Google
Câu hỏi: . 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 $z=\dfrac{{{z}_{1}}}{{{z}_{2}}}=a+bi\left( a,b\in \mathbb{R} \right).$ Khi đó $\left| b \right|$ bằng
A. $\dfrac{\sqrt{3}}{8}.$
B. $\dfrac{3\sqrt{3}}{8}.$
C. $\dfrac{\sqrt{2}}{4}.$
D. $\dfrac{\sqrt{5}}{4}.$
A. $\dfrac{\sqrt{3}}{8}.$
B. $\dfrac{3\sqrt{3}}{8}.$
C. $\dfrac{\sqrt{2}}{4}.$
D. $\dfrac{\sqrt{5}}{4}.$
+ Biểu diễn lượng giác của số phức
+ $\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|},{{z}_{2}}\ne 0$
Cách 1: Gọi A, B lần lượt là các điểm biểu diễn của số phức ${{z}_{1}},{{z}_{2}}$
Theo đề bài, ta có: $OA=3,OB=4,AB=\sqrt{41}\Rightarrow \cos \widehat{AOB}=\dfrac{{{3}^{2}}+{{4}^{2}}-41}{2.3.4}=-\dfrac{2}{3}$
Đặt ${{z}_{1}}=3\left( \cos \varphi +i\sin \varphi \right)\Rightarrow {{z}_{2}}=4\left( \cos (\varphi \pm AOB) \right)=4\left( \cos (\varphi \pm \alpha )+i\sin (\varphi \pm \alpha ) \right)$ $\left( \alpha =AOB \right)$
$\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{3\left( \cos \varphi +i\sin \varphi \right)}{4\left( \cos (\varphi \pm \alpha )+i\sin \left( \varphi \pm \alpha \right) \right)}=\dfrac{3}{4}\left( \cos \varphi +i\sin \varphi \right)\left( \cos (\varphi \pm \alpha )-i\sin \left( \varphi \pm \alpha \right) \right)$
$=\dfrac{3}{4}\left[ \left( \cos \varphi .\cos \left( \varphi \pm \alpha \right)+\sin \varphi .\sin \left( \varphi \pm \alpha \right) \right)+i\left( \sin \varphi .\cos (\varphi \pm \alpha ) \right)-\cos \varphi .\sin \left( \varphi \pm \alpha \right) \right]$
$=\dfrac{3}{4}\left[ \cos \left( \pm \alpha \right)+i\sin \left( \pm \alpha \right) \right]=\dfrac{3}{4}\left( \cos \alpha \pm i\sin \alpha \right)$
$\Rightarrow b=\pm \dfrac{3}{4}\sin \alpha \Rightarrow \left| b \right|=\dfrac{3}{4}\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}}=\dfrac{\sqrt{5}}{4}$.
Cách 2: Ta có:
$\left| {{z}_{1}} \right|=3,\left| {{z}_{2}} \right|=4,\left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{41}\Rightarrow \left\{ \begin{aligned}
& \dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{3}{4} \\
& \dfrac{\left| {{z}_{1}}-{{z}_{2}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\sqrt{41}}{4} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{3}{4} \\
& \left| \dfrac{{{z}_{1}}}{{{z}_{2}}}-1 \right|=\dfrac{\sqrt{41}}{4} \\
\end{aligned} \right.$
$z=\dfrac{{{z}_{1}}}{{{z}_{2}}}=a+bi,\left( a,b\in \mathbb{R} \right)\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}} \\
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}={{\left( \dfrac{\sqrt{41}}{4} \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=\dfrac{9}{16} \\
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}=\dfrac{41}{16} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{b}^{2}}=\dfrac{9}{16}-{{a}^{2}} \\
& {{\left( a-1 \right)}^{2}}+\dfrac{9}{16}-{{a}^{2}}=\dfrac{41}{16} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{b}^{2}}=\dfrac{5}{16} \\
& a=-\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| b \right|=\dfrac{\sqrt{5}}{4} \\
& a=-\dfrac{1}{2} \\
\end{aligned} \right.$.
Vậy $\left| b \right|=\dfrac{\sqrt{5}}{4}$.
+ $\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|},{{z}_{2}}\ne 0$
Cách 1: Gọi A, B lần lượt là các điểm biểu diễn của số phức ${{z}_{1}},{{z}_{2}}$
Theo đề bài, ta có: $OA=3,OB=4,AB=\sqrt{41}\Rightarrow \cos \widehat{AOB}=\dfrac{{{3}^{2}}+{{4}^{2}}-41}{2.3.4}=-\dfrac{2}{3}$
Đặt ${{z}_{1}}=3\left( \cos \varphi +i\sin \varphi \right)\Rightarrow {{z}_{2}}=4\left( \cos (\varphi \pm AOB) \right)=4\left( \cos (\varphi \pm \alpha )+i\sin (\varphi \pm \alpha ) \right)$ $\left( \alpha =AOB \right)$
$\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{3\left( \cos \varphi +i\sin \varphi \right)}{4\left( \cos (\varphi \pm \alpha )+i\sin \left( \varphi \pm \alpha \right) \right)}=\dfrac{3}{4}\left( \cos \varphi +i\sin \varphi \right)\left( \cos (\varphi \pm \alpha )-i\sin \left( \varphi \pm \alpha \right) \right)$
$=\dfrac{3}{4}\left[ \left( \cos \varphi .\cos \left( \varphi \pm \alpha \right)+\sin \varphi .\sin \left( \varphi \pm \alpha \right) \right)+i\left( \sin \varphi .\cos (\varphi \pm \alpha ) \right)-\cos \varphi .\sin \left( \varphi \pm \alpha \right) \right]$
$=\dfrac{3}{4}\left[ \cos \left( \pm \alpha \right)+i\sin \left( \pm \alpha \right) \right]=\dfrac{3}{4}\left( \cos \alpha \pm i\sin \alpha \right)$
$\Rightarrow b=\pm \dfrac{3}{4}\sin \alpha \Rightarrow \left| b \right|=\dfrac{3}{4}\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}}=\dfrac{\sqrt{5}}{4}$.
Cách 2: Ta có:
$\left| {{z}_{1}} \right|=3,\left| {{z}_{2}} \right|=4,\left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{41}\Rightarrow \left\{ \begin{aligned}
& \dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{3}{4} \\
& \dfrac{\left| {{z}_{1}}-{{z}_{2}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\sqrt{41}}{4} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{3}{4} \\
& \left| \dfrac{{{z}_{1}}}{{{z}_{2}}}-1 \right|=\dfrac{\sqrt{41}}{4} \\
\end{aligned} \right.$
$z=\dfrac{{{z}_{1}}}{{{z}_{2}}}=a+bi,\left( a,b\in \mathbb{R} \right)\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}} \\
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}={{\left( \dfrac{\sqrt{41}}{4} \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=\dfrac{9}{16} \\
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}=\dfrac{41}{16} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{b}^{2}}=\dfrac{9}{16}-{{a}^{2}} \\
& {{\left( a-1 \right)}^{2}}+\dfrac{9}{16}-{{a}^{2}}=\dfrac{41}{16} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{b}^{2}}=\dfrac{5}{16} \\
& a=-\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left| b \right|=\dfrac{\sqrt{5}}{4} \\
& a=-\dfrac{1}{2} \\
\end{aligned} \right.$.
Vậy $\left| b \right|=\dfrac{\sqrt{5}}{4}$.
Đáp án D.