ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} 2\left|\mathrm{z}+1\right|=\left|\mathrm{z}+4\right| \\ \mathrm{กำหนดให้} \mathrm{z}=\mathrm{a}+\mathrm{bi} \\ \mathrm{จะได้} 2\left|\left(\mathrm{a}+\mathrm{bi}\right)+1\right|=\left|\left(\mathrm{a}+\mathrm{bi}\right)+4\right| \\ 2\left|\left(\mathrm{a}+1\right)+\mathrm{bi}\right|=\left|\left(\mathrm{a}+4\right)+\mathrm{bi}\right| \end{gathered}$$
$$\begin{gathered} 2\sqrt{{\left(\mathrm{a}+1\right)}^2+{\mathrm{b}}^2}=\sqrt{{\left(\mathrm{a}+4\right)}^2+{\mathrm{b}}^2} \\ 2\sqrt{\left({\mathrm{a}}^2+2\mathrm{a}+1\right)+{\mathrm{b}}^2}=\sqrt{\left({\mathrm{a}}^2+8\mathrm{a}+16\right)+{\mathrm{b}}^2} \\ 4\left({\mathrm{a}}^2+2\mathrm{a}+1+{\mathrm{b}}^2\right)=\left({\mathrm{a}}^2+8\mathrm{a}+16+{\mathrm{b}}^2\right) \\ 4{\mathrm{a}}^2+\cancel{8\mathrm{a}}+4+4{\mathrm{b}}^2={\mathrm{a}}^2+\cancel{8\mathrm{a}}+16+{\mathrm{b}}^2 \\ 3{\mathrm{a}}^2+3{\mathrm{b}}^2=12 \\ {\mathrm{a}}^2+{\mathrm{b}}^2=4 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \left|\mathrm{z}\right|=\left|\mathrm{a}+\mathrm{bi}\right| \\ =\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{โดย} {\mathrm{a}}^2+{\mathrm{b}}^2=4 \\ =\sqrt{4} \\ =2 \end{gathered}$$