Question

Let \( S^1 = \{z \in \mathbb{C} : |z| = 1\} \) be the circle group under multiplication and \( i = \sqrt{-1}. \) Then the set \( \{\theta \in \mathbb{R} : (e^{i2\pi\theta}) \text{ is infinite}\} \) is
 

• A. empty
• B. non-empty and finite
• C. countably infinite
• D. uncountable

Hint:

For complex exponentials, rational multiples of \( 2\pi \) give periodic (finite) sets, while irrational multiples generate dense (uncountable) sets on the unit circle.

Answer 1

The Correct Option is D

Step 1: Interpretation of the set.
For \( e^{i2\pi\theta} \), the value depends on whether \( \theta \) is rational or irrational.

Step 2: If \( \theta \) is rational,
say \( \theta = \frac{p}{q}, \) then \[ (e^{i2\pi\theta})^q = e^{i2\pi p} = 1, \] hence the subgroup generated is finite.

Step 3: If \( \theta \) is irrational,
then the set \( \{e^{i2\pi n\theta} : n \in \mathbb{Z}\} \) is dense on the unit circle, so it has infinitely many distinct elements.

Step 4: Set of irrationals in \([0,1]\) is uncountable. Hence, the set of \(\theta\) for which \( e^{i2\pi\theta} \) generates an infinite subgroup is uncountable.

Final Answer: (D) uncountable.