Cho số phức z thoả mãn $\left| {{z}^{2}}+16 \right|+\left| z\left(...

Theo giả thiết, ta có $\left| {{z}^{2}}+16 \right|+\left| z\left( z+4i \right) \right|=4\left| z+4i \right|$
$\begin{aligned}
& \Leftrightarrow \left| z+4i \right|.\left| z-4i \right|+\left| z \right|.\left| z+4i \right|=4\left| z+4i \right| \\
& \Leftrightarrow \left[ \begin{aligned}
& \left| z+4i \right|=0 \\
& \left| z-4i \right|+\left| z \right|=4 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& z=-4i \\
& \left| z-4i \right|+\left| z \right|=4 \left( * \right) \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& z=-4i \\
& z=bi,b\in \left[ 0;4 \right] \\
\end{aligned} \right. \\
\end{aligned}$
Khi đó $\left| z+1-i \right|=\left[ \begin{aligned}
& \left| -4i+1-i \right|=\sqrt{26} \\
& \left| bi+1-i \right|=\sqrt{1+{{\left( b-1 \right)}^{2}}}\in \left[ 1;\sqrt{10} \right],\forall b\in \left[ 0;4 \right] \\
\end{aligned} \right.$
Do đó $m=1,M=\sqrt{26}\Rightarrow P=1+\sqrt{26}.$
* Chú ý: Với $M\left( z \right),A\left( 4i \right),B\left( 0 \right)\Rightarrow \left( * \right)\Leftrightarrow MA+MB=AB\Leftrightarrow M\in AB.$
Do đó $z=bi,b\in \left[ 0;4 \right].$