For any real number r, let A_r = {e^{i pi r n | vedclass

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(D) The set is defined as $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$.For $A_1$, we have $e^{i \pi n} = (e^{i \pi})^n = (-1)^n$. Since $n \in \mathbb

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$A_{0.3}$ is a finite set and $A_{1/\pi}$ is an infinite set

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(D) The set is defined as $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$.For $A_1$, we have $e^{i \pi n} = (e^{i \pi})^n = (-1)^n$. Since $n \in \mathbb{N}$, the set is $\{-1, 1\}$, which is finite.For $A_{0.3}$, we have $e^{i \pi (0.3) n} = e^{i \pi (3/10) n}$. This set is finite because $e^{i \pi (3/10) n}$ repeats every $20$ values of $n$ (since $e^{i \pi (3/10) n} = e^{i \pi (3/10) (n+20)}$).For $A_{1/\pi}$, we have $e^{i \pi (1/\pi) n} = e^{i n} = \cos(n) + i \sin(n)$. Since $n$ is a natural number and $1$ is not a rational multiple of $\pi$, the values of $e^{in}$ are distinct for all $n \in \mathbb{N}$, making the set infinite.Thus, $A_{0.3}$ is finite and $A_{1/\pi}$ is infinite.Therefore, option $(d)$ is correct.

For any real number $r$, let $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$ be a set of complex numbers. Then,

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