Câu hỏi 11

Gọi số phức đó là \(z=a+bi,\,\,a,b\in R\). Khi đó:

\(\begin{array}{l}\left| z \right| - 2\overline z  =  - 7 + 3i + z \Leftrightarrow \sqrt {{a^2} + {b^2}}  - 2\left( {a - bi} \right) =  - 7 + 3i + a + bi \Leftrightarrow \sqrt {{a^2} + {b^2}}  - 2a + 2bi =  - 7 + a + \left( {3 + b} \right)i\\ \Leftrightarrow \left\{ \begin{array}{l}\sqrt {{a^2} + {b^2}}  - 2a =  - 7 + a\\2b = 3 + b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sqrt {{a^2} + {3^2}}  - 2a =  - 7 + a\\b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sqrt {{a^2} + 9}  = 3a - 7\\b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge \frac{7}{3}\\{a^2} + 9 = {\left( {3a - 7} \right)^2}\\b = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a \ge \frac{7}{3}\\8{a^2} - 42a + 40 = 0\\b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge \frac{7}{3}\\\left[ \begin{array}{l}a = \frac{5}{4}\\a = 4\end{array} \right.\\b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\\b = 3\end{array} \right. \Rightarrow z = 4 + 3i \Rightarrow \left| z \right| = \sqrt {{4^2} + {3^2}}  = 5\end{array}\)

Chọn: B