ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} \left|z+1\right| = \left|z+i\right| \\ \mathrm{กำหนดให้} z = x+yi \\ \mathrm{จะได้} \left|x+yi+1\right| = \left|x+yi+i\right| \\ \left|\left(x+1\right)+yi\right| = \left|x+\left(y+i\right)i\right| \\ \sqrt{{\left(x+1\right)}^2+y^2} = \sqrt{x^2+{\left(y+1\right)}^2} \\ \cancel{ x^2}+2x\bcancel{+1}+\cancel{y^2} = \cancel{x^2}+\cancel{y^2}+2y\bcancel{+1} \\ 2x = 2y \\ x = y \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \left|z-2+i\right| = \left|z+2-2i\right| \\ \mathrm{จะได้} \left|x+yi-2+i\right| = \left|x+yi+2-2i\right| \\ \left|\left(x-2\right)+\left(y+1\right)i\right| = \left|\left(x+2\right)+\left(y-2\right)i\right| \\ -\mathrm{แทน} y=x \\ \mathrm{จะได้} \left|\left(x-2\right)+\left(x+1\right)i\right| = \left|\left(x+2\right)+\left(x-2\right)i\right| \\ \sqrt{\cancel{{\left(x-2\right)}^2}+{\left(x+1\right)}^2} = \sqrt{{\left(x+2\right)}^2+\cancel{{\left(x-2\right)}^2}} \\ {\left(x+1\right)}^2 = {\left(x+2\right)}^2 \end{gathered}$$

$$\begin{gathered} \cancel{x^2}+2x+1 = \cancel{x^2}+4x+4 \\ -2x = 3 \\ x = \frac{-3}{2} ; \mathrm{จาก} y=x \\ \mathrm{จะได้} y = \frac{-3}{2} \\ \mathrm{แสดงว่า} z = \frac{-3}{2}-\frac{3}{2}i \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} {\left|2z\right|}^2 = {\left|2\left(\frac{-3}{2}-\frac{3}{2}\right)\right|}^2 \\ = {\left|-3-3i\right|}^2 \\ = {\left(\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}\right)}^2 \\ = 9+9 = 18 \end{gathered}$$