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Câu hỏi: Số phức $z$ thỏa mãn $\left| z-2+2i \right|=\sqrt{2}\left| z-1+i \right|$. Môđun $z$ bằng:
A. $4$.
B. $\sqrt{2}$.
C. $2$.
D. $2\sqrt{2}$.
A. $4$.
B. $\sqrt{2}$.
C. $2$.
D. $2\sqrt{2}$.
Thay $z=a+bi$
$\left| z-2+2i \right|=\sqrt{2}\left| z-1+i \right|\Leftrightarrow {{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}}=2{{\left( a-1 \right)}^{2}}+2{{\left( b+1 \right)}^{2}}$
$\Leftrightarrow {{a}^{2}}-4a+4+{{b}^{2}}+4b+4=2\left( {{a}^{2}}-2a+1 \right)+2\left( {{b}^{2}}+2b+1 \right)$
$\Leftrightarrow {{a}^{2}}-2+{{b}^{2}}-2=0\Leftrightarrow {{a}^{2}}+{{b}^{2}}=4\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}=2\Leftrightarrow \left| z \right|=2$
$\left| z-2+2i \right|=\sqrt{2}\left| z-1+i \right|\Leftrightarrow {{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}}=2{{\left( a-1 \right)}^{2}}+2{{\left( b+1 \right)}^{2}}$
$\Leftrightarrow {{a}^{2}}-4a+4+{{b}^{2}}+4b+4=2\left( {{a}^{2}}-2a+1 \right)+2\left( {{b}^{2}}+2b+1 \right)$
$\Leftrightarrow {{a}^{2}}-2+{{b}^{2}}-2=0\Leftrightarrow {{a}^{2}}+{{b}^{2}}=4\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}=2\Leftrightarrow \left| z \right|=2$
Đáp án C.