Cho số z thỏa mãn $\left| z+8-3i \right|=\left| z-i \right|$ và...

Giả sử $z=x+yi\left( x,y\in \mathbb{R} \right)$
Ta có $\left| z+8-3i \right|=\left| z-i \right|\Leftrightarrow \left| \left( x+8 \right)+\left( y-3 \right)i \right|=\left| x+\left( y-1 \right)i \right|$
$\Leftrightarrow {{\left( x-8 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{x}^{2}}+{{\left( y-1 \right)}^{2}}\Leftrightarrow 16x-4y+72=0\Leftrightarrow 4x-y+18=0$.
Lại có $\left| z+8-7i \right|=\left| z+4-i \right|\Leftrightarrow \left| \left( x+8 \right)+\left( y-7 \right)i \right|=\left| \left( x+4 \right)+\left( y-1 \right)i \right|$
$\Leftrightarrow {{\left( x+8 \right)}^{2}}+{{\left( y-7 \right)}^{2}}={{\left( x-4 \right)}^{2}}+{{\left( y-1 \right)}^{2}}\Leftrightarrow 8x-12y+96=0\Leftrightarrow 2x-3y+24=0$
Giải hệ $\left\{ \begin{aligned}
& 4x-y+18=0 \\
& 2x-3y+24=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-3 \\
& y=6 \\
\end{aligned} \right.\Rightarrow \left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=3\sqrt{5}$.