JEE Mains · Maths · STD 11 - 4.1 complex nubers
For all \(z \in C\) on the curve \(C _1:| z |=4\), let the locus of the point \(z +\frac{1}{ z }\) be the curve \(C _2\). Then
- A the curves \(C_1\) and \(C_2\) intersect at \(4\) points
- B the curves \(C_1\) lies inside \(C_2\)
- C the curves \(C_1\) and \(C_2\) intersect at \(2\) points
- D the curves \(C_2\) lies inside \(C_1\)
Answer & Solution
Correct Answer
(A) the curves \(C_1\) and \(C_2\) intersect at \(4\) points
Step-by-step Solution
Detailed explanation
Let \(w = z +\frac{1}{ z }=4 e ^{ i \theta}+\frac{1}{4} e ^{- i \theta}\) \(\Rightarrow w =\frac{17}{4} \cos \theta+ i \frac{15}{4} \sin \theta\) So locus of \(w\) is ellipse \(\frac{x^2}{\left(\frac{17}{4}\right)^2}+\frac{y^2}{\left(\frac{15}{4}\right)^2}=1\) Locus of \(z\) is…
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