Cubic with complex roots I have a problem figuring out how exactly I find the cube roots of a cubic with complex numbers. 
I need solve the cubic equation $z^3 − 3z − 1 = 0$.
I've come so far as to calculate the two complex roots of the associated quadratic but then I'm stuck. I've got the solutions here and my lecture notes, have a look at this:

What I don't understand is how you go from $e^{i\pi/3}$ to $e^{i\pi/9}$. Because that root, as I understand it, should be the two conjugates roots added together which I believe do not add up to $e^{i\pi/9}$. What's the step going on here?
Any help is much appreciated!

Certainly $e^{\pi i/9}$ is one of the cube roots of $e^{\pi i/3}$. 
Note that the $3$ cube roots of $1$ are $1$, $e^{2\pi i/3}$, and $e^{4\pi i/3}$.
More familiarly, they are $1$, $\frac{-1+i\sqrt{3}}{2}$, and $\frac{-1-i\sqrt{3}}{2}$.
It follows that $e^{\pi i/9}e^{2\pi i/3}$ and $e^{\pi i/9}e^{4\pi i/3}$ are also cube roots of $e^{\pi i/3}$. If we simplify a bit, we get the $e^{7\pi i/9}$ and $e^{13\pi i/9}$ circled in the post. 
Remark: In general the $n$-th roots of unity are $e^{2\pi ik/n}$, where $k$ ranges from $0$ to $n-1$. For any $a\ne 0$, if we have found an $n$-th root $w$ of $a$, then the $n$-th roots of $a$ are given by $we^{2\pi ik/n}$, where $k$ ranges from $0$ to $n-1$. 
