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

Ta có
$1+\overline{z}={{\left| \overline{z}-i \right|}^{2}}+{{\left( iz-1 \right)}^{2}}\Leftrightarrow 1+a-bi={{a}^{2}}+{{\left( b+1 \right)}^{2}}-{{a}^{2}}+{{\left( b+1 \right)}^{2}}-2a\left( b+1 \right)i$
$\Leftrightarrow \left\{ \begin{aligned}
& 1+a=2{{\left( b+1 \right)}^{2}} \\
& -b=-2a\left( b+1 \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2{{\left( b+1 \right)}^{2}}-1 \\
& 1-\left( b+1 \right)=-2a\left( b+1 \right) \\
\end{aligned} \right..$
Thế $a=2{{\left( b+1 \right)}^{2}}-1$ vào phương trình dưới ta được
$4{{\left( b+1 \right)}^{3}}-3\left( b+1 \right)+1=0\Leftrightarrow \left[ \begin{aligned}
& b+1=-1 \\
& b+1=\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& b=-2\Rightarrow a=1\left( L \right) \\
& b=-\dfrac{1}{2}\Rightarrow a=-\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow \left| z \right|=\dfrac{\sqrt{2}}{2}.$