Question

[Phương Anh]

1>cho phương trình phức :\(\left(z+i\right)^2+3\left(z^2+3zi-2\right)+2\left(z^2+4zi-4\right)=0\) có 2 nghiệm z1,z2 (|z1|<|z2|),tính 2z1+3z2?

A.8i B.-8i C.\(\frac{-47i}{6}\) D.\(\frac{47i}{6}\)

2) cho pt phức \(z^2-z\left(4-i\right)+5+i=0\) có hai nghiệm z1,z2 (|z1|<|z2|). tính |z1-2z2|

A.\(\sqrt{21}\) B.\(\sqrt{17}\) C.\(2\sqrt{5}\) D.\(5\sqrt{2}\)

Answer 1

[Nguyễn Hoàng Việt]

1) Chọn B

\(\left(z+i\right)^2+3\left(z^2+3zi+2i^2\right)+2\left(z^2+4zi+4i^2\right)=0\\ \Leftrightarrow\left(z+i\right)^2+3\left(z+i\right)\left(z+2i\right)+2\left(z+2i\right)^2=0\\ \Leftrightarrow\left(2z+3i\right)\left(3z+5i\right)=0\)

\(\Rightarrow\left\{\begin{matrix}z_1=-3i:2\\z_2=-5i:3\end{matrix}\right.\)

Vậy \(2z_1+3z_2=2\left(\frac{-3i}{2}\right)+3\left(\frac{-5i}{3}\right)=-8i\)

Answer 2

[Nguyễn Hoàng Việt]

2) Chọn D

\(\Delta=\left(4-i\right)^2-4\left(5+i\right)=-5-12i\)

Ta có: \(\Delta=\left(2-3i\right)^2\Rightarrow\sqrt{\Delta}=\pm\left(2-3i\right)\)

Nghiệm của pt là:

\(z=\frac{4-i\pm\sqrt{\Delta}}{2}=\frac{4-i\pm\left(2-3i\right)}{2} \)

\(\Rightarrow\left[\begin{matrix}z=3-2i\\z=1+i\end{matrix}\right.\)

Vì \(\left|z_1\right|< \left|z_2\right|\Rightarrow\left\{\begin{matrix}z_1=1+i\\z_2=3-2i\end{matrix}\right.\)

Vậy \(\left|z_1-2z_2\right|=\left|i+1-6+4i\right|=5\sqrt{2}\)