6. In 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
6.A.

Since $|z|=1$, we can let $z=\cos \theta+\mathrm{i} \sin \theta(0 \leqslant \theta<2 \pi)$.
From $z^{3}+1=\bar{z}$
$$
\begin{array}{l}
\Rightarrow 1+\cos 3 \theta+\mathrm{i} \sin 3 \theta=\cos \theta-\mathrm{i} \sin \theta \\
\Rightarrow \cos 3 \theta=\cos \theta-1, \sin 3 \theta=-\sin \theta \\
\Rightarrow(\cos \theta-1)^{2}+(-\sin \theta)^{2}=\cos ^{2} 3 \theta+\sin ^{2} 3 \theta=1 \\
\Rightarrow(\cos \theta-1)^{2}+(-\sin \theta)^{2}=1 \Rightarrow 2 \cos \theta=1 \\
\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \text { or } \frac{5 \pi}{3} .
\end{array}
$$

We also know that $\theta=\frac{\pi}{3}$ or $\frac{5 \pi}{3}$ do not satisfy $1+\cos 3 \theta=\cos \theta$, so the two curves have no intersection points.