Có bao nhiêu số phức Z thỏa mãn |Z+2i-1| = |i|; |Z+3-i|

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} Đặt z=a+bi (a;b\in \mathrm{ℝ}). \mathrm{Ta} \mathrm{có}: \\ \left|\mathrm{z}+2\mathrm{i}-1\right|=\left|\mathrm{i}\right| \\ \iff \sqrt{{\left(\mathrm{a}-1\right)}^2+{(\mathrm{b}+2)}^2}=1 \\ \iff {\left(\mathrm{a}-1\right)}^2+{(\mathrm{b}+2)}^2=1\iff {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{a}+4\mathrm{b}+1+4=1 \\ \iff {\mathrm{a}}^2+{\mathrm{b}}^2=2\mathrm{a}-4\mathrm{b}-4 \left(1\right) \\ \mathrm{Và} \left|\mathrm{z}+3-\mathrm{i}\right|=4 \\ \iff {(\mathrm{a}+3)}^2+{(\mathrm{b}-1)}^2=16\iff {\mathrm{a}}^2+{\mathrm{b}}^2+6\mathrm{a}-2\mathrm{b}+9+1=16 \\ \iff {\mathrm{a}}^2+{\mathrm{b}}^2=-6\mathrm{a}+2\mathrm{b}+6 \left(2\right) \\ \mathrm{Từ} \left(1\right) \mathrm{và} \left(2\right) \Rightarrow 2\mathrm{a}-4\mathrm{b}-4=-6\mathrm{a}+2\mathrm{b}+6 \\ \iff 8\mathrm{a}-6\mathrm{b}=10 \\ \iff \mathrm{a}=\frac{10+6\mathrm{b}}{8}=\frac{5+3\mathrm{b}}{4}. \mathrm{Thay} \mathrm{vào} \left(1\right) \\ {\left(\frac{5+3\mathrm{b}}{4}\right)}^2+{\mathrm{b}}^2=2.\frac{5+3\mathrm{b}}{4}-4\mathrm{b}-4 \\ \iff {(5+3\mathrm{b})}^2+16{\mathrm{b}}^2=8(5\mathrm{b}+3)-64\mathrm{b}-64 \\ \iff \mathrm{b}=\frac{-7}{5} \Rightarrow \mathrm{a}=\frac{1}{5} \\ \mathrm{KL}: \mathrm{Có} 1 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{bài} \mathrm{toán} \end{gathered}$$