Xét số phức z thỏa mãn \(\left( {1 + 2i} \right)\l...

D

- Hướng dẫn giải

Ta có \({z^{ - 1}} = \frac{1}{{{{\left| z \right|}^2}}}\overline z .\)

Vậy \(\left( {1 + 2i} \right)\left| z \right| = \frac{{\sqrt {10} }}{z} - 2 + i  \Leftrightarrow \left( {\left| z \right| + 2} \right) + \left( {2\left| z \right| - 1} \right)i = \left( {\frac{{\sqrt {10} }}{{{{\left| z \right|}^2}}}} \right).\overline z \)

\( \Rightarrow {\left( {\left| z \right| + 2} \right)^2} + {\left( {2\left| z \right| - 1} \right)^2} = \left( {\frac{{10}}{{{{\left| z \right|}^4}}}} \right).{\left| z \right|^2} = \frac{{10}}{{{{\left| z \right|}^2}}}.\) Đặt \(\left| z \right| = a > 0.\)

\( \Rightarrow {\left( {a + 2} \right)^2} + {\left( {2a - 1} \right)^2} = \left( {\frac{{10}}{{{a^2}}}} \right) \Leftrightarrow {a^4} + {a^2} - 2 = 0 \Leftrightarrow \left[ \begin{array}{l}
{a^2} = 1\\
{a^2} =  - 2
\end{array} \right. \Rightarrow a = 1 \Rightarrow \left| z \right| = 1.\)