For all z C on the curve C 1 : | z | = 4 , let the locus of the point z + 1 z be the curve C 2 . Then: [2023]

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Solution

Q.
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: [2023]

1 the curve $C_{1}$ lies inside $C_{2}$

2 the curves $C_{1}$ and $C_{2}$ intersect at 4 points

3 the curve $C_{2}$ lies inside $C_{1}$

4 the curves $C_{1}$ and $C_{2}$ intersect at 2 points

Ans.
(2)

We have, curve $C_{1} :$ $| z |$ $= 4 ∀ z \in C$ , which is a circle with centre (0, 0) and radius 4, therefore,

$x^{2} + y^{2} = 16$ $\dots (i)$

$Now, z = 4 e^{i θ}$

$z + \frac{1}{z} = 4 e^{i θ} + \frac{1}{4 e^{i θ}} = 4 e^{i θ} + \frac{1}{4} e^{- i θ}$

$= 4 (\cos θ + i \sin θ) + \frac{1}{4} (\cos θ - i \sin θ) = \frac{17}{4} \cos θ + i \frac{15}{4} \sin θ$

$To eliminate θ, let α = \frac{17}{4} \cos θ and β = \frac{15}{4} \sin θ$

$\frac{α^{2}}{{(\frac{17}{4})}^{2}} + \frac{β^{2}}{{(\frac{15}{4})}^{2}} = \frac{{(\frac{17}{4})}^{2} \cos^{2} θ}{{(\frac{17}{4})}^{2}} + \frac{{(\frac{15}{4})}^{2} \sin^{2} θ}{{(\frac{15}{4})}^{2}} = 1$

$∴$ $The curve C_{2} is \frac{x^{2}}{{(\frac{17}{4})}^{2}} + \frac{y^{2}}{{(\frac{15}{4})}^{2}} = 1$ $\dots (i i)$

which is an ellipse with centre (0, 0).

From (i) and (ii),

Hence, the curves $C_{1}$ and $C_{2}$ intersect at 4 points.

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