$1$, $z$, and $z^{2019}$ are vertices of an equilateral triangle 
Find all complex numbers $z$, such that the points, in the complex plane, which correspond to $1$, $z$, and $z^{2019}$ form an equilateral triangle.

Thanks in advance.
So far I got $$\frac{z^{2019}-1}{z-1}-\cos(\pi/3)-i\sin(\pi/3)=0$$ and $$\frac{z^{2019}-1}{z-1}-\cos(\pi/3)+i\sin(\pi/3)=0$$ and my task will be completed if I knew how many different roots these two polynomials have?
 A: Let $z$ be any root of the polynomial
$$ \begin{align}f(X)&=\frac{(X^{2019}-1)^3+(X-1)^3}{(X^{2019}-1)+(X-1)}\\
&=(X^{2019}-1)^2-(X^{2019}-1)(X-1)+(X-1)^2\\
&=X^{4038}-X^{2020}-X^{2019}+X^2-X+1.\end{align}$$
Then  $a=1$, $b=z$, $c=z^{2019}$ form an equilateral triangle.
Indeed, if $z=1$, we have a degenerate equilateral triangle, which we may or may not accept as one of the possible solutoins.
In all other cases, let $w=\frac{a-c}{b-a}$ and observe that
$$ 0=(a-c)^2+(a-c)(b-a)+(b-a)^2=(w^2+w+1)(b-a)^2,$$
i.e., $w$ is a primitive third root of unity. It follows that $|a-c|=|b-a|$, but also
$$ c-b=-(a-c)-(b-a)=(-w-1)(b-a)=\overline w(b-a),$$
i.e., also $|c-b|=|b-a|$.
Hence for each of the roots of $f$ (with $z=1$ counted or not), we obtain an equilateral triangle, and there are no other solutions.
Note that $$f'(X)=4038X^{4037}-2020X^{2019}-2019X^{2019}+2X-1$$
and by a straightforward but laborious calculation, we find that
$$ \gcd(f(X),f'(X))=X-1,$$
i.e., $z=1$ is the only root of $f$ that has higher (namely, double) multiplicity. The remaining $4036$ simple roots of $f$ are proper solutions to the problem.
