Cho số phức $z=a+bi$ $\left( a,b\in \mathbb{R} \right)$ thỏa mãn...

Giả sử $z=a+bi\left( a,b\in \mathbb{R} \right)$
Ta có $\left\{ \begin{aligned}
& \left| z+1 \right|=2\sqrt{5} \\
& \left| z+5 \right|=2\sqrt{5} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| a+bi+1 \right|=2\sqrt{5} \\
& \left| a+bi+5 \right|=2\sqrt{5} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{\left( a+1 \right)}^{2}}+{{b}^{2}}}=2\sqrt{5} \\
& \sqrt{{{\left( a+5 \right)}^{2}}+{{b}^{2}}}=2\sqrt{5} \\
\end{aligned} \right.$
$\Rightarrow {{\left( a+1 \right)}^{2}}+{{b}^{2}}={{\left( a+5 \right)}^{2}}+{{b}^{2}}\Leftrightarrow a=-3\Rightarrow 4+{{b}^{2}}=20\Leftrightarrow {{b}^{2}}=16\Rightarrow a+{{b}^{2}}=13.$