Đáp án:
$A.\ P = -2^{2021}$
Giải thích các bước giải:
$\quad z^2 + 2z+ 4 = 0$
$\Leftrightarrow \left[\begin{array}{l}z_1 = - 1 + i\sqrt3\\z_2 = - 1 - i\sqrt3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}z_1 = 2\left(- \dfrac12 + i\dfrac{\sqrt3}{2}\right)\\z_2 =2\left( - \dfrac12 - i\dfrac{\sqrt3}{2}\right)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}z_1= 2\left(\cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\right)\\z_2= 2\left(\cos\dfrac{2\pi}{3} - i\sin\dfrac{2\pi}{3}\right)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}z_1^{2021}= 2^{2021}\left(\cos\dfrac{4042\pi}{3} + i\sin\dfrac{4042\pi}{3}\right)\\z_2^{2021}= 2^{2021}\left(\cos\dfrac{4042\pi}{3} - i\sin\dfrac{4042\pi}{3}\right)\end{array}\right.$ (Công thức $Moivre$)
$\Leftrightarrow \left[\begin{array}{l}z_1^{2021}= 2^{2021}\left(-\dfrac12 - i\dfrac{\sqrt3}{2}\right)\\z_2^{2021}= 2^{2021}\left(-\dfrac12+i\dfrac{\sqrt3}{2}\right)\end{array}\right.$
$\Leftrightarrow z_1^{2021} + z_2^{2021} = - 2^{2020} - 2^{2020}.i\sqrt3 - 2^{2020} + 2^{2020}.i\sqrt3$
$\Leftrightarrow z_1^{2021} + z_2^{2021} = - 2.2^{2020}$
$\Leftrightarrow z_1^{2021} + z_2^{2021} = -2^{2021}$
Vậy $P = -2^{2021}$
