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Câu hỏi: Trong các số phức $z$ thỏa mãn $\left| z-i \right|=\left| \bar{z}-2-3i \right|$. Hãy tìm $z$ có môđun nhỏ nhất.
A. $z=\dfrac{27}{5}+\dfrac{6}{5}i$.
B. $z=-\dfrac{6}{5}-\dfrac{27}{5}i$.
C. $z=-\dfrac{6}{5}+\dfrac{27}{5}i$.
D. $z=\dfrac{3}{5}-\dfrac{6}{5}i$.
A. $z=\dfrac{27}{5}+\dfrac{6}{5}i$.
B. $z=-\dfrac{6}{5}-\dfrac{27}{5}i$.
C. $z=-\dfrac{6}{5}+\dfrac{27}{5}i$.
D. $z=\dfrac{3}{5}-\dfrac{6}{5}i$.
Giả sử $z=x+yi$ $\left( x,y\in \mathbb{R} \right)$ $\Rightarrow \bar{z}=x-yi$.
Ta có $\left| x+yi-i \right|=\left| x-yi-2-3i \right|$ $\Leftrightarrow \left| x+\left( y-1 \right)i \right|=\left| \left( x-2 \right)-\left( y+3 \right)i \right|$
$\Leftrightarrow {{x}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}$ $\Leftrightarrow 1-2y=13-4x+6y\Leftrightarrow 4x=12+8y\Leftrightarrow x=2y+3$.
Do đó ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}={{\left( 2y+3 \right)}^{2}}+{{y}^{2}}=5{{y}^{2}}+12y+9={{\left( y\sqrt{5}+\dfrac{6}{\sqrt{5}} \right)}^{2}}+\dfrac{9}{5}\ge \dfrac{9}{5}$.
Dấu $''=''$ xảy ra $\Leftrightarrow y=-\dfrac{6}{5}$, khi đó $x=\dfrac{3}{5}\Rightarrow z=\dfrac{3}{5}-\dfrac{6}{5}i$.
Ta có $\left| x+yi-i \right|=\left| x-yi-2-3i \right|$ $\Leftrightarrow \left| x+\left( y-1 \right)i \right|=\left| \left( x-2 \right)-\left( y+3 \right)i \right|$
$\Leftrightarrow {{x}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}$ $\Leftrightarrow 1-2y=13-4x+6y\Leftrightarrow 4x=12+8y\Leftrightarrow x=2y+3$.
Do đó ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}={{\left( 2y+3 \right)}^{2}}+{{y}^{2}}=5{{y}^{2}}+12y+9={{\left( y\sqrt{5}+\dfrac{6}{\sqrt{5}} \right)}^{2}}+\dfrac{9}{5}\ge \dfrac{9}{5}$.
Dấu $''=''$ xảy ra $\Leftrightarrow y=-\dfrac{6}{5}$, khi đó $x=\dfrac{3}{5}\Rightarrow z=\dfrac{3}{5}-\dfrac{6}{5}i$.
Đáp án D.