4 On the complex plane, the number of intersection points between the curve $\left|z-\frac{1}{z}\right|=1$ and the circle $|z|=1$ is
A. 2
B. 3
C. 4
D. 5
4 C. From $|z|=1$ we get
$$
z=\cos \theta+i \cdot \sin \theta(0 \leqslant \theta<2 \pi),
$$

Substituting into $\left|z^{2}-1\right|=1$ we get
$$
|(-1+\cos 2 \theta)+i \cdot \sin 2 \theta|=1,
$$

Thus, we have $2-2 \cdot \cos 2 \theta=1$, $\cos 2 \theta=\frac{1}{2}$, hence $2 \theta=\frac{\pi}{3}$, or $2 \pi-\frac{\pi}{3}$, or $2 \pi+\frac{\pi}{3}$, or $4 \pi-\frac{\pi}{3}$, i.e., $\theta=\frac{\pi}{6}, \pi \pm \frac{\pi}{6}, 2 \pi-\frac{\pi}{6}$, so there are four intersection points.