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Câu hỏi: Cho hai số phức ${{z}_{1}}=1+3i$ và ${{z}_{2}}=3-4i$. Môđun của số phức $\text{w}=\dfrac{{{z}_{1}}}{{{z}_{2}}}$ là
A. $\left| \text{w} \right|=\dfrac{\sqrt{10}}{2}$
B. $\left| \text{w} \right|=\dfrac{-9}{25}+\dfrac{13}{25}i$
C. $\left| \text{w} \right|=\dfrac{\sqrt{5}}{10}$
D. $\left| \text{w} \right|=\dfrac{\sqrt{10}}{5}$
A. $\left| \text{w} \right|=\dfrac{\sqrt{10}}{2}$
B. $\left| \text{w} \right|=\dfrac{-9}{25}+\dfrac{13}{25}i$
C. $\left| \text{w} \right|=\dfrac{\sqrt{5}}{10}$
D. $\left| \text{w} \right|=\dfrac{\sqrt{10}}{5}$
Cách 1:
Ta có: $\text{w}=\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{1+3i}{3-4i}=\dfrac{\left( 1+3i \right)\left( 3+4i \right)}{25}=\dfrac{-9}{25}+\dfrac{13}{25}i$
Do đó $\left| \text{w} \right|=\left| \dfrac{-9}{25}+\dfrac{13}{25}i \right|=\sqrt{{{\left( \dfrac{-9}{25} \right)}^{2}}+{{\left( \dfrac{13}{25} \right)}^{2}}}=\dfrac{\sqrt{10}}{5}$
Cách 2: Ta có: $\left| \text{w} \right|=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\sqrt{10}}{5}$
Ta có: $\text{w}=\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{1+3i}{3-4i}=\dfrac{\left( 1+3i \right)\left( 3+4i \right)}{25}=\dfrac{-9}{25}+\dfrac{13}{25}i$
Do đó $\left| \text{w} \right|=\left| \dfrac{-9}{25}+\dfrac{13}{25}i \right|=\sqrt{{{\left( \dfrac{-9}{25} \right)}^{2}}+{{\left( \dfrac{13}{25} \right)}^{2}}}=\dfrac{\sqrt{10}}{5}$
Cách 2: Ta có: $\left| \text{w} \right|=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{\sqrt{10}}{5}$
Đáp án D.