Cho số phức $z=a+bi$ $\left( a,b\in \mathbb{R},a<0 \right)$ thỏa...

Ta có $1+a-bi={{\left| a-bi-i \right|}^{2}}+{{\left[ i\left( a+bi \right)-1 \right]}^{2}}$
$\Rightarrow 1+a-bi={{a}^{2}}+{{\left( b+1 \right)}^{2}}+{{\left( -b-1+ai \right)}^{2}}=2{{\left( b+1 \right)}^{2}}-2a\left( b+1 \right)i$
$\Rightarrow \left\{ \begin{array}{*{35}{l}}
1+a=2{{\left( b+1 \right)}^{2}} \\
-b=-2a\left( b+1 \right) \\
\end{array} \right.\Rightarrow b=2\left( b+1 \right)\left[ 2{{\left( b+1 \right)}^{2}}-1 \right]=2\left( b+1 \right)\left( 2{{b}^{2}}+4b+1 \right)$
$\Rightarrow b=2\left( 2{{b}^{3}}+6{{b}^{2}}+5b+1 \right)\Rightarrow 4{{b}^{3}}+12{{b}^{2}}+9b+2=0\Rightarrow \left[ \begin{array}{*{35}{l}}
b=-2\Rightarrow a=1 \\
b=-\dfrac{1}{2}\Rightarrow a=-\dfrac{1}{2} \\
\end{array} \right.$
Bài ra $a<0\Rightarrow a=-\dfrac{1}{2};b=-\dfrac{1}{2}\Rightarrow \left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=\dfrac{\sqrt{2}}{2}.$ Chọn A.