4. Let $r>1$ be a real number. If the moving point $z$ on the complex plane satisfies $|z|=r$, then the trajectory of the moving point $\omega=z+\frac{1}{z}$ is an ellipse with a focal distance of ( ).
(A) 4
(B) $\frac{4}{r}$
(C) 2
(D) $\frac{2}{r}$
4. A.

Let $z=r(\cos \theta+\mathrm{i} \sin \theta)(\theta \in[0,2 \pi))$. Then $\omega=z+\frac{1}{z}=\left(r+\frac{1}{r}\right) \cos \theta+\mathrm{i}\left(r-\frac{1}{r}\right) \sin \theta$. Let $\omega=x+\mathrm{i} y$. Thus,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{r+\frac{1}{r}}=\cos \theta, \\
\frac{y}{r-\frac{1}{r}}=\sin \theta
\end{array}\right. \\
\Rightarrow \frac{x^{2}}{\left(r+\frac{1}{r}\right)^{2}}+\frac{y^{2}}{\left(r-\frac{1}{r}\right)^{2}}=1 .
\end{array}
$$

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