates is same – Archis Welankar Nov 05 '15 at 10:56
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    What fails is $\sqrt {ab}=\sqrt a,\sqrt b $. Not true in general for complex numbers. – Martin Argerami Nov 05 '15 at 10:56
  • Are people seriously still answering this question? If you can't be bothered with looking for one of the many duplicates, don't answer at all. It doesn't help anyone. – Git Gud Nov 05 '15 at 11:12
  • @GitGud Next time I'll check for an available answer. Would it be better to delete this question now? – Omur Nov 05 '15 at 11:21
  • @Omur You should always check for duplicates before asking, but I'm not blaming you as you seem to be a new user. I would delete the question, but that's my personal opinion, I'm no authority here. – Git Gud Nov 05 '15 at 11:27
  • $\sqrt x$ returns the +ve square root, but complex numbers are unordered: http://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number/1117231#1117231 – JMP Nov 05 '15 at 11:42

1 Answers1

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In general (at least as soon as complex numbers are involved), $\sqrt{ab}\ne\sqrt a\sqrt b$. Similarly $\sqrt{\frac1{-1}}\ne\frac{\sqrt 1}{\sqrt{-1}}$

Hagen von Eitzen
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