smc

Let $P(x)=x^5+a{x}^4+b{x}^3+c{x}^2+dx+2020$. We first notice that $\frac{-1+i\sqrt{3}}{2}=e^{2\pi i/3}$. That is because of Euler's Formula : $e^{ix}=\cos (x)+i\cdot \sin (x)$. $\frac{-1+i\sqrt{3}}{2}$ = $-\frac{1}{2}+i\cdot \frac{\sqrt{3}}{2}$ = $\cos (120^{\circ })+i\cdot \sin (120^{\circ })=e^{2\pi i/3}$. In order $r$ to be a root of $P$, $r{e}^{2\pi i/3}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$, $r{e}^{i\frac{2\pi }{3}}$, $r{e}^{i\frac{4\pi }{3}}$. However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$, if $w$ is a root, then $w{e}^{2\pi i/3}$ and $w{e}^{4\pi i/3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$, $r{e}^{2\pi i/3}$, or $r{e}^{4\pi i/3}$. Thus we know that the polynomial $P$ can be written in the form $(x-r{)}^m(x-r{e}^{2\pi i/3}{)}^n(x-r{e}^{4\pi i/3}{)}^p$. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r|{|}^5=2020$, meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$. In order for the coefficients of the polynomial to all be real, $n=p$ due to $r{e}^{2\pi i/3}$ and $r{e}^{4\pi i/3}$ being conjugates and since $m+n+p=5$, (as the polynomial is 5th degree) we have two possible solutions for $(m,n,p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{2}$.