ข้อสอบคณิต 9 สามัญ - ปี 2562

$$\begin{gathered} \mathrm{สูตร} \left(a-bi\right)\left(a+bi\right) = a^2+b^2 \\ {\left(\frac{1+i}{2}-\frac{1}{1+i}\right)}^3 = {\left(\frac{1+i}{2}-\frac{1}{1+i}\frac{\left(1-i\right)}{\left(1-i\right)}\right)}^3 \\ = {\left(\frac{1+i}{2}-\frac{1-i}{1^2+1^2}\right)}^3 \\ = {(\frac{1+i}{2}-\frac{1-i}{2})}^3 \end{gathered}$$

$$\begin{gathered} = {\left(\frac{1+i-\left(1-i\right)}{2}\right)}^3 \\ = {\left(\frac{\cancel{1}+i\cancel{-1}+i}{2}\right)}^3 \\ = {\left(\frac{\cancel{2}i}{\cancel{2}}\right)}^3 \\ = i^3 = -\mathrm{i} \end{gathered}$$