cho phương trình z^3-(2i-1)z^2+(3-2i)z+3 06/12/2021 Bởi Kylie cho phương trình z^3-(2i-1)z^2+(3-2i)z+3
$z^3-(2i-1)z^2+(3-2i)z+3=0$ $⇔ (z+1)(z^2-2iz+3)=0$ $⇔$ \(\left[ \begin{array}{l}z=-1\\z^2-2iz+3=0\end{array} \right.\) Ta có: $z^2-2iz+3=0$ $Δ’=i^2-3=-4=4i^2$ $⇒$ \(\left[ \begin{array}{l}z=3i\\z=-i\end{array} \right.\) Bình luận
$z^3-(2i-1)z^2+(3-2i)z+3=0$
$⇔ (z+1)(z^2-2iz+3)=0$
$⇔$ \(\left[ \begin{array}{l}z=-1\\z^2-2iz+3=0\end{array} \right.\)
Ta có:
$z^2-2iz+3=0$
$Δ’=i^2-3=-4=4i^2$
$⇒$ \(\left[ \begin{array}{l}z=3i\\z=-i\end{array} \right.\)