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

Giả thiết trở thành $1+a-bi={{\left| a-bi-i \right|}^{2}}+{{\left[ i.(a+bi)-1 \right]}^{2}}$
$\Leftrightarrow 1+a=bi={{\left| a-(b+1)i \right|}^{2}}+{{(-b-1+ai)}^{2}}$
$\Leftrightarrow 1+a-bi={{a}^{2}}+{{(b+1)}^{2}}+{{(b+1)}^{2}}-2a.(b+1).i-{{a}^{2}}$
$\Leftrightarrow 1+a-bi=2.{{(b+1)}^{2}}-2a.(b+1).i\Leftrightarrow \left\{ \begin{aligned}
& 1+a=2.{{(b+1)}^{2}} \\
& -b=-2a(b+1) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{b}{2b+2} \\
& 1+\dfrac{b}{2b+2}=2.{{(b+1)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{b}{2b+2} \\
& 4.{{(b+1)}^{2}}=3b+2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& b=-\dfrac{1}{2} \\
& b=-2 \\
\end{aligned} \right. \\
& a=\dfrac{b}{2b+2} \\
\end{aligned} \right.$
Mặt khác $a<0\Rightarrow a=b=-\dfrac{1}{2}\Rightarrow z=-\dfrac{1}{2}-\dfrac{1}{2}i\Rightarrow \left| z \right|=\dfrac{\sqrt{2}}{2}$