Có bao nhiêu số phức z thỏa mãn hệ |z-1+2i|=|z+3-i|

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Hệ} \left\{\begin{array}{l}\left|\mathrm{z}-1+2\mathrm{i}\right|=\left|\mathrm{z}+3-\mathrm{i}\right| \\ \left|\mathrm{z}-1+3\mathrm{i}\right|=4 \end{array}\right. \\ \iff \left\{\begin{array}{l}\left|\mathrm{a}-1+(\mathrm{b}+2)\mathrm{i}\right|=\left|\mathrm{a}+3+(\mathrm{b}-1)\mathrm{i}\right|\\ \left|\mathrm{a}-1+(\mathrm{b}+3)\mathrm{i}\right|=4\end{array}\right. \\ \iff \left\{\begin{array}{l}{\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+2\right)}^2={\left(\mathrm{a}+3\right)}^2+{\left(\mathrm{b}-1\right)}^2\\ {\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+3\right)}^2=16\end{array}\right. \\ \iff \left\{\begin{array}{l}8\mathrm{a}-6\mathrm{b}+5=0 (*)\\ {\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+3\right)}^2=16 (**)\end{array}\right. \\ \mathrm{Từ} (*)\iff \mathrm{b}=\frac{5+8\mathrm{a}}{6} \mathrm{thay} \mathrm{vào} (**): \\ {\left(\mathrm{a}-1\right)}^2+{\left(\frac{5+8\mathrm{a}}{6}+3\right)}^2=16 \\ \iff {\left(\mathrm{a}-1\right)}^2+{\left(\frac{8\mathrm{a}+23}{6}\right)}^2=16 \\ \iff 36{\mathrm{a}}^2-72\mathrm{a}+36+64{\mathrm{a}}^2+368\mathrm{a}+529-576=0 \\ \iff 100{\mathrm{a}}^2+296\mathrm{a}-11=0 \left(1\right) \\ \mathrm{PT}\left(1\right) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{a} \mathrm{phân} \mathrm{biệt} \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \end{gathered}$$