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

Given:

$C_1:\left|z\right|=4, C_2:z+\frac{1}{z}$

Here, $\left|z\right|=4$ is a circle $x^2+y^2=16$.

Now, let $z=4e^{i\theta }$, 

So, $z+\frac{1}{z}=4e^{i\theta }+\frac{e^{-i\theta }}{4}$

$\Rightarrow x+iy=4\cos \theta +i4\sin \theta +\frac{\cos \theta }{4}-\frac{i\sin \theta }{4} \left\{\text{taking }z+\frac{1}{z}=x+iy\right\}$

Now on comparing both side we get,

$x=\frac{17}{4}\cos \theta \& y=\frac{15}{4}\sin \theta$

Now on solving ${\cos }^2\theta +{\sin }^2\theta =1$ we get,

 $\frac{x^2}{{\left({\displaystyle \frac{17}{4}}\right)}^2}+\frac{y^2}{{\displaystyle {\left({\displaystyle \frac{15}{4}}\right)}^2}}=1$

Which is a equation of ellipse,

Therefore, curves $C_1$ and $C_2$ intersect at $4$ points.