For n ∈ N , let S n = z ∈ C : z - 3 + 2 i = n 4 and T n = z ∈ C : z - 2 + 3 i = 1 n . Then the number of elements in the set n ∈ N : S n ∩ T n = ϕ is

Step-by-step solution

Here S n : z - 3 - 2 i = n 4 represents a circle with center C 1 3 , - 2 and radius n 4 and T n : z - 2 - 3 i = 1 n represents a circle with center C 2 2 , - 3 and radius 1 n For S n ∩ T n = ϕ , both circles do not intersect each other. When C 1 C 2 > n 4 + 1 n i.e. 2 > n 4 + 1 n then possible values of n = 1 , 2 , 3 , 4 When C 1 C 2 < n 4 - 1 n ⇒ 2 < n 2 - 4 4 n then n has infinite solutions for n ∈ N Hence, there are total four values possible.