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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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    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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First and foremost, $\sqrt{1}$ is not $1$ or $-1$. When we discuss the square root of a number $\sqrt{n}$ this denotes the positive or principal square root. Therefore, $\sqrt{1} = 1$ and $1$ alone. This is different, however, than the fact that $x^2 = 1$ has two solutions, namely $1$ and $-1$.

Secondly, square roots and complex numbers do not behave as square roots and real numbers. This is because to properly define the square root function in the complex plane, we need to take a branch cut of the function $f(z) = \sqrt{z}$.

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    @Arturo: That's fair. Thank you for the comment.2011-11-18