For $$n \in \mathbb{N}$$, let $$S_n = \{z \in \mathbb{C} : |z - 3 + 2i| = \dfrac{n}{4}\}$$ and $$T_n = \{z \in \mathbb{C} : |z - 2 + 3i| = \dfrac{1}{n}\}$$. Then the number of elements in the set $$\{n \in \mathbb{N} : S_n \cap T_n = \phi\}$$ is

We are given $$ S_n = \{z \in \mathbb{C} : |z - 3 + 2i| = \frac{n}{4}\} $$ and $$ T_n = \{z \in \mathbb{C} : |z - 2 + 3i| = \frac{1}{n}\} $$.

$$ S_n $$ is a circle centered at $$ (3, -2) $$ with radius $$ r_1 = \frac{n}{4} $$, and $$ T_n $$ is a circle centered at $$ (2, -3) $$ with radius $$ r_2 = \frac{1}{n} $$.

$$d = \sqrt{(3-2)^2 + (-2+3)^2} = \sqrt{2}$$

Two circles are disjoint when $$ d > r_1 + r_2 $$, i.e., the circles are too far apart to touch each other externally. This requires:

$$\frac{n}{4} + \frac{1}{n} < \sqrt{2}$$

Multiplying through by $$ 4n $$:

$$n^2 + 4 < 4\sqrt{2} \cdot n$$

$$n^2 - 4\sqrt{2}\,n + 4 < 0$$

The roots of $$ n^2 - 4\sqrt{2}\,n + 4 = 0 $$ are:

$$n = \frac{4\sqrt{2} \pm \sqrt{32 - 16}}{2} = \frac{4\sqrt{2} \pm 4}{2} = 2\sqrt{2} \pm 2$$

So $$ n \in (2\sqrt{2} - 2, \, 2\sqrt{2} + 2) \approx (0.828, \, 4.828) $$.

The natural numbers in this interval are $$ n = 1, 2, 3, 4 $$.

$$ n = 1 $$: $$ r_1 + r_2 = 1.25 < 1.414 $$. Disjoint. ✓

$$ n = 2 $$: $$ r_1 + r_2 = 1.0 < 1.414 $$. Disjoint. ✓

$$ n = 3 $$: $$ r_1 + r_2 \approx 1.083 < 1.414 $$. Disjoint. ✓

$$ n = 4 $$: $$ r_1 + r_2 = 1.25 < 1.414 $$. Disjoint. ✓

$$ n = 5 $$: $$ r_1 + r_2 = 1.45 > 1.414 $$. Not disjoint by this condition. ✗

There are exactly 4 natural numbers for which the sum of radii is less than the distance between centers, making $$ S_n \cap T_n = \emptyset $$.

The answer is Option D.