Trên tập hợp số phức, xét phương trình ${{z}^{2}}+az+b=0$ ( $a...

Ta có .
- TH1: thì .
${{z}_{1}}-3=\left( 1-\left| {{z}_{2}} \right| \right)i\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}-3=0 \\
& 0=1-\left| {{z}_{2}} \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}=3 \\
& \left| {{z}_{2}} \right|=1 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}=3 \\
& {{z}_{2}}=1 \\
\end{aligned} \right. \left\{ \begin{aligned}
& {{z}_{1}}=3 \\
& {{z}_{2}}=-1 \\
\end{aligned} \right.\left\{ \begin{aligned}
& {{z}_{1}}=3 \\
& {{z}_{2}}=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& S=4 \\
& P=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -a=4 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-4 \\
& b=3 \\
\end{aligned} \right.\left\{ \begin{aligned}
& {{z}_{1}}=3 \\
& {{z}_{2}}=-1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& S=2 \\
& P=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -a=2 \\
& b=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=-3 \\
\end{aligned} \right.{\Delta }'<0\Leftrightarrow {{a}^{2}}-4b<0{{z}_{1}}, {{z}_{2}}\notin \mathbb{R}\Rightarrow {{z}_{1}}=\overline{{{z}_{2}}}{{z}_{1}}-3=\left( 1-\left| {{z}_{2}} \right| \right)i\Leftrightarrow {{z}_{1}}=3+\left( 1-\left| {{z}_{1}} \right| \right)i\Rightarrow \left| {{z}_{1}} \right|=\sqrt{9+{{\left( 1-\left| {{z}_{1}} \right| \right)}^{2}}}\Leftrightarrow {{\left| {{z}_{1}} \right|}^{2}}=10-2\left| {{z}_{1}} \right|+{{\left| {{z}_{1}} \right|}^{2}}\Leftrightarrow \left| {{z}_{1}} \right|=5\Rightarrow {{z}_{1}}=3-4i\Rightarrow {{z}_{2}}=3+4i\Rightarrow \left\{ \begin{aligned}
& S=6 \\
& P=25 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -a=6 \\
& b=25 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-6 \\
& b=25 \\
\end{aligned} \right.\left( a; b \right)$ thỏa.