Cho số phức $z$ thỏa mãn $\left| z+\overline{z} \right|\le 2$ và...

Đặt \)">z=x+yi\left( x;y\in \mathbb{R} \right)\Rightarrow \overline{z}=x-yi\left\{ \begin{aligned}
& \left| z+\overline{z} \right|\le 2 \\
& \left| z-\overline{z} \right|\le 2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| x+yi+x-yi \right|\le 2 \\
& \left| x+yi-x+yi \right|\le 2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| x \right|\le 1 \\
& \left| y \right|\le 1 \\
\end{aligned} \right.M\left( z \right)ABCDA\left( 1;1 \right),B\left( -1;1 \right),C\left( -1;-1 \right),D\left( 1;-1 \right)T=MEE\left( 0;2 \right)\left\{ \begin{aligned}
& M{{E}_{\min }}=EN \\
& M{{E}_{\max }}=EC=ED \\
\end{aligned} \right.NAB\left\{ \begin{aligned}
& M={{T}_{\max }}=\sqrt{10} \\
& m={{T}_{\min }}=11 \\
\end{aligned} \right.\Rightarrow M+n=1+\sqrt{10}$.