Câu hỏi 45

Ta có: \(z = \dfrac{1}{{1 + i}} + \dfrac{2}{{1 - i}} = \dfrac{{1 - i}}{{1 - {i^2}}} + \dfrac{{2\left( {1 + i} \right)}}{{1 - {i^2}}}\)\( = \dfrac{{1 - i}}{2} + \dfrac{{2 + 2i}}{2} = \dfrac{{3 + i}}{2}\)

\( \Rightarrow \left| z \right| = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^2}}  = \dfrac{{\sqrt {10} }}{2}.\)

Chọn B.