Mathematics - Complex Number Question with Solution | TestHub

(1) $\begin{array}{rl}& (-1+\sqrt{2}{)}^{\mathrm{n}}=\mathrm{m}+\sqrt{2}\mathrm{n},\mathrm{m},\text{ where, }\mathrm{n}\in \mathbb{Z}\\ & (1+\sqrt{2}{)}^{\mathrm{n}}={\mathrm{m}}_1+\sqrt{2}{\mathrm{n}}_1,\text{ where, }{\mathrm{m}}_1,{\mathrm{n}}_1\in \mathbb{Z}\\ & \Rightarrow \mathbb{Z}\cup {\mathrm{T}}_1\cup {\mathrm{T}}_2\subseteq \mathrm{S}\end{array}$ but $b\sqrt{2}\in S$ for negative $b\in \mathbb{Z}$. So $\mathbb{Z}\cup {\mathrm{T}}_1\cup {\mathrm{T}}_2\subset \mathrm{S}$ (2) $\begin{array}{rl}& (\sqrt{2}-1{)}^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1{)}^{\mathrm{n}}}<\frac{1}{2024}\\ & \Rightarrow 2024<(\sqrt{2}+1{)}^{\mathrm{n}},\text{ where, }\exists \mathrm{n}\in \mathbb{N}\\ & \Rightarrow {\mathrm{T}}_1\cap \left(0,\frac{1}{2024}\right)\ne \varphi \end{array}$ (3) $\begin{array}{rl}& (1+\sqrt{2}{)}^{\mathrm{n}}>2024,\text{ where, }\exists \mathrm{n}\in \mathbb{N}\\ & \Rightarrow {\mathrm{T}}_2\cap (2024,\infty )\ne \varphi \end{array}$ (4) $\sin (\pi (a+b\sqrt{2})=0)\Rightarrow b=0,\text{ where, }a\in \mathbb{Z}$ $\Rightarrow$ Options (1), (3), (4) are Correct.