令 $z=r(\cos\theta+\mathrm{i}\sin\theta),\theta\in[0,2\pi)$，则$$\omega=z+\dfrac1z=\left(r\cos\theta+\dfrac1r\cos\theta\right)+\left(r\sin\theta-\dfrac1r\sin\theta\right)\mathrm{i}.$$设 $\omega=x+\mathrm{i}y$，则有$$\begin{cases}x=r\cos\theta+\dfrac1r\cos\theta,\\y=r\sin\theta-\dfrac1r\sin\theta,\end{cases}$$由此进一步得$$\begin{cases}\dfrac{x}{r+\frac1r}=\cos\theta,\\\dfrac{y}{r-\frac1r}=\sin\theta,\end{cases}$$两式分别平方相加，得$$\dfrac{x^2}{\left(r+\frac1r\right)^2}+\dfrac{y^2}{\left(r-\frac1r\right)^2}=1,$$其中 $r>1$，焦距为$$2\sqrt{\left(r+\dfrac1r\right)^2-\left(r-\dfrac1r\right)^2}=4.$$