The error is from the 3rd step to the 4th step. But why is this an error? Can't $i$ be interchangeable with $\sqrt{-1}$?
$-1 = i\cdot i = \sqrt{-1}\cdot \sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$
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1The third equality. – T.J. Gaffney Nov 28 '14 at 22:03
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2And the second equality. Square root of $x$ is uniquely defined for positive real numbers. Not so for complex numbers. – Simon S Nov 28 '14 at 22:09
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@Eoin I'm not sure what you're getting at. When I read your statement it appears true to me. – Klik Nov 28 '14 at 22:22
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@Eoin Wait, you're saying $-i^2 = -1$?? I thought $(-i)^2=-1$ but $-i^2 = 1$. Hmm... – Klik Nov 29 '14 at 00:37
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@Klik That's what I meant. I didn't add the parantheses on the last one. My mistake – Eoin Nov 29 '14 at 00:38
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What happens is that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is valid for positive real numbers, only.
Ivo Terek
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