For all z ∈ C on the curve C _ 1 : | z | = 4 , let the locus of the...

Given:

${C}_{1}: \left|z\right| = 4, {C}_{2}: z + \dfrac{1}{z}$

Here, \left|z\right| = 4$is a circle${x}^{2} + {y}^{2} = 16$.</p><p>Now, let$z = 4{e}^{i\theta}$, </p><p>So,$z + \dfrac{1}{z} = 4{e}^{i\theta} + \dfrac{{e}^{- i\theta}}{4}$</p><p>$⇒ x + iy = 4cos\theta + i4sin\theta + \dfrac{cos\theta}{4} - \dfrac{isin\theta}{4}\left\{\text{taking}z + \dfrac{1}{z} = x + iy\right\}$</p><p>Now on comparing both side we get,</p><p>$x = \dfrac{17}{4}cos\theta \text{ and } y = \dfrac{15}{4}sin\theta

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

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

Which is a equation of ellipse,

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