JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Complex Numbers

Put $z=4e^{i\theta }$ on $C_1$. Then $w=z+\frac{1}{z}=4e^{i\theta }+\frac{1}{4}{e}^{-i\theta }=\frac{17}{4}\cos \theta +i\frac{15}{4}\sin \theta$. With $w=x+iy$ the locus $C_2$ is the ellipse $\frac{x^2}{(17/4{)}^2}+\frac{y^2}{(15/4{)}^2}=1$, whose semi-axes are $\frac{17}{4}=4.25$ and $\frac{15}{4}=3.75$. The circle $C_1$ has radius $4$, a value lying between these semi-axes, so the circle falls outside the ellipse along the major axis and inside it along the minor axis. Crossing once in each quadrant, the two curves meet at four points.