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Câu hỏi: Cho hai số phức $z_1, z_2$ thỏa mãn $\left|z_1\right|=1,\left|z_2\right|=4,\left|2 z_1+z_2\right|=\sqrt{15}$. Giá trị lớn nhất của biểu thức $\left|4 z_1+z_2+3 i\right|$ bằng
A. $2 \sqrt{22}+3$.
B. $4 \sqrt{22}+3$.
C. $\sqrt{22}-3$.
D. $\sqrt{22}+3$.
A. $2 \sqrt{22}+3$.
B. $4 \sqrt{22}+3$.
C. $\sqrt{22}-3$.
D. $\sqrt{22}+3$.
$
(*) \Leftrightarrow\left\{\begin{array} { l }
{ a ^ { 2 } + b ^ { 2 } = 1 6 } \\
{ ( a + 2 ) ^ { 2 } + b ^ { 2 } = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a ^ { 2 } + b ^ { 2 } = 1 6 } \\
{ a = - \dfrac { 5 } { 4 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a = - \dfrac { 5 } { 4 } } \\
{ b = \pm \dfrac { \sqrt { 2 3 1 } } { 4 } }
\end{array} \Rightarrow \left[\begin{array}{l}
\dfrac{z_2}{z_1}=-\dfrac{5}{4}+\dfrac{\sqrt{231}}{4} i \\
\dfrac{z_2}{z_1}=-\dfrac{5}{4}-\dfrac{\sqrt{231}}{4} i
\end{array}\right.\right.\right.\right.
$
Ta có: $\left|4 z_1+z_2+3 i\right|=\left|4+\dfrac{z_2}{z_1}+\dfrac{3 i}{z_1}\right| \leq\left|4+\dfrac{z_2}{z_1}\right|+\left|\dfrac{3 i}{z_1}\right|=\sqrt{22}+3$.
(*) \Leftrightarrow\left\{\begin{array} { l }
{ a ^ { 2 } + b ^ { 2 } = 1 6 } \\
{ ( a + 2 ) ^ { 2 } + b ^ { 2 } = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a ^ { 2 } + b ^ { 2 } = 1 6 } \\
{ a = - \dfrac { 5 } { 4 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a = - \dfrac { 5 } { 4 } } \\
{ b = \pm \dfrac { \sqrt { 2 3 1 } } { 4 } }
\end{array} \Rightarrow \left[\begin{array}{l}
\dfrac{z_2}{z_1}=-\dfrac{5}{4}+\dfrac{\sqrt{231}}{4} i \\
\dfrac{z_2}{z_1}=-\dfrac{5}{4}-\dfrac{\sqrt{231}}{4} i
\end{array}\right.\right.\right.\right.
$
Ta có: $\left|4 z_1+z_2+3 i\right|=\left|4+\dfrac{z_2}{z_1}+\dfrac{3 i}{z_1}\right| \leq\left|4+\dfrac{z_2}{z_1}\right|+\left|\dfrac{3 i}{z_1}\right|=\sqrt{22}+3$.
Đáp án D.