Cho số phức z thảo mãn 

D

- Hướng dẫn giải

Đáp án D

Ta có 

$\begin{array}{l}\left|z+\frac{1}{z}\right|=3\iff {\left|z+\frac{1}{z}\right|}^2=9\\ \iff \left(z+\frac{1}{z}\right)\left(\overline{z}+\frac{1}{\overline{z}}\right)=9\\ \iff \frac{\left(z^2+1\right)\left({\overline{z}}^2+1\right)}{z.\overline{z}}=9\end{array}$

$\begin{array}{l}\iff z^2.{\overline{z}}^2+z^2+{\overline{z}}^2+1=9z\overline{z}=9{\left|z\right|}^2\\ \iff {\left|z\right|}^4+{\left(z+\overline{z}\right)}^2-2{\left|z\right|}^2+1=9{\left|z\right|}^2\end{array}$ 

Do ${\left(z+\overline{z}\right)}^2\ge 0$ nên 

$\begin{array}{l}-{\left|z\right|}^4+11{\left|z\right|}^2-1\ge 0\\ \iff {\left|z\right|}^4-11{\left|z\right|}^2+1\le 0\end{array}$

$\begin{array}{l}\iff \frac{11-3\sqrt{13}}{2}\le {\left|z\right|}^2\le \frac{11+3\sqrt{13}}{2}\\ \iff \frac{-3+\sqrt{13}}{2}\le \left|z\right|\le \frac{3+\sqrt{13}}{2}.\end{array}$

Vậy

$\begin{array}{l}\max \left|z\right|+\min \left|z\right|\\ =\frac{-3+\sqrt{13}}{2}+\frac{3+\sqrt{13}}{2}=\sqrt{13}\end{array}$