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…

Here $S_n:\left|z-\left(3-2i\right)\right|=\frac{n}{4}$ represents a circle with center $C_1\left(3,-2\right)$ and radius $\frac{n}{4}$

and $T_{\mathit{n}}:\left|z-\left(2-3i\right)\right|=\frac{1}{n}$ represents a circle with center $C_2\left(2,-3\right)$ and radius $\frac{1}{n}$

For $S_n\cap T_n=\varphi$, both circles do not intersect each other.

When $C_1{C}_2>\frac{n}{4}+\frac{1}{n}$

i.e. $\sqrt{2}>\frac{n}{4}+\frac{1}{n}$
then possible values of $n=1,2,3,4$

When $C_1{C}_2<\left|\frac{n}{4}-\frac{1}{n}\right|$

$\Rightarrow \sqrt{2}<\left|\frac{n^2-4}{4n}\right|$

then $n$ has infinite solutions for $n\in N$

Hence, there are total four values possible.