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Possible Duplicate:
-1 is not 1, so where is the mistake?
Significance of $\displaystyle\sqrt[n]{a^n} $?

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

hence,

$i^2 = 1$ What is wrong with the steps shown below? I start with i = sqrt(-1) but then I end up with i = sqrt(1). What is the solution out of this paradox?

Is it that by definition, i is that number whose square is -1. or is it because,

by taking the denominator '1' inside the sqrt, I am losing some information because:

$\sqrt{1} = \pm 1$

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    Very clever. Yes the problem is that the square root is only well-defined up to $\pm 1$, so you're equation shows $i=\pm \dfrac{1}{i}$ which is not a paradox.2011-11-18
  • 0
    And many more of the same spirit are linked to [here](http://math.stackexchange.com/questions/3210/simple-complex-number-problem-1-1)2011-11-18

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