Why does $i^3$ equal $-i$ if you multiply the numbers inside a radical? $i^3$ 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 $i$?
 A: 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.
A: 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 $2\pi.$
A: 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.
