ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \left|{\mathrm{z}}_1+{\mathrm{z}}_2\right|=3 \\ \mathrm{จะได้} {\left|{\mathrm{z}}_1+{\mathrm{z}}_2\right|}^2=9 \\ {\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2+{\mathrm{z}}_1\overline{{\mathrm{z}}_2}+\overline{{\mathrm{z}}_1}{\mathrm{z}}_2=9\to \boxed{1} \\ \mathrm{จากโจทย์} \left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|=1 \\ \mathrm{จะได้} {\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|}^2=1 \\ {\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}-\overline{{\mathrm{z}}_1}{\mathrm{z}}_2=1\to \boxed{2} \\ \mathrm{นำ} \boxed{1}+\boxed{2} 2{\left|{\mathrm{z}}_1\right|}^2+2{\left|{\mathrm{z}}_2\right|}^2=10 \\ {\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2=5 \end{gathered}$$