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Câu hỏi: Cho các số phức $z$ và $w$ thỏa mãn $\left( 3-i \right)\left| z \right|=\dfrac{z}{w-1}+1-i$. Tìm giá trị lớn nhất $T=\left| w+i \right|$.
A. $\dfrac{3\sqrt{2}}{2}$.
B. 2.
C. $\dfrac{\sqrt{2}}{2}$.
D. $\dfrac{1}{2}$.
A. $\dfrac{3\sqrt{2}}{2}$.
B. 2.
C. $\dfrac{\sqrt{2}}{2}$.
D. $\dfrac{1}{2}$.
Điều kiện: $w\ne 1$.
Ta có:
$\left( 3-i \right)\left| z \right|=\dfrac{z}{w-1}+1-i\Leftrightarrow 3\left| z \right|-1+\left( 1-\left| z \right| \right)i=\dfrac{z}{w-1}$ $\Rightarrow \sqrt{{{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( \left( 1-\left| z \right| \right) \right)}^{2}}}=\dfrac{\left| z \right|}{\left| w-1 \right|}$ $\Rightarrow \left| w-1 \right|=\dfrac{\left| z \right|}{\sqrt{{{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( \left( 1-\left| z \right| \right) \right)}^{2}}}}=\dfrac{\left| z \right|}{\sqrt{10{{\left| z \right|}^{2}}-8\left| z \right|+2}}$
Đặt $t=\left| z \right|, t>0$ vì $\left| w-1 \right|\ne 0$.
Xét $f\left( t \right)=\dfrac{t}{\sqrt{10{{t}^{2}}-8t+2}}=\dfrac{1}{\sqrt{10-8\dfrac{1}{t}+2\dfrac{1}{{{t}^{2}}}}}=\dfrac{1}{\sqrt{2{{\left( \dfrac{1}{t}-2 \right)}^{2}}+2}}\le \dfrac{\sqrt{2}}{2}$
Dấu $''=''$ xảy ra khi $t=\dfrac{1}{2}\Leftrightarrow \left| z \right|=\dfrac{1}{2}$.
Suy ra $\left| w-1 \right|\le \dfrac{\sqrt{2}}{2}$.
$T=\left| w+i \right|=\left| w-1+1+i \right|\le \left| w-1 \right|+\left| 1+i \right|\le \dfrac{\sqrt{2}}{2}+\sqrt{2}=\dfrac{3\sqrt{2}}{2}$.
Dấu $''=''$ xảy ra khi $\left\{ \begin{aligned}
& w=\dfrac{3}{2}+\dfrac{1}{2}i \\
& z=\dfrac{1}{2}i \\
\end{aligned} \right. $. Vậy giá trị lớn nhất của $ \left| w+i \right| $ bằng $ \dfrac{3\sqrt{2}}{2}$.
Ta có:
$\left( 3-i \right)\left| z \right|=\dfrac{z}{w-1}+1-i\Leftrightarrow 3\left| z \right|-1+\left( 1-\left| z \right| \right)i=\dfrac{z}{w-1}$ $\Rightarrow \sqrt{{{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( \left( 1-\left| z \right| \right) \right)}^{2}}}=\dfrac{\left| z \right|}{\left| w-1 \right|}$ $\Rightarrow \left| w-1 \right|=\dfrac{\left| z \right|}{\sqrt{{{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( \left( 1-\left| z \right| \right) \right)}^{2}}}}=\dfrac{\left| z \right|}{\sqrt{10{{\left| z \right|}^{2}}-8\left| z \right|+2}}$
Đặt $t=\left| z \right|, t>0$ vì $\left| w-1 \right|\ne 0$.
Xét $f\left( t \right)=\dfrac{t}{\sqrt{10{{t}^{2}}-8t+2}}=\dfrac{1}{\sqrt{10-8\dfrac{1}{t}+2\dfrac{1}{{{t}^{2}}}}}=\dfrac{1}{\sqrt{2{{\left( \dfrac{1}{t}-2 \right)}^{2}}+2}}\le \dfrac{\sqrt{2}}{2}$
Dấu $''=''$ xảy ra khi $t=\dfrac{1}{2}\Leftrightarrow \left| z \right|=\dfrac{1}{2}$.
Suy ra $\left| w-1 \right|\le \dfrac{\sqrt{2}}{2}$.
$T=\left| w+i \right|=\left| w-1+1+i \right|\le \left| w-1 \right|+\left| 1+i \right|\le \dfrac{\sqrt{2}}{2}+\sqrt{2}=\dfrac{3\sqrt{2}}{2}$.
Dấu $''=''$ xảy ra khi $\left\{ \begin{aligned}
& w=\dfrac{3}{2}+\dfrac{1}{2}i \\
& z=\dfrac{1}{2}i \\
\end{aligned} \right. $. Vậy giá trị lớn nhất của $ \left| w+i \right| $ bằng $ \dfrac{3\sqrt{2}}{2}$.
Đáp án A.