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Câu hỏi: Cho hai số phức ${{z}_{1}}=2+i,{{z}_{2}}=1-2i$. Môđun của số phức $w=\dfrac{z_{1}^{2019}}{z_{2}^{2020}}$ là
A. $\left| w \right|=5$.
B. $\left| w \right|=\dfrac{\sqrt{5}}{5}$.
C. $\left| w \right|=\dfrac{1}{5}$.
D. $\left| w \right|=\sqrt{5}$.
A. $\left| w \right|=5$.
B. $\left| w \right|=\dfrac{\sqrt{5}}{5}$.
C. $\left| w \right|=\dfrac{1}{5}$.
D. $\left| w \right|=\sqrt{5}$.
Ta có $\left| w \right|=\left| \dfrac{z_{1}^{2019}}{z_{2}^{2020}} \right|=\dfrac{\left| z_{1}^{2019} \right|}{\left| z_{2}^{2020} \right|}=\dfrac{{{\left| {{z}_{1}} \right|}^{2019}}}{{{\left| {{z}_{2}} \right|}^{2020}}}$.
Mà $\left| {{z}_{1}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}}=\sqrt{5};\left| {{z}_{2}} \right|=\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{5}$. Vậy $\left| w \right|=\dfrac{{{\left( \sqrt{5} \right)}^{2019}}}{{{\left( \sqrt{5} \right)}^{2020}}}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}$.
Mà $\left| {{z}_{1}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}}=\sqrt{5};\left| {{z}_{2}} \right|=\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{5}$. Vậy $\left| w \right|=\dfrac{{{\left( \sqrt{5} \right)}^{2019}}}{{{\left( \sqrt{5} \right)}^{2020}}}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}$.
Đáp án B.