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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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,

Similar Questions

If $z = x + iy$ and $|z - zi| = 1,$ then

Let $a, b, c, d$ be real numbers such that $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$. Then,the sum of all possible values of $|a-d|$ is

If $z$ is a complex number,then the minimum value of $|z| + |z - 1|$ is

The maximum value of the modulus of $e^{z^2}$ on the set $\{z \in \mathbb{C} : 0 \leq \operatorname{Re}(z) \leq 1, 0 \leq \operatorname{Im}(z) \leq 1\}$ is

$A$ rectangle is constructed in the complex plane with its sides parallel to the axes and its centre situated at the origin. If one of the vertices of the rectangle is $a + ib\sqrt{3}$,then the area of the rectangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module