Đáp án:
$\left|z_1\right|+\left|z_2\right|+\left|z_3\right|+\left|z_4\right| =4 +2\sqrt3$
Giải thích các bước giải:
$\quad z^4 - z^2 - 12 = 0$
$\Leftrightarrow (z^2 +3)(z^2 -4)=0$
$\Leftrightarrow (z-i\sqrt3)(z +i\sqrt3)(z-2)(z+2)=0$
$\Leftrightarrow \left[\begin{array}{l}z_1 =i\sqrt3\\z_2 = -i\sqrt3\\z_3 = 2\\z_4 = -2\end{array}\right.$
$\Rightarrow \left|z_1\right|+\left|z_2\right|+\left|z_3\right|+\left|z_4\right| = \sqrt3 +\sqrt3 + 2 + 2 = 4 +2\sqrt3$
