I was asked to find the 3rd root of -i. Now, since $i^3 = -i$ I thought that I can just extract the 3rd root out of both sides and get $\root3\of{i^3} = \root3\of{-i} \implies i = \root3\of{-i}$. Searching Google and Wolfram Alpha, I found that $\root3\of{-i}=0.866025404-0.5i$.
The question is: is $i$ a cube root of $-i$ as well? (P.S.: This might be a silly question, but I have just started studying complex numbers).
On
It happens that every non-zero complex numbers has $3$ cube roots.
In your case, since $i=\cos\left(\frac\pi2\right)+\sin\left(\frac\pi2\right)i$, those cube roots will be$$\cos\left(\frac\pi6\right)+\sin\left(\frac\pi6\right)i,\ \cos\left(\frac{5\pi}6\right)+\sin\left(\frac{5\pi}6\right)i\text{ and }\cos\left(\frac{3\pi}2\right)+\sin\left(\frac{3\pi}2\right)i(=-i).$$
On
Yes, $i$ is one of the three cube roots of $-i$. To find the others, you can factor $$x^3+i=x^3-i^3=(x-i)(x^2+ix+i^2)=(x-i)(x^2+ix-1)$$ and now you can find the other two roots by solving the quadratic equation $x^2+ix-1=0$.
More simply, the three cube roots of $-i$ are $i$, $i\omega$, and $i\omega^2$, where $$\omega=\cos120^\circ+i\sin120^\circ=-\frac12+i\frac{\sqrt3}2$$ is a primitive cube root of $1$.
Let simply check that
$$i\times i = -1 \implies i\times i \times i= -1\times i=-i$$
indeed $z^3=-i$ has exactly three solutions in $\mathbb{C}$ and one is $z=i$.