Cho số phức

D

- Hướng dẫn giải

Ta có 

$$\begin{gathered} z=\frac{-1+\sqrt{3}i}{2}\Rightarrow 2z+1=\sqrt{3}i \\ \Rightarrow {\left(2x+1\right)}^2=-3 \end{gathered}$$

hay $z^2+z+1=0\iff z+\frac{1}{z}=-1$

Khi đó:

$$\begin{gathered} z^2=\frac{1}{z^2}={\left(z+\frac{1}{z}\right)}^2-2=-1 \\ {z}^3=\frac{1}{z^3}={\left(z+\frac{1}{z}\right)}^3-3\left(z+\frac{1}{z}\right)=2 \\ {z}^4=\frac{1}{z^4}=\left(z^2+\frac{1}{z^2}\right)-2=-1 \end{gathered}$$

Như vậy

$$\begin{gathered} P={\left(-1\right)}^{2016}+{\left(-1\right)}^{2017}+2^{2018} \\ +{\left(-1\right)}^{2019}-2^{2018}=-1 \end{gathered}$$

Đáp án D