Có bao nhiêu số phức z thỏa mãn z^2=3(z+zngang)+4 và z-1-i=z-3+3i ?

Đáp án B

Đặt $z=a+bi$  .

Khi đó ta có hệ phương trình$\left\{\begin{array}{l}{a}^2+b^2=4\left|a\right|+4\\ \sqrt{{\left(a-1\right)}^2+{\left(b-1\right)}^2}=\sqrt{{\left(a-3\right)}^2+{\left(b+3\right)}^2}\end{array}\right.$

.$\iff \left\{\begin{array}{l}{a}^2+b^2=4\left|a\right|+4\\ a^2+b^2-2a-2b+2=a^2+b^2-6a-6b+18\end{array}\right.\iff \left\{\begin{array}{l}{a}^2+b^2=4\left|a\right|+4\\ 4a=8b+16\end{array}\right.$

$\iff \left\{\begin{array}{l}{\left(2b+4\right)}^2+b^2=4\left|2b+4\right|+4\\ a=2b+4\end{array}\right.\iff \left\{\begin{array}{l}a=2b+4\\ 5b^2+16b+12=\left|8b+16\right|\end{array}\right.$

$\iff \left\{\begin{array}{l}a=2b+4\\ \left[\begin{array}{l}5b^2+16b+12=8b+16\\ 5b^2+16b+12=-8b-16\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}1=2b+4\\ \left[\begin{array}{l}b=\frac{2}{5}\\ b=-2\\ b=-\frac{14}{5}\end{array}\right.\end{array}\right.$

Vậy ta có các số phức$z_1=-2i;z_2=\frac{24}{5}+\frac{2}{5}i;z_3=-\frac{8}{5}-\frac{14}{5}i$  (thỏa mãn).