Cho 2 số phức

C

- Hướng dẫn giải

Đáp án C

Ta có 

$\begin{array}{l}\left\{\begin{array}{l}{z}_1+z_2=3\\ z_1.z_2=4\end{array}\right.\\ \iff \left\{\begin{array}{l}{z}_2=3-z_1\\ z_1\left(3-z_1\right)=4\end{array}\right.\\ \iff \left\{\begin{array}{l}{z}_2=3-z_1\\ z_1^2-3z_1+4=0\end{array}\right.\end{array}$

$\begin{array}{l}\iff \left\{\begin{array}{l}{z}_2=3-z_1\\ \left[\begin{array}{l}{z}_1=\frac{3}{2}+\frac{\sqrt{7}}{2}i\\ z_2=\frac{3}{2}-\frac{\sqrt{7}}{2}i\end{array}\right.\end{array}\right.\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}{z}_1=\frac{3}{2}+\frac{\sqrt{7}}{2}i\\ z_2=\frac{3}{2}-\frac{\sqrt{7}}{2}i\end{array}\right.\\ \left\{\begin{array}{l}{z}_1=\frac{3}{2}-\frac{\sqrt{7}}{2}i\\ z_2=\frac{3}{2}+\frac{\sqrt{7}}{2}i\end{array}\right.\end{array}\right.\\ \Rightarrow \left|z_1\right|+\left|z_2\right|=2\sqrt{\frac{9}{4}+\frac{7}{4}}=4\end{array}$