Question

For n ∈ N let \(S_n\)={z∈C:|z−3+2i|=\(\frac{n}{4}\)} and \(T_n\)={z ∈ C:|z−2+3i|=\(\frac{1}{n}\)}.Then the number of elements in the set

• A. 0
• B. 2
• C. Infinite
• D. 4

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze the nature of the sets \( S_n \) and \( T_n \), which are defined in the complex plane with the help of complex numbers.

1. Understanding the Set \( S_n \):

   The set \( S_n \) is defined as:

   S_n = \{z \in \mathbb{C} : |z - 3 + 2i| \leq \frac{n}{4}\}

   Here, \( |z - 3 + 2i| \leq \frac{n}{4} \) represents a circle centered at \( (3, -2) \) in the complex plane with radius \( \frac{n}{4} \). As \( n \) can take any natural number, the size of the circle increases with increasing \( n \).

2. Understanding the Set \( T_n \):

   The set \( T_n \) is defined as:

   T_n = \{z \in \mathbb{C} : |z - 2 + 3i| = \frac{1}{n}\}

   This represents a circle centered at \( (2, -3) \) with radius \( \frac{1}{n} \). As \( n \) increases, the radius of the circle decreases.

3. Intersection of \( S_n \) and \( T_n \):

   We need to determine how many elements are there in the intersection of \( S_n \) and \( T_n \), i.e., \( S_n \cap T_n \).

   The circle \( T_n \) is a very small circle whose radius becomes smaller as \( n \) increases, while \( S_n \) forms a relatively larger circle whose radius increases with \( n \). For any given \( n \), \( T_n \) (a small circle) will always lie entirely within \( S_n \) as \( n \) becomes sufficiently large due to the different rates at which their radii change.

   This ensures that there are infinite elements \( z \) that satisfy both conditions, as \( S_n \) subsumes \( T_n \) when \( n \) is large.

4. Conclusion:

   Therefore, the number of elements in \( S_n \cap T_n \) is infinite.

Thus, the correct answer is: Infinite.