Kí hiệu \({z_1};{z_2};{z_3};{z_4}\) là các nghiệm...

D

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Phương trình: \({\left( {\dfrac{{z - 1}}{{2z - i}}} \right)^4} = 1\) điều kiện \(z \ne \dfrac{i}{2}\)               

\(\begin{array}{l} \Leftrightarrow {\left( {z - 1} \right)^4} = {\left( {2{\rm{z}} - i} \right)^4} \Leftrightarrow {\left( {z - 1} \right)^4} - {\left( {2{\rm{z}} - i} \right)^4} = 0\\ \Leftrightarrow \left[ {{{\left( {z - 1} \right)}^2} - {{\left( {2{\rm{z}} - i} \right)}^2}} \right]\left[ {{{\left( {z - 1} \right)}^2} + {{\left( {2{\rm{z}} - i} \right)}^2}} \right] = 0\\ \Leftrightarrow \left( {z - 1 + 2{\rm{z}} - i} \right)\left( {z - 1 - 2{\rm{z}} + i} \right)\left( {{z^2} - 2{\rm{z}} + 1 + 4{{\rm{z}}^2} - 4iz + {i^2}} \right) = 0\\ \Leftrightarrow (3{\rm{z}} - 1 - i)( - z - 1 + i)(5{{\rm{z}}^2} - 2{\rm{z}} - 4iz) = 0\\ \Leftrightarrow \left[ \begin{array}{l}3{\rm{z}} - 1 - i = 0\\ - z - 1 + i = 0\\5{{\rm{z}}^2} - 2{\rm{z}} - 4iz = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3{\rm{z}} - 1 - i = 0\\ - z - 1 + i = 0\\z(5{\rm{z}} - 2 - 4i) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{z_1} = \dfrac{{1 + i}}{3}\\{z_2} = - 1 + i\\{z_3} = 0\\{z_4} = \dfrac{{2 + 4i}}{5}\end{array} \right.\end{array}\)

Khi đó:

\(\begin{array}{l}{z_1}^2 + 1 = {\left( {\dfrac{{1 + i}}{3}} \right)^2} + 1 = \dfrac{{9 + 2i}}{9}\\{z_2}^2 + 1 = {\left( { - 1 + i} \right)^2} + 1 = 1 - 2i\\{z_3}^2 + 1 = {0^2} + 1 = 1\\{z_4}^2 + 1 = {\left( {\dfrac{{2 + 4i}}{5}} \right)^2} + 1 = \dfrac{{4 + 16i + 16{i^2}}}{{25}} + 1 = \dfrac{{13 + 16i}}{{25}}\\ \Rightarrow T = \left( {{z_1}^2 + 1} \right)\left( {{z_2}^2 + 1} \right)\left( {{z_3}^2 + 1} \right)\left( {{z_4}^2 + 1} \right) = \left( {\dfrac{{9 + 2i}}{9}} \right).(1 - 2i).1.\left( {\dfrac{{13 + 16i}}{{25}}} \right)\\ = \dfrac{{\left( {9 - 18i + 2i - 4{i^2}} \right)\left( {13 + 16i} \right)}}{{225}} = \dfrac{{\left( {13 - 16i} \right)\left( {13 + 16i} \right)}}{{225}} = \dfrac{{17}}{9}\end{array}\)

Chọn D