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JEE Mains · Maths · STD 11 - 4.1 complex nubers

For \(n \in N\), let \(S _{ n }=\left\{ z \in C :| z -3+2 i |=\frac{ n }{4}\right\}\) and \(T _{ n }=\left\{ z \in C :| z -2+3 i |=\frac{1}{ n }\right\}\) Then the number of elements in the set \(\left\{ n \in N : S _{ a } \cap T _{ n }=\phi\right\}\) is.

  1. A \(0\)
  2. B \(2\)
  3. C \(3\)
  4. D \(4\)
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Correct Answer

(D) \(4\)

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Detailed explanation

\(Bonus\) \(S_{n}:|z-(3-2 i)|=\frac{n}{4}\) is a circle center \(C_{1}(3,-2)\) and radius \(n / 4\) \(T _{ n }:| z -(2-3 i )|=\frac{1}{ n }\) is a circle center \(C _{2}(2,-3)\) and radius \(1 / n\) Here \(S_{ a } \cap T _{ n }=\phi\) Both circles do not intersect each other…
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