Here Sn:∣z−(3−2i)∣=4n represents a circle with center C1(3,−2) and radius 4n
and Tn:∣z−(2−3i)∣=n1 represents a circle with center C2(2,−3) and radius n1
For Sn∩Tn=ϕ, both circles do not intersect each other.
When C1C2>4n+n1
i.e. 2>4n+n1
then possible values of n=1,2,3,4
When C1C2<∣4n−n1∣
⇒2<∣4nn2−4∣
then n has infinite solutions for n∈N
Hence, there are total four values possible.