Câu hỏi: Cho hai số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|=1.$ Tính $\left| {{z}_{1}}+{{z}_{2}} \right|$
A. $\sqrt{3}.$
B. $2\sqrt{3}.$
C. $1.$
D. $\dfrac{\sqrt{3}}{2}.$
A. $\sqrt{3}.$
B. $2\sqrt{3}.$
C. $1.$
D. $\dfrac{\sqrt{3}}{2}.$
Ta có: ${{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}-{{z}_{2}}} \right)=\left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}}-\overline{{{z}_{2}}} \right)={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-\left( {{z}_{1}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}} \right)$
$\Rightarrow {{z}_{1}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}}=1$
$\Rightarrow {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}+{{z}_{2}}} \right)=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+\overline{{{z}_{2}}} \right)={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+\left( {{z}_{1}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}} \right)=3.$
Từ đó suy ra $\left| {{z}_{1}}+{{z}_{2}} \right|=\sqrt{3}.$
$\Rightarrow {{z}_{1}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}}=1$
$\Rightarrow {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}+{{z}_{2}}} \right)=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+\overline{{{z}_{2}}} \right)={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+\left( {{z}_{1}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}} \right)=3.$
Từ đó suy ra $\left| {{z}_{1}}+{{z}_{2}} \right|=\sqrt{3}.$
Đáp án A.