Google
Câu hỏi: Cho hai số phức $z$ và $\text{w}$ thỏa mãn $\left| z \right|=4,\left| \text{w} \right|=2$. Khi $\left| z+\overline{\text{w}}+5+12i \right|$ đạt giá trị lớn nhất, phần thực của $z+iw$ bằng
A. $\dfrac{30}{13}$.
B. $-\dfrac{4}{13}$.
C. $\dfrac{44}{13}$.
D. $\dfrac{58}{13}$.
A. $\dfrac{30}{13}$.
B. $-\dfrac{4}{13}$.
C. $\dfrac{44}{13}$.
D. $\dfrac{58}{13}$.
Ta có $\left| \text{w} \right|=2\Rightarrow \left| \overline{\text{w}} \right|=2$.
Ta lại có $\left| z+\overline{\text{w}}+5+12i \right|\le \left| z+\overline{\text{w}} \right|+\left| 5+12i \right|\le \left| z \right|+\left| \overline{\text{w}} \right|+13$.
Suy ra $\left| z+\overline{\text{w}}+5+12i \right|\le 19$. Dấu $''=''$ xảy ra khi $\left\{ \begin{aligned}
& z=k\overline{\text{w}} \\
& z+\overline{\text{w}}=h(5+12i) \\
\end{aligned} \right.\left( k,h\in \mathbb{R};k,h>0 \right)$
$\Rightarrow \left\{ \begin{aligned}
& k=2 \\
& h=\dfrac{6}{13} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \overline{\text{w}}=\dfrac{10}{13}+\dfrac{24}{13}i \\
& z=\dfrac{20}{13}+\dfrac{48}{13}i \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{w}=\dfrac{10}{13}-\dfrac{24}{13}i \\
& z=\dfrac{20}{13}+\dfrac{48}{13}i \\
\end{aligned} \right.\Rightarrow z+iw=\dfrac{44}{13}+\dfrac{58}{13}i$.
Vậy phần thực của $z+iw$ bằng $\dfrac{44}{13}$.
Ta lại có $\left| z+\overline{\text{w}}+5+12i \right|\le \left| z+\overline{\text{w}} \right|+\left| 5+12i \right|\le \left| z \right|+\left| \overline{\text{w}} \right|+13$.
Suy ra $\left| z+\overline{\text{w}}+5+12i \right|\le 19$. Dấu $''=''$ xảy ra khi $\left\{ \begin{aligned}
& z=k\overline{\text{w}} \\
& z+\overline{\text{w}}=h(5+12i) \\
\end{aligned} \right.\left( k,h\in \mathbb{R};k,h>0 \right)$
$\Rightarrow \left\{ \begin{aligned}
& k=2 \\
& h=\dfrac{6}{13} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \overline{\text{w}}=\dfrac{10}{13}+\dfrac{24}{13}i \\
& z=\dfrac{20}{13}+\dfrac{48}{13}i \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{w}=\dfrac{10}{13}-\dfrac{24}{13}i \\
& z=\dfrac{20}{13}+\dfrac{48}{13}i \\
\end{aligned} \right.\Rightarrow z+iw=\dfrac{44}{13}+\dfrac{58}{13}i$.
Vậy phần thực của $z+iw$ bằng $\dfrac{44}{13}$.
Đáp án C.