Câu hỏi 2

Giả sử \(z=a+bi\left( a,b\in R \right)\), ta có:

\(5\overline{z}+3-i=(-2+5i)z\)

\(\begin{array}{l} \Leftrightarrow 5(a - bi) + 3 - i = ( - 2 + 5i)(a + bi)\\ \Leftrightarrow 5a - 5bi + 3 - i =  - 2a - 2bi + 5ai - 5b\end{array}\)

\( \Leftrightarrow \left\{ \begin{array}{l}5a + 3 =  - 2a - 5b\\ - 5b - 1 = 5a - 2b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}7a + 5b =  - 3\\5a + 3b =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b =  - 2\end{array} \right.\)

\(\Rightarrow z=1-2i\)

\(\Rightarrow P=\left| 3i{{\left( z-1 \right)}^{2}} \right|=\left| 3i{{(1-2i-1)}^{2}} \right|=\left| -12i \right|=12\)

Chọn B