Three. (20 points) Determine all real numbers $\theta$ such that the complex sequence $\{\cos n \theta+\mathrm{i} \sin n \theta\}$ forms an arithmetic sequence.


Three, let $a_{n}=\cos n \theta+\mathrm{i} \sin n \theta$. Then $\left\{a_{n}\right\}$ is an arithmetic sequence, and $a_{n}-a_{n-1}$ is a constant value independent of $n$. Also,
$$
\begin{array}{l}
a_{n}-a_{n-1} \\
=(\cos n \theta-\cos (n-1) \theta)+\mathrm{i}(\sin n \theta-\sin (n-1) \theta) \\
=2 \sin \frac{\theta}{2}\left(-\sin \frac{2 n-1}{2} \theta+\mathrm{i} \cos \frac{2 n-1}{2} \theta\right),
\end{array}
$$

It is clear that $f(\theta)=-\sin \frac{2 n-1}{2} \theta+\mathrm{i} \cos \frac{2 n-1}{2} \theta$ must be independent of $n$. Taking $n=1$ and $n=2$, we get
$$
\sin \frac{\theta}{2}=\sin \frac{3}{2} \theta, \cos \frac{\theta}{2}=\cos \frac{3}{2} \theta \text {. }
$$

Thus, $\frac{3 \theta}{2}=\frac{\theta}{2}+2 k \pi$, which means $\theta=2 k \pi$.
It is easy to verify that when $\theta=2 k \pi$, $a_{n}=1$.
Therefore, the $\theta$ that makes $a_{n}$ an arithmetic sequence is $\theta=2 k \pi, k \in \mathbf{Z}$.