Cho số phức

C

- Hướng dẫn giải

Đáp án C.

Đặt$z=a+bi,\left(a,b\in ℝ\right)$. Ta có $\left(1-i\right){\left|z\right|}^2+\left(2+2i\right)z^2+2z\left(z+i\right)=0$.

Với $a^2+b^2>0\Rightarrow z\ne 0;{\left|z\right|}^2=z.\overline{z}$. Ta có

$\begin{array}{l}\left(1\right)\iff \left(1-i\right).z.\overline{z}+\left(2+2i\right)z^2+2z\left(z+i\right)=0\\ \iff \left(1-i\right)\overline{z}+\left(2+2i\right)z+2\left(z+i\right)=0\end{array}$

$\iff \left(1-i\right)\left(a-bi\right)+\left(2+2i\right)\left(a+bi\right)$$+2\left(a+\left(b+1\right)i\right)=0$

$\iff a-b-\left(a+b\right)i+2a-2b$$+\left(2a+2b\right)i+2a+\left(2b+2\right)i=0$

$\begin{array}{l}\iff 5a-3b+\left(a+3b+\right)i=0\\ \iff \left\{\begin{array}{l}5a-3b=0\\ a+3b=-2\end{array}\right.\\ \iff \left\{\begin{array}{l}a=-\frac{1}{3}\\ b=-\frac{5}{9}\end{array}\right.\Rightarrow F=\frac{3}{5}\end{array}$