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Câu hỏi: Có bao nhiêu số phức $z$ thỏa $\left| \dfrac{z+1}{i-z} \right|=1$ và $\left| \dfrac{z-i}{2+z} \right|=1?$
A. $1$
B. $2$
C. $3$
D. $4$
A. $1$
B. $2$
C. $3$
D. $4$
Ta có : $\left\{ \begin{aligned}
& \left| \dfrac{z+1}{i-z} \right|=1 \\
& \left| \dfrac{z-i}{2+z} \right|=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| z+1 \right|=\left| i-z \right| \\
& \left| z-i \right|=\left| 2+z \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-y \\
& 4x+2y=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-\dfrac{3}{2} \\
& y=\dfrac{3}{2} \\
\end{aligned} \right.\Rightarrow z=-\dfrac{3}{2}+\dfrac{3}{2}i.$
& \left| \dfrac{z+1}{i-z} \right|=1 \\
& \left| \dfrac{z-i}{2+z} \right|=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| z+1 \right|=\left| i-z \right| \\
& \left| z-i \right|=\left| 2+z \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-y \\
& 4x+2y=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-\dfrac{3}{2} \\
& y=\dfrac{3}{2} \\
\end{aligned} \right.\Rightarrow z=-\dfrac{3}{2}+\dfrac{3}{2}i.$
Đáp án A.