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Câu hỏi: Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn $\left| z-2 \right|=\left| z \right|$ và $\left( z+1 \right)\left( \bar{z}-i \right)$ là số thực. Tính $a+b.$
A. 2.
B. $-2.$
C. 1.
D. $-1.$
A. 2.
B. $-2.$
C. 1.
D. $-1.$
Giả sử $z=a+bi\left( a,b\in R \right)$
Ta có $\left\{ \begin{array}{*{35}{l}}
\left| z-2 \right|=\left| z \right| \\
\left( z+1 \right)\left( \bar{z}-i \right)\in R \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
\left| a-2+bi \right|=\left| a+bi \right| \\
\left( a+1+bi \right)\left[ a-\left( b+1 \right)i \right]\in R \\
\end{array} \right.$
$\Rightarrow \left\{ \begin{array}{*{35}{l}}
{{\left( a-2 \right)}^{2}}+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\
a\left( a+1 \right)+b\left( b+1 \right)-\left( a+b+1 \right)i\in \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=1 \\
a+b+1=0 \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=1 \\
b=-2 \\
\end{array} \right.\Rightarrow a+b=-1.$
Ta có $\left\{ \begin{array}{*{35}{l}}
\left| z-2 \right|=\left| z \right| \\
\left( z+1 \right)\left( \bar{z}-i \right)\in R \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
\left| a-2+bi \right|=\left| a+bi \right| \\
\left( a+1+bi \right)\left[ a-\left( b+1 \right)i \right]\in R \\
\end{array} \right.$
$\Rightarrow \left\{ \begin{array}{*{35}{l}}
{{\left( a-2 \right)}^{2}}+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\
a\left( a+1 \right)+b\left( b+1 \right)-\left( a+b+1 \right)i\in \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=1 \\
a+b+1=0 \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=1 \\
b=-2 \\
\end{array} \right.\Rightarrow a+b=-1.$
Đáp án D.