For all $z\in C$ on the curve $C_1:|z|=4$, let the locus of the point $z+\dfrac{1}{z}$ be the curve $C_2$. Then:
the curves $C_1$ and $C_2$ intersect at 4 points
the curve $C_1$ lies inside $C_2$
the curves $C_1$ and $C_2$ intersect at 2 points
the curve $C_2$ lies inside $C_1$
Step-by-Step Solution
Key Concept: $z=4e^{i\theta}$: $z+1/z=(4+1/4)\cos\theta+i(4-1/4)\sin\theta=(17\cos\theta/4,\,15\sin\theta/4)$. $C_2$: ellipse $\frac{x^2}{(17/4)^2}+\frac{y^2}{(15/4)^2}=1$.
Step 1:
To solve this problem, we first need to understand what the curves $C_1$ and $C_2$ represent and how they are related. The curve $C_1$ is defined by the equation $|z| = 4$, which represents a circle centered at the origin with radius 4 in the complex plane.
Step 2:
The curve $C_2$ is defined as the locus of the point $z + \frac{1}{z}$. To find the equation of $C_2$, we start with the expression $z + \frac{1}{z}$. Let's express $z$ in polar form as $z = re^{i\theta}$, where $r = |z|$ and $\theta = \text{Arg}(z)$. For points on $C_1$, $r = 4$, so $z = 4e^{i\theta}$.
Step 3:
Substituting $z = 4e^{i\theta}$ into the expression $z + \frac{1}{z}$ gives us $4e^{i\theta} + \frac{1}{4e^{i\theta}} = 4e^{i\theta} + \frac{1}{4}e^{-i\theta}$. Using Euler's formula, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$, we can rewrite this as $4(\cos(\theta) + i\sin(\theta)) + \frac{1}{4}(\cos(-\theta) + i\sin(-\theta))$.
Step 4:
Simplifying the expression from Step 3, we note that $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, so we get $4\cos(\theta) + 4i\sin(\theta) + \frac{1}{4}\cos(\theta) - \frac{1}{4}i\sin(\theta)$. Combining like terms yields $\left(4 + \frac{1}{4}\right)\cos(\theta) + i\left(4 - \frac{1}{4}\right)\sin(\theta)$, which simplifies to $\frac{17}{4}\cos(\theta) + i\frac{15}{4}\sin(\theta)$.
Step 5:
The expression $\frac{17}{4}\cos(\theta) + i\frac{15}{4}\sin(\theta)$ represents an ellipse in the complex plane, with semi-major axis $\frac{17}{4}$ and semi-minor axis $\frac{15}{4}$. This is the equation of the curve $C_2$.
Step 6:
To determine how $C_1$ and $C_2$ intersect, we need to consider the geometric shapes of these curves. $C_1$ is a circle with radius 4, and $C_2$ is an ellipse. The intersection points occur where the circle and ellipse cross. Given the symmetry and the fact that the ellipse's axes are aligned with the real and imaginary axes, we can reason about the number of intersections.
Step 7:
Since $C_1$ is a circle with a radius of 4 and $C_2$ is an ellipse with axes of lengths $\frac{17}{2}$ and $\frac{15}{2}$, and considering the relative sizes and shapes, the circle $C_1$ and the ellipse $C_2$ will intersect at points where the circle cuts through the ellipse. Given the geometry, there will be intersections in each quadrant of the complex plane.
Step 8:
Concluding the analysis, because the circle $C_1$ and the ellipse $C_2$ have the described geometric relationship and considering the symmetrical nature of both curves, they intersect at 4 points. This matches Option 1.
The final answer is: $\boxed{1}$
Correct Answer: 1