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Câu hỏi: Trong các số phức thỏa mãn điều kiện $\left| z+3i \right|=\left| z+2-i \right|.$ Tìm số phức có môđun nhỏ nhất?
A. $z=1-2i$.
B. $z=-\dfrac{1}{5}+\dfrac{2}{5}i$.
C. $z=\dfrac{1}{5}-\dfrac{2}{5}i$.
D. $z=-1+2i$.
A. $z=1-2i$.
B. $z=-\dfrac{1}{5}+\dfrac{2}{5}i$.
C. $z=\dfrac{1}{5}-\dfrac{2}{5}i$.
D. $z=-1+2i$.
Phương pháp tự luận
Giả sử $z=x+yi \left( x,y\in \mathbb{R} \right)$
$\left| z+3i \right|=\left| z+2-i \right|\Leftrightarrow \left| x+\left( y+3 \right)i \right|=\left| \left( x+2 \right)+\left( y-1 \right)i \right|\Leftrightarrow {{x}^{2}}+{{\left( y+3 \right)}^{2}}={{\left( x+2 \right)}^{2}}+{{\left( y-1 \right)}^{2}}$
$\Leftrightarrow 6y+9=4x+4-2y+1\Leftrightarrow 4x-8y-4=0\Leftrightarrow x-2y-1=0\Leftrightarrow x=2y+1$
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{\left( 2y+1 \right)}^{2}}+{{y}^{2}}}=\sqrt{5{{y}^{2}}+4y+1}=\sqrt{5{{\left( y+\dfrac{2}{5} \right)}^{2}}+\dfrac{1}{5}}\ge \dfrac{\sqrt{5}}{5}$
Suy ra ${{\left| z \right|}_{\min }}=\dfrac{\sqrt{5}}{5}$ khi $y=-\dfrac{2}{5}\Rightarrow x=\dfrac{1}{5}$
Vậy $z=\dfrac{1}{5}-\dfrac{2}{5}i.$
Giả sử $z=x+yi \left( x,y\in \mathbb{R} \right)$
$\left| z+3i \right|=\left| z+2-i \right|\Leftrightarrow \left| x+\left( y+3 \right)i \right|=\left| \left( x+2 \right)+\left( y-1 \right)i \right|\Leftrightarrow {{x}^{2}}+{{\left( y+3 \right)}^{2}}={{\left( x+2 \right)}^{2}}+{{\left( y-1 \right)}^{2}}$
$\Leftrightarrow 6y+9=4x+4-2y+1\Leftrightarrow 4x-8y-4=0\Leftrightarrow x-2y-1=0\Leftrightarrow x=2y+1$
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{\left( 2y+1 \right)}^{2}}+{{y}^{2}}}=\sqrt{5{{y}^{2}}+4y+1}=\sqrt{5{{\left( y+\dfrac{2}{5} \right)}^{2}}+\dfrac{1}{5}}\ge \dfrac{\sqrt{5}}{5}$
Suy ra ${{\left| z \right|}_{\min }}=\dfrac{\sqrt{5}}{5}$ khi $y=-\dfrac{2}{5}\Rightarrow x=\dfrac{1}{5}$
Vậy $z=\dfrac{1}{5}-\dfrac{2}{5}i.$
Đáp án C.