Câu hỏi 9

Đặt \(z = a + bi\,\,\left( {a;b \in \mathbb{R}} \right) \Rightarrow \overline z  = a - bi\).

Theo bài ra ta có: \(\left( {2 - i} \right)z + 3 + 16i = 2\left( {\overline z  + i} \right).\)

\(\begin{array}{l} \Leftrightarrow \left( {2 - i} \right)\left( {a + bi} \right) + 3 + 16i = 2\left( {a - bi + i} \right)\\ \Leftrightarrow 2a + 2bi - ai + b + 3 + 16i = 2a - 2bi + 2i\\ \Leftrightarrow \left\{ \begin{array}{l}2a + b + 3 = 2a\\2b - a + 16 =  - 2b + 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b =  - 3\\4b - a =  - 14\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b =  - 3\\a = 2\end{array} \right.\\ \Rightarrow z = 2 - 3i \Rightarrow \left| z \right| = \sqrt {{2^2} + {{\left( { - 3} \right)}^2}}  = \sqrt {13} \end{array}\)

Chọn C