For all z C on the curve C_1: |z|=4, let the locus of the point z+1z be the curve C_2.

Question

For all $z\in\mathbb C$ on the curve $C_1: |z|=4$, let the locus of the point $z+\dfrac1z$ be the curve $C_2$. Then:

Accepted Answer

Let $z=4e^{i\theta}$. Then:
$$w=z+\frac1z=4e^{i\theta}+\frac14e^{-i\theta}=\left(4+\frac14\right)\cos\theta+i\left(4-\frac14\right)\sin\theta = \frac{17}4\cos\theta+i\frac{15}4\sin\theta.$$

So $C_2$ is the ellipse $\dfrac{x^2}{(17/4)^2}+\dfrac{y^2}{(15/4)^2}=1$, with semi-axes $4.25$ and $3.75$.

$C_1$ is the circle $x^2+y^2=16$ (radius 4). Since $4.25>4>3.75$, the ellipse extends beyond the circle along the major axis and lies inside it along the minor axis --- so the two curves cross at 4 points.

Solutions · 1