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Câu hỏi: Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn $\left| \dfrac{z-1}{z-i} \right|=\left| \dfrac{z-3i}{z+i} \right|=1$. Tính $a+b.$
A. 4.
B. 1.
C. 2.
D. 3.
A. 4.
B. 1.
C. 2.
D. 3.
Giả sử $z=a+bi\ \left( a,b\in \mathbb{R} \right)$
Ta có $\left\{ \begin{aligned}
& \left| z-1 \right|=\left| z-i \right| \\
& \left| z-3i \right|=\left| z+i \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}={{a}^{2}}+{{\left( b-1 \right)}^{2}} \\
& {{a}^{2}}+{{\left( b-3 \right)}^{2}}={{a}^{2}}+{{\left( b+1 \right)}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& -2\text{a}+1=-2b+1 \\
& -6b+9=2b+1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=1 \\
\end{aligned} \right.\Rightarrow a+b=2$.
Ta có $\left\{ \begin{aligned}
& \left| z-1 \right|=\left| z-i \right| \\
& \left| z-3i \right|=\left| z+i \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}={{a}^{2}}+{{\left( b-1 \right)}^{2}} \\
& {{a}^{2}}+{{\left( b-3 \right)}^{2}}={{a}^{2}}+{{\left( b+1 \right)}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& -2\text{a}+1=-2b+1 \\
& -6b+9=2b+1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=1 \\
\end{aligned} \right.\Rightarrow a+b=2$.
Đáp án C.