Cube Roots are Complex? I am in DE and finding singular points for power series. For a singular point at $x^3=-1$, I said $x=-1$ and called it a day. Turns out this is wrong (and can be shown via Euler's formula). 
When are you supposed to learn this stuff in math? Euler's formula was always given as a quick hand-wavey tool for turning $e^{a+ib}$ into a function of sine and cosine. (I understand the intuition behind it from my own extra googling, but there was never a point at which it was formally introduced.) Is the usage, including the implication that a cube root of a negative has complex roots, something that should have been taught to me, or something I should have just known by familiarity with Euler's? 
I'm ok with things becoming more complex (not in the math sense), but feel like some of this stuff is hidden and difficult to figure out on one's own. All of math up until this point has taught that $\sqrt[3]{x} == -\sqrt[3]{|x|} , x<0$.
 A: $z^n -1 = 0$ has n roots.  This is a consequence of the Fundamental Theorem of Algebra.
In $n$ is odd one of them is real $z = 1$.  If n is even 2 are real $z=-1$ is also a root
The rest are complex.
This example is called "the roots of unity" 
Frequently, we only care about the real roots, so we can get away with saying 
$\sqrt[3]{-1} = -1$
When we use the $\sqrt[3]{}$ symbol, it is generally and indication that we are looking for the real roots.  
However,  $z = \frac 12 + \frac {\sqrt 3}{2} i, z = \frac 12 - \frac {\sqrt 3}{2} i$ are also roots of $z^3 = -1.$
When were you supposed to have learned this?  It was discussed briefly in Trig class.  You leaned DeMovire's theorem, and learned how to calculate these.  Then it wasn't used again for all of calculus 1, and you forgot it.  It is usually comes up again sometime in your second year of calculus, and you say, "oh yah, I forgot about that."
And then, you take Complex Analysis, and this along with Euler's Identity is drilled into your head.
A: I learned that the equation $x = \sqrt[n]{a}$ has $n $ distinct solutions when I was 17, in school, so I guess you should have already heard about it taking into account you are now in a DE course.
But since you haven't, allow me:
I will assume you are familiar with the polar representation of a complex number as $z = \rho e^{i\theta} $ where $\rho $ is the distance from $z $ to the origin and $\theta $ is its phase or the angle it makes with the real axis.
Now if $z =  \rho_z e^{i\theta_z}$ and $w =  \rho_w e^{i\theta_w}$ then the product $zw = (\rho_z e^{i\theta_z})(\rho_w e^{i\theta_w}) = (\rho_z\rho_w)e^{i (\theta_z + \theta_w)} $
That is way nicer than multiplying two complex numbers in their cartesian form ($z = x + iy $)
Now imagine you are asked to solve $x = \sqrt[n]{z} $ for a given $z = \rho e^{i\theta}$. Then you know $$x = \rho_x e^{i\theta_x} \rightarrow x^n = (\rho_x e^{i\theta_x})^n = (\rho_x)^ne^{i(n\theta_x)} =  \rho e^{i\theta} = z$$
After thinking for a second you realize that can only happen if


*

*$\rho = (\rho_x)^n \iff \rho_x = \sqrt[n]{\rho} $

*$\theta = n \theta_x $


But both $\rho $ and $\rho_x $ are non-negative reals so you can compute that root easily. Then all it takes is to solve the second point.
$\theta = n \theta_x \iff \frac {\theta}{n} = \theta_x $ and we are done! 
Wrong!
We must remember that $e^{i\theta} = e^{i (\theta + 2k\pi)}$ for any integer $k $ so we should instead have
$$\theta + 2k\pi = n \theta_x \iff \frac {\theta + 2k\pi}{n} \iff \theta_x = \frac {\theta}{n} + \frac {2k\pi}{n}$$
Now looking into your formula you see that $k \in \{0, 1, 2, \cdots, n-1\} $ gives $n $ different possible values of $\theta_x $ thus meaning $z $ has $n $ n-roots.
If you try to understand the geometry behind this, you see that all roots are distributed evenly in a circumference with center in the origin and radius $\sqrt[n]{\rho}$.
