Đặt z = ((1+i)/(căn 2))^(2n) với n thuộc {1,2,3,....,20} thì: z thuộc cộng trừ

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{z}={\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)}^{2\mathrm{n}}={\left[{\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)}^2\right]}^{\mathrm{n}}={\left(\frac{2\mathrm{i}}{2}\right)}^{\mathrm{n}}={\mathrm{i}}^{\mathrm{n}} \\ \mathrm{Với}: \\ \mathrm{n}=1 \mathrm{thì} \mathrm{z}=\mathrm{i} \\ \mathrm{n}=2 \mathrm{thì} \mathrm{z}={\mathrm{i}}^2=-1 \\ \mathrm{n}=3 \mathrm{thì} \mathrm{z}={\mathrm{i}}^3=-\mathrm{i} \\ \mathrm{n}=4 \mathrm{thì} \mathrm{z}={\mathrm{i}}^4=1 \\ \mathrm{NX}: {\mathrm{i}}^{2\mathrm{n}+2}={\mathrm{i}}^2.{\left({\mathrm{i}}^2\right)}^{\mathrm{n}}=(-1).{(-1)}^{\mathrm{n}}={(-1)}^{\mathrm{n}+1} \\ và {\mathrm{i}}^{2\mathrm{n}+1}=\mathrm{i}.{\mathrm{i}}^{2\mathrm{n}}=\mathrm{i}.{(-1)}^{\mathrm{n}} \\ \mathrm{KL}: \mathrm{Với} \mathrm{n}\in \left\{1;2;...;20\right\} \mathrm{thì} \mathrm{z}\in \left\{\pm 1;\pm \mathrm{i}\right\} \end{gathered}$$