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$i^3$ equals $-i$. Since $i$ is $\sqrt{-1}$, and you can multiply the number inside radicals that are being multiplied together, wouldn't $i^3$ equal $\sqrt{-1×-1×-1}$, which is $\sqrt{-1}$, which is $i$?

Golemwire
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  • Square roots of negative numbers are not uniquely defined – J. W. Tanner Feb 08 '21 at 22:14
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    sqrt(-1) = ±i the same way sqrt(4) = ±2 – DaveBensonPhillips Feb 08 '21 at 22:14
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    Great question. The use of $\sqrt{-1}$ as shorthand for $i$ is common but unfortunate. In general, you cannot perform algebra of non-positive real numbers both within and outside of square roots without making choices and breaking the properties of the square root that make the notation useful in the real case. e.g. $\sqrt{ab}$ isn't going to be $\sqrt{a} \sqrt{b}$ all of the time and $\sqrt{-1} \sqrt{-1} \sqrt{-1}$ can't be expected to be $\sqrt{(-1)(-1)(-1)}$. For this reason many books avoid the notation $\sqrt{-1}$ for $i$ at all, although for historical reasons many other books don't. – leslie townes Feb 08 '21 at 22:16
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    It is less likely to lead to problems if we think that the main property of $i$ is $i^2=-1$. If that is the first thing we think of when we see $i$, then $i^3 = -i$ is the result we get. – GEdgar Feb 08 '21 at 22:23
  • The whole world of complex analysis can be conjugated and works well as before since we can’t distinguish $i$ and $-i$. – Seewoo Lee Feb 09 '21 at 00:29

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In complex numbers it is wrong to write $i$ as $\sqrt{-1}$. For any complex number $z \neq 0$ there are two distinct values for $\sqrt{z}$ and, for $z \notin [0, \infty)$ there is no choice which is compatible with the algebraic properties you learned about roots.

If you want to use roots, square root is a multivalued function over the complex numbers and $$ \sqrt{-1} = \pm i $$ which makes your formula correct.

N. S.
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In complex analysis, as the other response indicates, the square root function is ambiguous.

In complex analysis, $i = e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2).$

Using DeMoivre's theorem, $i^3 = e^{3i\pi/2}.$

This equates to $-i = e^{-i\pi/2}$ precisely because

$3\pi/2 \equiv -\pi/2 \pmod{2\pi}.$

Note that $(-i)$ is the complex conjugate of $(i)$, which explains why arg$(-i)$ = -arg$(i)$, within a modulus of $2\pi.$

user2661923
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If $a=\sqrt{b}\sqrt{c}$ with $a,\,b,\,c\in\Bbb C\supseteq\Bbb R$ then $a^2=\sqrt{b}\sqrt{c}\sqrt{b}\sqrt{c}=\sqrt{b}\sqrt{b}\sqrt{c}\sqrt{c}=bc$ so $a=\pm\sqrt{bc}$. If $b,\,c\ge0$, we define the square roots of $b,\,c,\,bc$ to be non-negative, so the $\pm$ sign can be dropped. In complex numbers, we can't do this. Famously $i^2=-1\ne\sqrt{(-1)^2}$; you've just encountered a more complicated counterexample.

J.G.
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