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Câu hỏi: Tìm số phức $z$ thỏa mãn $\left| z-2 \right|=\left| z \right|$ và $\left( z+1 \right)\left( \bar{z}-i \right)$ là số thực.
A. $z=2-i.$
B. $z=1-2i.$
C. $z=1+2i.$
D. $z=-1-2i.$
A. $z=2-i.$
B. $z=1-2i.$
C. $z=1+2i.$
D. $z=-1-2i.$
Gọi $z=x+iy$ với $x,y\in \mathbb{R}$ ta có hệ phương trình $\left\{ \begin{aligned}
& \left| z-2 \right|=\left| z \right| \\
& \left( z+1 \right)\left( \bar{z}-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-2 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{y}^{2}} \\
& \left( x+1+iy \right)\left( x-iy-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-2 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{y}^{2}} \\
& \left( x+1+iy \right)\left( x-iy-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& \left( -x-1 \right)\left( y+1 \right)+xy=0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& y=-2 \\
\end{aligned} \right.$
& \left| z-2 \right|=\left| z \right| \\
& \left( z+1 \right)\left( \bar{z}-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-2 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{y}^{2}} \\
& \left( x+1+iy \right)\left( x-iy-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-2 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{y}^{2}} \\
& \left( x+1+iy \right)\left( x-iy-i \right)\in \mathbb{R} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& \left( -x-1 \right)\left( y+1 \right)+xy=0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& y=-2 \\
\end{aligned} \right.$
Đáp án B.