Cho 3 số phức ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ thỏa mãn${{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0,\left| {{{z}_{1}}} \right|=\left| {{{z}_{2}}} \right|=\left| {{{z}_{3}}} \right|=1.$ Khi đó tổng$z_{1}^{2}+z_{2}^{2}+z_{3}^{2}$ bằng? A. 0. B. 1. C. 2. D. 3.
Đáp án đúng: A Đặt ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i,{{z}_{2}}={{a}_{2}}+{{b}_{2}}i+{{z}_{3}}={{a}_{3}}+{{b}_{3}}i\left( {{{a}_{1}},{{a}_{2}},{{a}_{3}},{{b}_{1}},{{b}_{2}},{{b}_{3}}\in R} \right).$ Ta có$A=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=\left( {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \right)-\left( {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}} \right)+2\left( {{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}} \right)i.$ Từ giả thiết ta có$\left\{ \begin{array}{l}{{a}_{1}}+{{a}_{2}}+{{a}_{3}}={{b}_{1}}+{{b}_{2}}+{{b}_{3}}=0=>\overline{{{{z}_{1}}}}+\overline{{{{z}_{2}}}}+\overline{{{{z}_{3}}}}=0\\a_{1}^{2}+b_{1}^{2}=a_{2}^{2}+b_{2}^{2}=a_{3}^{2}+b_{3}^{2}=1\end{array} \right..$ Khi đó$A=3+2\left[ {{{a}_{1}}{{b}_{1}}i+{{a}_{2}}{{b}_{2}}i+{{a}_{3}}{{b}_{3}}i-b_{1}^{2}-b_{2}^{2}-b_{3}^{2}} \right]=3+2\left[ {{{b}_{1}}i\left( {{{a}_{1}}+{{b}_{1}}i} \right)+{{b}_{2}}i\left( {{{a}_{2}}+{{b}_{2}}i} \right)+{{b}_{3}}i\left( {{{a}_{3}}+{{b}_{3}}i} \right)} \right]=3+2.\left( {-\frac{3}{2}} \right)=0.$