Finding the cube roots of $-i$. As the title suggests, I am having trouble finding the cube roots of $-i$. I looked at the similar post of finding the cube roots of just $i$ but I am having some confusion.
Complex numbers are of the form $z=a+bi$.
$$-i= 0+(-1)i$$
Giving a modulus of $|z|=\sqrt{0^2+(-1)^2}=1$
Then finding $\arg(-1)=\tan^{-1}\left(\large\frac{b}{a}\right)$, right? But our $a=0$, so how can we continue from here? 
I'm sure there is some small detail I am overlooking. Thanks in advance.
 A: You can say $\tan^{-1}(1/0)=\tan^{-1}(\infty)= 90^\circ$ (or you can say $\pi/2$ radians), and this "$\infty$" is neither $+\infty$ nor $-\infty$, but rather is the single $\infty$ that is at both ends of the real line.  That is the "$\infty$" that is appropriate when talking about the ranges of trigonometric functions or about either the domains or the ranges of rational functions, and it makes all such functions continuous at every point in their domains.
So a cube root is $\cos30^\circ+i\sin30^\circ$ (or $\pi/6$ radians if you like).
(Then you also get the two other cube roots that come from adding $\pm360^\circ/3=\pm120^\circ$.)
PS: I had read the question is asking for cube roots of $i$.  I now see that it now says $-i$.  With $-i$, just take the complex conjugates of the numbers above. Or, working from scratch: for $-i$, we need the other angle whose tangent is $\infty$, namely $-90^\circ$.  The answer is then $\cos(-30^\circ)+i\sin(-30^\circ)$, which is the same as $\cos30^\circ-i\sin30^\circ$, and then add $\pm120^\circ$ to get the other two cube roots.
A: There's a simpler and more elementary way of determining the cube roots of $i$.
We have
$$a+bi=\sqrt[3]{-i}$$
Cubing both sides give
$$a^3+3a^2bi-3ab^2-b^3i=-i$$
so
$$a^3-3ab^2=0$$
$$3a^2b-b^3=-1$$
And from there it's easy solving the system.
A: Here's a generalized formula for powers of $i$, taken directly from Euler's formula:
\begin{align}
i &= e^{i\pi \over 2}\\
e^{ix} &= \cos x + i\sin x
\end{align}
Combining the two, we get:
$$
i^x = \left(e^{i\pi \over 2}\right)^x = e^{i{\pi x \over 2}}
$$
Thus:
$$
i^x = \cos\left(\pi x \over 2\right) + i\sin\left(\pi x \over 2\right)
$$
In this specific case, we see that one cube root of $i$ is given by:
$$
i^{1 \over 3} = \cos{\pi \over 6} + i \sin{\pi \over 6} = {\sqrt 3 + i \over 2}
$$
As was mentioned, the argument here is $\pi \over 6$. You can find the remaining two cube roots by adding and subtracting $2\pi \over 3$ from the argument.
A: You can find one solution by inspection: $i^3=(i^2)i=-i$.
So $z=i$ is one solution of the equation $z^3+i=0$.
If you happen to know the cube roots of $1$, you can find all three cube roots of $-i$ by multiplying each of the cube roots of $1$ by $i$:
$i\cdot1=i,\ \ i\cdot(-\frac12+\frac{\sqrt3}2i)=-\frac{\sqrt3}2-\frac12i,\ \ i\cdot(-\frac12-\frac{\sqrt3}2i)=\frac{\sqrt3}2-\frac12i$.
Alternatively, since $z-i$ is a factor of $z^3+i$, you can use long division to find the factorization $z^3+i=(z-i)(z^2+iz-1$) and then use the quadratic formula to solve the quadratic equation $z^2+iz-1=0$.
