Let  $P,Q\in R$  and  ${(1-\sqrt{3}i)}^{200}=2^{199}(P+iQ)$ then  $P-Q+Q^2$  ,  $P+Q+Q^2$  are roots of the equator

$$\begin{gathered} {(1-\sqrt{3}i)}^{200}=2^{199}(P+iQ) \\ \Rightarrow \frac{1}{2}(P+iQ)={(\frac{1}{2}-\frac{\sqrt{3}}{2}i)}^{200} \\ \Rightarrow \begin{array}{r}2{\left(-\frac{1}{2}-\mathrm{i}\frac{\sqrt{3}}{2}\right)}^{200}=\mathrm{p}+\mathrm{iq}\\ \Rightarrow 2{\left(-{\omega }^2\right)}^{200}=\mathrm{p}+\mathrm{iq}\\ \Rightarrow 2\omega =\mathrm{p}+\mathrm{iq}\Rightarrow \mathrm{p}=-1,\mathrm{q}=\sqrt{3}\\ \mathrm{p}=-1,\mathrm{q}=-\sqrt{3}\\ \alpha =\mathrm{p}+\mathrm{q}+{\mathrm{q}}^2=2-\sqrt{3}\\ \beta =\mathrm{p}-\mathrm{q}+{\mathrm{q}}^2=2+\sqrt{3}\\ \alpha +\beta =4\\ \alpha \cdot \beta =1\\ \text{ equation }{\mathrm{x}}^2-4\mathrm{x}+\mathrm{l}=0\end{array} \end{gathered}$$