Google
Câu hỏi: Gọi ${{{z}_{1}}}$, ${{{z}_{2}}}$ là hai nghiệm phức của phương trình ${{{z}^{2}}-z+1=0}$. Tính giá trị của ${P=\left| z_{1}^{2017}-z_{2}^{2017} \right|}$.
A. ${P=\sqrt{3}}$.
B. ${P=2\sqrt{3}}$.
C. ${P=3}$.
D. ${P=0}$.
${{{z}^{2}}-z+1=10}$ ${\Rightarrow \left[ \begin{aligned}
& {{z}_{1}}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\
& {{z}_{2}}=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\
\end{aligned} \right.}$.
Ta có: ${{{\left( 1+\sqrt{3}i \right)}^{2017}}}$ ${={{\left[ {{\left( 1+\sqrt{3}i \right)}^{3}} \right]}^{672}}\left( 1+\sqrt{3}i \right)}$ ${={{\left( -8 \right)}^{672}}\left( 1+\sqrt{3}i \right)}$.
${{{\left( 1-\sqrt{3}i \right)}^{2017}}={{\left[ {{\left( 1-\sqrt{3}i \right)}^{3}} \right]}^{672}}\left( 1-\sqrt{3}i \right)}$ ${={{\left( -8 \right)}^{672}}\left( 1-\sqrt{3}i \right)}$.
Suy ra: ${P=\left| z_{1}^{2017}-z_{2}^{2017} \right|=\dfrac{1}{{{2}^{2017}}}.\left| {{\left( -8 \right)}^{672}}\left( 2\sqrt{3}i \right) \right|=\sqrt{3}}$.
A. ${P=\sqrt{3}}$.
B. ${P=2\sqrt{3}}$.
C. ${P=3}$.
D. ${P=0}$.
${{{z}^{2}}-z+1=10}$ ${\Rightarrow \left[ \begin{aligned}
& {{z}_{1}}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\
& {{z}_{2}}=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\
\end{aligned} \right.}$.
Ta có: ${{{\left( 1+\sqrt{3}i \right)}^{2017}}}$ ${={{\left[ {{\left( 1+\sqrt{3}i \right)}^{3}} \right]}^{672}}\left( 1+\sqrt{3}i \right)}$ ${={{\left( -8 \right)}^{672}}\left( 1+\sqrt{3}i \right)}$.
${{{\left( 1-\sqrt{3}i \right)}^{2017}}={{\left[ {{\left( 1-\sqrt{3}i \right)}^{3}} \right]}^{672}}\left( 1-\sqrt{3}i \right)}$ ${={{\left( -8 \right)}^{672}}\left( 1-\sqrt{3}i \right)}$.
Suy ra: ${P=\left| z_{1}^{2017}-z_{2}^{2017} \right|=\dfrac{1}{{{2}^{2017}}}.\left| {{\left( -8 \right)}^{672}}\left( 2\sqrt{3}i \right) \right|=\sqrt{3}}$.
Đáp án A.