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

Ta có: $\left| z=4 \right|i+\left| z-2i \right|=\sqrt{5}\left( 1+i \right)\Leftrightarrow \left| a-4+bi \right|i+\left| a+\left( b-2 \right)i \right|=\sqrt{5}+\sqrt{5}i$
$\Leftrightarrow i\sqrt{{{\left( a-4 \right)}^{2}}+{{b}^{2}}}+\sqrt{{{a}^{2}}+{{\left( b-2 \right)}^{2}}}=\sqrt{5}i+\sqrt{5}\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{\left( b-2 \right)}^{2}}=5 \\
& {{\left( a-4 \right)}^{2}}+{{b}^{2}}=5 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}-4b-1=0 \\
& {{a}^{2}}+{{b}^{2}}-8a+11=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 4b+1=8a-11 \\
& {{a}^{2}}+{{b}^{2}}-4b-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2a-3 \\
& {{a}^{2}}+{{\left( 2a-3 \right)}^{2}}-4\left( 2a-3 \right)-1=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=2a-3 \\
& 5{{a}^{2}}-20a+20=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2a-3 \\
& {{\left( a-2 \right)}^{2}}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=1 \\
\end{aligned} \right. $. Vậy $ T=a+b=2+1=3$.