Câu hỏi: Cho hai số phức thỏa mãn $\left| 2z-i \right|=\left| 2+iz \right|$ biết $\left| {{z}_{1}}-{{z}_{2}} \right|=1$. Tính giá trị của biểu thức $P=\left| {{z}_{1}}+{{z}_{2}} \right|$.
A. $P=\dfrac{\sqrt{3}}{2}$.
B. $P=\sqrt{2}$.
C. $P=\dfrac{\sqrt{2}}{2}$.
D. $P=\sqrt{3}$.
A. $P=\dfrac{\sqrt{3}}{2}$.
B. $P=\sqrt{2}$.
C. $P=\dfrac{\sqrt{2}}{2}$.
D. $P=\sqrt{3}$.
Giả sử $z=a+bi \left( a,b\in \mathbb{R} \right)$.
Ta có: $\left| 2z-i \right|=\left| 2+iz \right|\Leftrightarrow \left| 2\left( a+bi \right)-i \right|=\left| 2+i\left( a+bi \right) \right|$
$\Leftrightarrow \left| 2a+\left( 2b-1 \right)i \right|=\left| \left( 2-b \right)+ai \right|\Leftrightarrow 4{{a}^{2}}+{{\left( 2b-1 \right)}^{2}}={{\left( 2-b \right)}^{2}}+{{a}^{2}}\Leftrightarrow 3{{a}^{2}}+3{{b}^{2}}=3\Leftrightarrow {{a}^{2}}+{{b}^{2}}=1$
Đặt ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i\left( {{a}_{1}},{{b}_{1}}\in \mathbb{R} \right),{{z}_{2}}={{a}_{2}}+{{b}_{2}}i\left( {{a}_{2}},{{b}_{2}}\in \mathbb{R} \right)$.
Vì ${{z}_{1}},{{z}_{2}}$ là hai số phức thỏa phương trình $\left| 2z-i \right|=\left| 2+iz \right|$ nên $a_{1}^{2}+b_{1}^{2}=1,a_{2}^{2}+b_{2}^{2}=1$.
Ta có: $\left| {{z}_{1}}-{{z}_{2}} \right|=1\Leftrightarrow \left| \left( {{a}_{1}}-{{a}_{2}} \right)+\left( {{b}_{1}}-{{b}_{2}} \right)i \right|=1$ $\Leftrightarrow {{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}=1\Leftrightarrow a_{1}^{2}+b_{1}^{2}+a_{2}^{2}+b_{2}^{2}-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)=1\Leftrightarrow 2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)=1$
Vậy $P=\left| {{z}_{1}}+{{z}_{2}} \right|=\left| \left( {{a}_{1}}+{{a}_{2}} \right)+\left( {{b}_{1}}+{{b}_{2}} \right)i \right|$
$=\sqrt{{{\left( {{a}_{1}}+{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}} \right)}^{2}}}=\sqrt{a_{1}^{2}+b_{1}^{2}+a_{2}^{2}+b_{2}^{2}+2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}=\sqrt{3}$.
Ta có: $\left| 2z-i \right|=\left| 2+iz \right|\Leftrightarrow \left| 2\left( a+bi \right)-i \right|=\left| 2+i\left( a+bi \right) \right|$
$\Leftrightarrow \left| 2a+\left( 2b-1 \right)i \right|=\left| \left( 2-b \right)+ai \right|\Leftrightarrow 4{{a}^{2}}+{{\left( 2b-1 \right)}^{2}}={{\left( 2-b \right)}^{2}}+{{a}^{2}}\Leftrightarrow 3{{a}^{2}}+3{{b}^{2}}=3\Leftrightarrow {{a}^{2}}+{{b}^{2}}=1$
Đặt ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i\left( {{a}_{1}},{{b}_{1}}\in \mathbb{R} \right),{{z}_{2}}={{a}_{2}}+{{b}_{2}}i\left( {{a}_{2}},{{b}_{2}}\in \mathbb{R} \right)$.
Vì ${{z}_{1}},{{z}_{2}}$ là hai số phức thỏa phương trình $\left| 2z-i \right|=\left| 2+iz \right|$ nên $a_{1}^{2}+b_{1}^{2}=1,a_{2}^{2}+b_{2}^{2}=1$.
Ta có: $\left| {{z}_{1}}-{{z}_{2}} \right|=1\Leftrightarrow \left| \left( {{a}_{1}}-{{a}_{2}} \right)+\left( {{b}_{1}}-{{b}_{2}} \right)i \right|=1$ $\Leftrightarrow {{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}=1\Leftrightarrow a_{1}^{2}+b_{1}^{2}+a_{2}^{2}+b_{2}^{2}-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)=1\Leftrightarrow 2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)=1$
Vậy $P=\left| {{z}_{1}}+{{z}_{2}} \right|=\left| \left( {{a}_{1}}+{{a}_{2}} \right)+\left( {{b}_{1}}+{{b}_{2}} \right)i \right|$
$=\sqrt{{{\left( {{a}_{1}}+{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}} \right)}^{2}}}=\sqrt{a_{1}^{2}+b_{1}^{2}+a_{2}^{2}+b_{2}^{2}+2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}=\sqrt{3}$.
Đáp án D.