Question

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

• A. $A_1,A_{\frac{1}{\pi }},A_{0.3}$ are all infinite sets
• B. $A_1$ is a finite set and $A_{\frac{1}{\pi }},A_{0.3}$ are infinite sets
• C. $A_1,A_{\frac{1}{\pi }},A_{0.3}$ are all finite sets
• D. $A_1,A_{0.3}$ are finite sets and $A_{\frac{1}{\pi }}$ is an infinite set

Answer 1

Answer: (D)

$A_1,A_{0.3}$ are finite sets and $A_{\frac{1}{\pi }}$ is an infinite set

We have to examine the sets $A_r$ for $r=1,0.3,\frac{1}{\pi }$.

$r=1\Rightarrow A_1=e^{i\pi n}=\cos \pi n+i\sin \pi n=(-1{)}^n,n\in N$

So, the set $A_1$ is clearly a finite set as it contains $-1,1$ as its only elements .

$r=0.3\Rightarrow A_{0.3}=e^{i3\pi n/10}=\cos \left(3\pi n/10\right)+i\sin \left(3\pi n/10\right),n\in N$

So, the set $A_{0.3}$ is clearly a set of $10th$ roots of unity and it contains only $10$ possible distinct values for $n=1,2,...,10$. So this set is also a finite set.

$r=\frac{1}{\pi }\Rightarrow A_{\frac{1}{\pi }}=e^{in}=\cos \left(n\right)+i\sin \left(n\right),n\in N$

Clearly, $e^{in}$ can take infinitely many values, so the set $A_{\frac{1}{\pi }}$ is an infinite set.

Therefore option D is the correct answer.