Cho hai số phức \[{z_1}\] và \[{z_2}\] thỏa mãn \[\left| {{z_1}} \right| = 3,\left| {{z_2}} \right| = 4;\left| {{z_1} - {z_2}} \right| = \sqrt {41} .\] Xét các số phức \[z = \frac{{{z_1}}}{{{z_2}}} = a + bi{\mkern 1mu} \left( {a,b \in \mathbb{R}} \right).\] Khi đó \[\left| b \right|\] bằng

Đáp án D

+ Biểu diễn lượng giác của số phức

+ \(\frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}},{z_2} \ne 0\)

Cách 1: Gọi A, B lần lượt là các điểm biểu diễn của số phức \({z_1},{z_2}\)

Theo đề bài, ta có: \(OA = 3,OB = 4,AB = \sqrt {41} \Rightarrow \cos \widehat {AOB} = \frac{{{3^2} + {4^2} - 41}}{{2.3.4}} = - \frac{2}{3}\)

Đặt \({z_1} = 3\left( {\cos \varphi + i\sin \varphi } \right) \Rightarrow {z_2} = 4\left( {\cos (\varphi \pm AOB)} \right) = 4\left( {\cos (\varphi \pm \alpha ) + i\sin (\varphi \pm \alpha )} \right)\) \(\left( {\alpha = AOB} \right)\)

\( \Rightarrow \frac{{{z_1}}}{{{z_2}}} = \frac{{3\left( {\cos \varphi + i\sin \varphi } \right)}}{{4\left( {\cos (\varphi \pm \alpha ) + i\sin \left( {\varphi \pm \alpha } \right)} \right)}} = \frac{3}{4}\left( {\cos \varphi + i\sin \varphi } \right)\left( {\cos (\varphi \pm \alpha ) - i\sin \left( {\varphi \pm \alpha } \right)} \right)\)

\( = \frac{3}{4}\left[ {\left( {\cos \varphi .\cos \left( {\varphi \pm \alpha } \right) + \sin \varphi .\sin \left( {\varphi \pm \alpha } \right)} \right) + i\left( {\sin \varphi .\cos (\varphi \pm \alpha )} \right) - \cos \varphi .\sin \left( {\varphi \pm \alpha } \right)} \right]\)

\( = \frac{3}{4}\left[ {\cos \left( { \pm \alpha } \right) + i\sin \left( { \pm \alpha } \right)} \right] = \frac{3}{4}\left( {\cos \alpha \pm i\sin \alpha } \right)\)

\( \Rightarrow b = \pm \frac{3}{4}\sin \alpha \Rightarrow \left| b \right| = \frac{3}{4}\sqrt {1 - {{\left( {\frac{2}{3}} \right)}^2}} = \frac{{\sqrt 5 }}{4}\).

Cách 2: Ta có: \(\left| {{z_1}} \right| = 3,{\rm{ }}\left| {{z_2}} \right| = 4,{\rm{ }}\left| {{z_1} - {z_2}} \right| = \sqrt {41} \Rightarrow \left\{ \begin{array}{l}\frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{3}{4}\\\frac{{\left| {{z_1} - {z_2}} \right|}}{{\left| {{z_2}} \right|}} = \frac{{\sqrt {41} }}{4}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{3}{4}\\\left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right| = \frac{{\sqrt {41} }}{4}\end{array} \right.\)

\(z = \frac{{{z_1}}}{{{z_2}}} = a + bi,\left( {a,b \in \mathbb{R}} \right) \Rightarrow \left\{ \begin{array}{l}{a^2} + {b^2} = {\left( {\frac{3}{4}} \right)^2}\\{\left( {a - 1} \right)^2} + {b^2} = {\left( {\frac{{\sqrt {41} }}{4}} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{a^2} + {b^2} = \frac{9}{{16}}\\{\left( {a - 1} \right)^2} + {b^2} = \frac{{41}}{{16}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{b^2} = \frac{9}{{16}} - {a^2}\\{\left( {a - 1} \right)^2} + \frac{9}{{16}} - {a^2} = \frac{{41}}{{16}}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{b^2} = \frac{5}{{16}}\\a = - \frac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| b \right| = \frac{{\sqrt 5 }}{4}\\a = - \frac{1}{2}\end{array} \right.\).

Vậy \(\left| b \right| = \frac{{\sqrt 5 }}{4}\).