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:

$${C}_{1}:|z|=4,{C}_{2}:z+\frac{1}{z}$$C1​:∣z∣=4,C2​:z+z1​

Here, $|z|=4$∣z∣=4 is a circle ${x}^{2}+{y}^{2}=16$x2+y2=16.

Now, let $z=4{e}^{i\theta }$z=4eiθ,

So, $z+\frac{1}{z}=4{e}^{i\theta }+\frac{{e}^{-i\theta }}{4}$z+z1​=4eiθ+4e−iθ​

$$\Rightarrow x+iy=4\mathrm{cos}\theta +i4\mathrm{sin}\theta +\frac{\mathrm{cos}\theta }{4}-\frac{i\mathrm{sin}\theta }{4}{\text{taking }z+\frac{1}{z}=x+iy}$$⇒x+iy=4cosθ+i4sinθ+4cosθ​−4isinθ​taking z+z1​=x+iy

Now on comparing both side we get,

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

Now on solving ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$cos2θ+sin2θ=1 we get,

$$\frac{{x}^{2}}{{(\frac{17}{4})}^{2}}+\frac{{y}^{2}}{{(\frac{15}{4})}^{2}}=1$$(417​)2x2​+(415​)2y2​=1

Which is a equation of ellipse,

Therefore, curves ${C}_{1}$C1​ and ${C}_{2}$C2​ intersect at $4$4 points.