Cho số phức z = a + bi ( a, b thuộc R, a < 0 thỏa mãn Tính z

Ta có

      $1+\overline{z}={\left|\overline{z}-i\right|}^2+{\left(iz-1\right)}^2\iff 1+a-bi=a^2+{\left(b+1\right)}^2-a^2+{\left(b+1\right)}^2-2a\left(b+1\right)i$

      $\iff \left\{\begin{array}{l}1+a=2{\left(b+1\right)}^2\\ -b=-2a\left(b+1\right)\end{array}\right.\iff \left\{\begin{array}{l}a=2{\left(b+1\right)}^2-1\\ 1-\left(b+1\right)=-2a\left(b+1\right)\end{array}\right..$

Thế $a=2{\left(b+1\right)}^2-1$ vào phương trình dưới ta được

      $4{\left(b+1\right)}^3-3\left(b+1\right)+1=0\iff \left[\begin{array}{l}b+1=-1\\ b+1=\frac{1}{2}\end{array}\right.\iff \left[\begin{array}{l}b=-2\Rightarrow a=1\left(L\right)\\ b=-\frac{1}{2}\Rightarrow a=-\frac{1}{2}\end{array}\right.\Rightarrow \left|z\right|=\frac{\sqrt{2}}{2}.$

Chọn đáp án A.