Question

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

• A. 0
• B. 2
• C. Infinite
• D. 4

Answer 1

Answer: (C)

Infinite

Answer 2

$S_n$={z∈C:|z−3+2i|=$\frac{n}{4}$}
represents a circle with centre C 1(3, –2) and radius
r1=$\frac{n}{4}$
Similarly,
T n represents circle with centre C 2(2, –3) and radius
r2=$\frac{1}{n}$
As _S n _∩ T n = φ
C 1 C 2>r 1 + r 2 OR C 1 C 2< |r 1 – r 2|
$\sqrt2$>$\frac{n}{4}+\frac{1}{n}$
OR
$\sqrt2$<|$\frac{n}{4}-\frac{1}{n}$|
n = 1, 2, 3, 4 _n _may take infinite values.

The Correct Option is (C): Infinite