Phương pháp giải:

+) Biểu diễn lượng giác của số phức.

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

Lời giải chi tiết:

Cách 1:

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

Theo đề bài, ta có: \(OA = 3,\,OB = 4,\,AB = \sqrt {41} \)

\( \Rightarrow \cos \widehat {AOB} = \dfrac{{{3^2} + {4^2} - 41}}{{2.3.4}} =  - \dfrac{2}{3}\)

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

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

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

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

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

Cách 2:

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

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

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

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

Chọn: D