For all $z\in C$ on the curve $C_1:\left|z\right|=4$ let the locus of the point $z+\frac{1}{z}$be the curve  $C_2$Then:

$C_1:\left|z\right|=4\text{ let }z=4(\cos \theta +i\sin \theta )\text{ let }\omega =z+\frac{1}{z}=x+iy$
$\Rightarrow x+iy=4(\cos \theta +i\sin \theta )+\left(\frac{\cos \theta -i\sin \theta }{4}\right)$
$$\begin{gathered} \therefore x=\frac{17}{4}\cos \theta ;y=\frac{15}{4}\sin \theta \\ \Rightarrow \frac{16x^2}{289}+\frac{16y^2}{225}=1 \end{gathered}$$
is the equation to the locus of the curve :
$C_2\frac{x^2}{\left(\frac{289}{16}\right)}+\frac{y^2}{\left(\frac{225}{16}\right)}=1\text{ and }{x}^2+y^2=16$
intersect at 4 points $\left(\because \frac{225}{16}<16<\frac{289}{16}\right)$