Xét số phức \(z = a + bi,\left( {a,b \in \mathbb{R...

C

- Hướng dẫn giải

Phương pháp giải:

\(\left| {{z_1}} \right|.\left| {{z_2}} \right| = \left| {{z_1}.{z_2}} \right|\)

Giải chi tiết:

\(\left| z \right| = 1 \Rightarrow \left| {\overline z } \right| = 1 \Rightarrow z.\overline z  = {\left| z \right|^2} = 1\)

\(z = a + bi \Rightarrow {z^2} = {\left( {a + bi} \right)^2} = {a^2} - {b^2} + 2abi\)

Ta có:

\(\begin{array}{l}\left| {{z^3} - z + 2} \right| = \left| {{z^3} - z + 2} \right|.\left| {\overline z } \right| = \left| {{z^3}.\overline z  - z.\overline z  + 2\overline z } \right| = \left| {{z^2}.\left( {z.\overline z } \right) - z.\overline z  + 2\overline z } \right|\\ = \left| {\left( {{a^2} - {b^2} + 2abi} \right).1 - 1 + 2\left( {a - bi} \right)} \right| = \left| {{a^2} - {b^2} + 2abi - 1 + 2a - 2bi} \right|\\ = \left| {{a^2} - {b^2} + 2a - 1 + 2b\left( {a - 1} \right)i} \right| = \left| {2{a^2} + 2a - 2 + 2b\left( {a - 1} \right)i} \right|\end{array}\)

 (do \(\left| z \right| = 1\)\( \Leftrightarrow {a^2} + {b^2} = 1\))

\( = 2\sqrt {{{\left( {{a^2} + a - 1} \right)}^2} + {b^2}{{\left( {a - 1} \right)}^2}}  = 2\sqrt {{{\left( {{a^2} + a - 1} \right)}^2} + \left( {1 - {a^2}} \right){{\left( {a - 1} \right)}^2}} \)

Xét hàm số \(f\left( a \right) = {\left( {{a^2} + a - 1} \right)^2} + \left( {1 - {a^2}} \right){\left( {a - 1} \right)^2} = 4{a^3} - {a^2} - 4a + 2\), \(a \in \left[ { - 1;1} \right]\)

\(f'\left( a \right) = 12{a^2} - 2a - 4,\,\,f'\left( a \right) = 0 \Leftrightarrow \left[ \begin{array}{l}a = \dfrac{2}{3}\\a =  - \dfrac{1}{2}\end{array} \right.\)

\(f\left( {\dfrac{2}{3}} \right) = \dfrac{2}{{27}};f\left( { - \dfrac{1}{2}} \right) = \dfrac{{13}}{4};\,\,f\left( { - 1} \right) = 1;\,\,f\left( 1 \right) = 1\)

\( \Rightarrow {\left| {{z^3} - z + 2} \right|_{\max }} = \sqrt {13} \) khi và chỉ khi \(a =  - \dfrac{1}{2} \Rightarrow {b^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4}\,\,\)

\( \Rightarrow P = 2a + 4{b^2} = 2.\dfrac{{ - 1}}{2} + 4.\dfrac{3}{4} =  - 1 + 3 = 2\).

Chọn: C