ข้อสอบคณิต 9 สามัญ - ปี 2563

$$\begin{gathered} \mathrm{จากคุณสมบัติของจำนวนเชิงซ้อน} \\ z·\overline{z} = {\left|z\right|}^2 \\ \mathrm{จะได้} {\left(\overline{z}\left|z\right|\right)}^2+2{\left(\overline{z}\right)}^3+z+2 = 0 \\ {\left(\overline{z}\right)}^2{\left|z\right|}^2+2{\left(\overline{z}\right)}^3+z+2 = 0 \\ {\left(\overline{z}\right)}^2\left(z·\overline{z}\right)+2{\left(\overline{z}\right)}^3+z+2 = 0 \end{gathered}$$

$$\begin{gathered} {\left(\overline{z}\right)}^{3}z+2{\left(\overline{z}\right)}^3+z+2 = 0 \\ {\left(\overline{z}\right)}^3\left(z+2 \right)+\left(z+2 \right) = 0 \\ \left(z+2 \right)\left({\left(\overline{z}\right)}^3+1\right) = 0 \\ \left(z+2 \right)\left(\overline{z}+1\right)\left({\left(\overline{z}\right)}^2-\overline{z}+1\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} z = -2 \mathrm{หรือ} \overline{z} = -1 \\ z = -1 \\ \overline{z} = \frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

$$\begin{gathered} = \frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)} \\ = \frac{1\pm \sqrt{3}i}{2} = \frac{1}{2}\pm \frac{\sqrt{3}}{2}i \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 3 \end{gathered}$$