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In complex numbers is $\sqrt{-1}$ equal to $i$ or $\pm i$ ?

In both cases how do we explain it?

The question arose when I saw it in Lathi's book (Linear Systems and Signals).

user2692669
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    Doesn't matter. Although $i$ and $-i$ are additive(and multiplicative) inverses of each other, there is no algebraic difference. You can choose anyone to be $\sqrt{-1}$ – Guy Mar 15 '14 at 11:42
  • I think the most usual definition is $;\sqrt{-1}:=i;$ , and this helps only to make things standard. Just like by definition $;\sqrt4=2;$ and not $;\sqrt4=-2;$ : no "natural" reason to choose one over the other, just simplicity and international (more or less..) standard – DonAntonio Mar 15 '14 at 12:10

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$i^2=-1$ and $(-i)^2=-1$. In $\mathbb C$ these are the two distinct roots of equation $z^2+1=0$.

Based on that it might be tempting to say that $\sqrt{-1}=i$ and/or $\sqrt{-1}=-i$, but we deal here with a function on $[0,\infty)\subset \mathbb R$ so $\sqrt{-1}$ is not defined properly here. Extension of the function to $\mathbb C$ leads to 'branches'.

drhab
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