Cho hai số phức $z,w$ thỏa mãn $\left| z+2w \right|=3$, $\left|...

Đặt ${{\left| z \right|}^{2}}=a;{{\left| w \right|}^{2}}=b$ với $a;b\ge 0$. Ta có:
$\left| z+2w \right|=3\Rightarrow \left( z+2w \right)\left( \overline{z}+2\overline{w} \right)=9\Rightarrow z.\overline{z}+2\left( z.\overline{w}+\overline{z}.w \right)+4.w.\overline{w}=9\Rightarrow a+2P+4b=9.$
$\begin{aligned}
& \left| 2z+3w \right|=5\Rightarrow \left( 2z+3w \right)\left( 2\overline{z}+3\overline{w} \right)=25\Rightarrow 4.z.\overline{z}+6\left( z.\overline{w}+\overline{z}.w \right)+9.w.\overline{w}=25 \\
& \Rightarrow 4a+6P+9b=25. \\
\end{aligned}$
$\left| z+3w \right|=4\Rightarrow \left( z+3w \right)\left( \overline{z}+3\overline{w} \right)=16\Rightarrow z.\overline{z}+3\left( z.\overline{w}+\overline{z}.w \right)+9.w.\overline{w}=16\Rightarrow a+3P+9b=16.$.
Vậy ta có hệ sau: $\left\{ \begin{aligned}
& a+2P+4b=9 \\
& 4a+6P+9b=25 \\
& a+3P+9b=16 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& P=2 \\
& b=1 \\
\end{aligned} \right.. $ Do đó $ P=2$.