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When trying to simplify $\frac{1}{\sqrt{-1}}$, you could rationalize it: $$\frac{1}{\sqrt{-1}}\cdot\frac{\sqrt{-1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{-1}=-\sqrt{-1}$$ Or you could simplify it as one radical: $$\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}$$ These are obviously different answers. I'm pretty sure the first method is algebraically sound and the second method might not be because of maybe a branch cut. Could someone please explain why the answers are different, and which is correct?

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marked as duplicate by JMoravitz, Chris Culter, 6005, Grigory M, colormegone Aug 13 '15 at 7:54

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  • $\begingroup$ I think the first method is correct. The second has problem when you identify $1 = \sqrt{1}$. Note that $1 = -\sqrt{-1}\sqrt{-1}$. $\endgroup$ – GAVD Aug 13 '15 at 4:17
  • $\begingroup$ @GAVD $1=\sqrt{1}$ is fine. $\endgroup$ – whacka Aug 13 '15 at 4:54
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Note that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is true only for $a \geq 0$ or $b \geq 0$. On the other hand, if $a,b < 0$, then $$\sqrt{ab} = \sqrt{(-a)(-b)} = \sqrt{-a}\sqrt{-b}$$ $$\sqrt{a}\sqrt{b} = i\sqrt{-a}\ \cdot i\sqrt{-b} = -\sqrt{-a}\sqrt{-b}$$

Therefore the formula $\sqrt{a} = \sqrt{\frac{a}{b}} \sqrt{b}$ holds if and only if $b \neq 0$ and ($\frac{a}{b} \geq 0$ or $b \geq 0$) if and only if $ab > 0$ or $b > 0$. In your second proof, $a = 1, b = -1$ does not satisfy the above condition.

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Let $$\frac{1}{\mathsf i}=a+b\mathsf i,$$ where $a$ and $b$ are real numbers to be determined. Multiplying both sides by $\mathsf i$ yields $$1=a\mathsf i+b\mathsf i^2=-b+a\mathsf i.$$ Now, matching the real and imaginary parts of both sides, one has $-b=1$ and $a=0$. Hence, $$\frac{1}{\mathsf i}=a+b\mathsf i=-\mathsf i.$$

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