12-36 On the complex plane, the number of intersection points between the curve $z^{4}+z=1$ and the circle $|z|=1$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(High School Mathematics Competition in Suzhou City, Jiangsu Province, China, 1993)
[Solution] From $|z|=1$, we know $z=\cos \theta + i \sin \theta (0 \leqslant \theta < 2 \pi)$. Substituting this into
$$
z^{3}+1=\frac{1}{z}=\frac{\bar{z}}{z \cdot \bar{z}}=\frac{\bar{z}}{|z|^2}
$$

we get
$$
1+\cos 3 \theta + i \sin 3 \theta = \cos \theta - i \sin \theta.
$$

By the definition of equality of complex numbers, we have
$$
1+\cos 3 \theta = \cos \theta, \quad \sin 3 \theta = -\sin \theta.
$$

Therefore,
$$
\begin{aligned}
1 & = \cos^2 3 \theta + \sin^2 3 \theta \\
& = (\cos \theta - 1)^2 + (-\sin \theta)^2 \\
& = 2 - 2 \cos \theta,
\end{aligned}
$$

which implies $\cos \theta = \frac{1}{2}$, so $\theta$ equals $\frac{\pi}{3}$ or $2 \pi - \frac{\pi}{3}$. However, neither satisfies $1+\cos 3 \theta = \cos \theta$.
Thus, the curve $z^{4} + z = 1$ does not intersect the circle $|z| = 1$.
Therefore, the answer is $(A)$.