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Possible Duplicate:
$i^2$ why is it $-1$ when you can show it is $1$?

Try to find what's wrong there:
nb: the squareroot can be defined for all complex numbers as $\exp(1/2\cdot\log(z))$

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

edit:
$ \sqrt{i} = \exp(1/2\cdot\log(i)) $

to find $\log(i)$:
$\begin{align} &x+iy = \log(i)\\ &\exp(x)\exp(iy) = i\\ &\exp(x)\exp(iy) = i\\ &\exp(x)(\cos(y)+i\sin(y)) = i \end{align}$ we arrive at $y=\pi/2$ and $x=0$

so $ \sqrt{i} = \exp(1/2\cdot i\cdot\pi/2) $

and $ \sqrt{i}^2 = i $

where are the multiple values?

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    well had to search far in the answers, gary's answer seems good, tx2012-05-13

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