1. Parametrize the curve $C_1$

Given $|z|=4,$ we can write $z=4e^{i\theta}=4(\cos\theta+i\sin\theta), \quad 0\le \theta<2\pi.$

2. Find the image of $z+\frac{1}{z}$

Let $w=z+\frac{1}{z}.$ Since $z=4e^{i\theta}$, $\frac{1}{z}=\frac{1}{4}e^{-i\theta}=\frac{1}{4}(\cos\theta-i\sin\theta).$ Therefore, $w=4e^{i\theta}+\frac{1}{4}e^{-i\theta}.$ Writing in real and imaginary parts, $w=\left(4+\frac14\right)\cos\theta+i\left(4-\frac14\right)\sin\theta.$ So, $w=\frac{17}{4}\cos\theta+i\frac{15}{4}\sin\theta.$

Hence the locus $C_2$ is the ellipse $\frac{x^2}{(17/4)^2}+\frac{y^2}{(15/4)^2}=1.$

3. Compare $C_1$ and $C_2$

The curve $C_1$ is the circle $x^2+y^2=16.$ The curve $C_2$ is the ellipse $\frac{x^2}{289/16}+\frac{y^2}{225/16}=1.$

To find intersection points, solve both together.

From the ellipse parametrization, $x=\frac{17}{4}\cos\theta, \qquad y=\frac{15}{4}\sin\theta.$ For this point to also lie on $C_1$, $x^2+y^2=16.$ Substitute: $\left(\frac{17}{4}\right)^2\cos^2\theta+\left(\frac{15}{4}\right)^2\sin^2\theta=16.$ Multiply by 16: $289\cos^2\theta+225\sin^2\theta=256.$ Using $\sin^2\theta=1-\cos^2\theta$, $289\cos^2\theta+225(1-\cos^2\theta)=256,$ $64\cos^2\theta=31,$ $\cos^2\theta=\frac{31}{64}.$ This gives valid values, and for $0\le \theta<2\pi$, there are 4 corresponding points: $\cos\theta=\pm \frac{\sqrt{31}}{8}, \qquad \sin\theta=\pm \frac{\sqrt{33}}{8},$ with independent sign combinations determined by quadrants, yielding 4 distinct intersection points.

So the curves intersect at 4 points.

4. Check the options

• A: $C_1$ and $C_2$ intersect at 4 points — True
• B: $C_2$ lies inside $C_1$ — False, since ellipse has point $\left(\frac{17}{4},0\right)$ with distance $\frac{17}{4}>4$
• C: $C_1$ lies inside $C_2$ — False, since ellipse has point $\left(0,\frac{15}{4}\right)$ with distance $\frac{15}{4}<4$, so circle is not fully inside ellipse
• D: intersect at 2 points — False

Therefore, the correct option is A.