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} = 3left( {cos varphi + i,sin varphi } right) Rightarrow {z_2} = 4left( {cos left( {varphi pm widehat {AOB}} right) + i,sin left( {varphi pm widehat {AOB}} right)} right) = 4left( {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{{3left( {cos varphi + i,sin varphi } right)}}{{4left( {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) + ileft( {,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 isin 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