JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 13)
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 $$\mathrm{C}_{2}$$. Then :
the curves $$C_{1}$$ and $$C_{2}$$ intersect at 4 points
the curve $$C_{2}$$ lies inside $$C_{1}$$
the curve $$C_{1}$$ lies inside $$C_{2}$$
the curves $$C_{1}$$ and $$C_{2}$$ intersect at 2 points
Explanation
Let $\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} \mathrm{e}^{-\mathrm{i} \theta}$
$\Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$
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 $\mathrm{z}$ is circle $\mathrm{x}^{2}+\mathrm{y}^{2}=16$
So intersect at 4 points.
$\Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$
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 $\mathrm{z}$ is circle $\mathrm{x}^{2}+\mathrm{y}^{2}=16$
So intersect at 4 points.
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