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Câu hỏi: Cho các số phức $z,w$ thỏa mãn $\left| z-i \right|=2$ và $\left| w-2 \right|=1$. Khi $P=\left| z+\overline{w}+1+3i \right|$ đạt giá trị lớn nhất, $\left| z-w+1-\dfrac{12}{5}i \right|$ bằng
A. $\dfrac{11}{5}$.
B. $\dfrac{5}{11}$.
C. $\dfrac{\sqrt{29}}{5}$.
D. $\dfrac{\sqrt{13}}{5}$.
A. $\dfrac{11}{5}$.
B. $\dfrac{5}{11}$.
C. $\dfrac{\sqrt{29}}{5}$.
D. $\dfrac{\sqrt{13}}{5}$.
Ta có: $P=\left| z-i+\overline{w}-2+3+4i \right|\le \left| z-i \right|+\left| \overline{w}-2 \right|+\left| 3+4i \right|=\left| z-i \right|+\left| w-2 \right|+\left| 3+4i \right|=8$
Dấu “=” xảy ra khi:
$\left\{ \begin{aligned}
& z-i=t\left( 3+4i \right) \\
& \overline{w}-2={t}'\left( 3+4i \right) \\
& \left| z-i \right|=2; \left| \overline{w}-2 \right|=1 \\
\end{aligned} \right., \forall t,{t}'\ge 0 $ $ \Leftrightarrow \left\{ \begin{aligned}
& z-i=\dfrac{2}{5}\left( 3+4i \right) \\
& \overline{w}-2=\dfrac{1}{5}\left( 3+4i \right) \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& z=\dfrac{6}{5}+\dfrac{13}{5}i \\
& \overline{w}=\dfrac{13}{5}+\dfrac{4}{5}i \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& z=\dfrac{6}{5}+\dfrac{13}{5}i \\
& w=\dfrac{13}{5}-\dfrac{4}{5}i \\
\end{aligned} \right.$
Khi đó: $\left| z-w+1-\dfrac{12}{5}i \right|=\left| -\dfrac{2}{5}+i \right|=\dfrac{\sqrt{29}}{5}$.
Dấu “=” xảy ra khi:
$\left\{ \begin{aligned}
& z-i=t\left( 3+4i \right) \\
& \overline{w}-2={t}'\left( 3+4i \right) \\
& \left| z-i \right|=2; \left| \overline{w}-2 \right|=1 \\
\end{aligned} \right., \forall t,{t}'\ge 0 $ $ \Leftrightarrow \left\{ \begin{aligned}
& z-i=\dfrac{2}{5}\left( 3+4i \right) \\
& \overline{w}-2=\dfrac{1}{5}\left( 3+4i \right) \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& z=\dfrac{6}{5}+\dfrac{13}{5}i \\
& \overline{w}=\dfrac{13}{5}+\dfrac{4}{5}i \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& z=\dfrac{6}{5}+\dfrac{13}{5}i \\
& w=\dfrac{13}{5}-\dfrac{4}{5}i \\
\end{aligned} \right.$
Khi đó: $\left| z-w+1-\dfrac{12}{5}i \right|=\left| -\dfrac{2}{5}+i \right|=\dfrac{\sqrt{29}}{5}$.
Đáp án C.