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
- 0
- 2
- Infinite
- 4
The Correct Option is C
Solution and Explanation
\(S_n\)={z∈C:|z−3+2i|=\(\frac{n}{4}\)}
represents a circle with centre C1(3, –2) and radius
r1=\(\frac{n}{4}\)
Similarly,
Tn represents circle with centre C2(2, –3) and radius
r2=\(\frac{1}{n}\)
As Sn ∩ Tn = φ
C1C2>r1 + r2 OR C1C2< |r1 – r2|
\(\sqrt2\)>\(\frac{n}{4}+\frac{1}{n}\)
OR
\(\sqrt2\)<|\(\frac{n}{4}-\frac{1}{n}\)|
n = 1, 2, 3, 4 n may take infinite values.
The Correct Option is (C): Infinite
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