(4) Let $r>1$ be a real number. If a moving point $z$ in the complex plane satisfies $|z|=r$, then the trajectory of the moving point $\omega=z+\frac{1}{z}$ is ( ).
(A) An ellipse with a focal distance of 4
(B) An ellipse with a focal distance of $\frac{4}{r}$
(C) An ellipse with a focal distance of 2
(D) An ellipse with a focal distance of $\frac{2}{r}$
(4) A Hint: Let $z=r(\cos \theta+\mathrm{i} \sin \theta), \theta \in[0,2 \pi)$, then
$$
\begin{aligned}
\omega & =z+\frac{1}{z}=r(\cos \theta+\mathrm{i} \sin \theta)+\frac{1}{r(\cos \theta+\mathrm{i} \sin \theta)} \\
& =\left(r \cos \theta+\frac{1}{r} \cos \theta\right)+\left(r \sin \theta-\frac{1}{r} \sin \theta\right) \mathrm{i} .
\end{aligned}
$$
$$
\left\{\begin{array}{l}
\frac{x}{r+\frac{1}{r}}=\cos \theta, \\
\frac{y}{r-\frac{1}{r}}=\sin \theta .
\end{array}\right.
$$

Squaring and adding the two equations, we get
$$
\frac{x^{2}}{\left(r+\frac{1}{r}\right)^{2}}+\frac{y^{2}}{\left(r-\frac{1}{r}\right)^{2}}=1,
$$

where $r>1$, focal distance $=2 \sqrt{\left(r+\frac{1}{r}\right)^{2}-\left(r-\frac{1}{r}\right)^{2}}=4$.