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Does $\sqrt{-1}=\pm\sqrt{-1}$ since it has 2 solutions for roots? Note the link says there are two complex square roots for -1 which are $i$ and $-i$ https://en.m.wikipedia.org/wiki/Imaginary_unit.

msm
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W. G.
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    Yes, but avoid this notation. Fractionary exponents should be used only with positive real numbers. – Bernard Oct 05 '16 at 21:18
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    @Bernard Why so? If we're dealing with complex functions/numbers there's no problem at all. Of course, a branch for the complex square root must be chosen, but in any case it has to be that way. – DonAntonio Oct 05 '16 at 21:21
  • I appreciate the response Bernard. Put it in the answer next time so I can give you credit! – W. G. Oct 05 '16 at 21:22
  • Of course, when one knows well complex function theory (for example, DonAntonio mentions a "branch of the function square root"), but if the O.P. is a newcomer in the field $\mathbb C$ :), I do think as Bernard that it is a misleading notation that should be short-circuited by speaking about the two roots of equation $z^2=-1=e^{i\pi}$, i.e. $\pm e^{i\frac{\pi}{2}}=\pm i.$ – Jean Marie Oct 05 '16 at 21:30
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    @JeanMarie You have a point there, yet I think it is one of the first things students are said when learning for the first time complex numbers: there nis no order as in the real numbers, and then I think the confussion between positive or negative root is moot. – DonAntonio Oct 05 '16 at 21:34
  • @DonAntonio I understand your point : ${i,-i}$ is a kind of "package" with no one discernable from the other. It's almost quantic... – Jean Marie Oct 05 '16 at 21:36
  • I don't think it is correct. Do you mean $i=\pm i$? – msm Oct 05 '16 at 21:41
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    IMO, the above is abuse of notation. As @msm mentions, it implies that $i=-i$, which is false. Rather, it is better to say that $\sqrt{-1}=i$ and $\pm i$ are the solutions to $x^2=-1$. When you state $x^2=-1$, it is much more clear what you are doing. – Simply Beautiful Art Oct 05 '16 at 21:43
  • https://math.stackexchange.com/q/13801/321264 – StubbornAtom Jul 02 '20 at 18:27

2 Answers2

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Your notation is $$i=\pm i$$ which implies $$i=-i \Rightarrow i=0$$ which is wrong.

Your confusion is probably from $$z^2=-1=e^{\pm i\pi}\Rightarrow z=e^{\pm i\pi/2}=0+i \sin(\pm \pi/2)=\pm i$$ But in the above equation $+i$ and $-i$ are two distinct solutions of the equation. They are not equal. However, they are two square roots of $-1$.

msm
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If $\sqrt{-1} = -\sqrt{-1}$, then $$\frac{\sqrt{-1}}{\sqrt{-1}} = -1.$$ But, $$ \frac{\sqrt{-1}}{\sqrt{-1}} =1.$$ So $1=-1$, a clear contradiction.

Oiler
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