$1 \cdot 14$ Let $S$ be the unit circle in the complex plane (i.e., the set of all complex numbers with modulus 1), and $f$ be a mapping from $S$ to $S$. For any $z \in S$, define
$$
\begin{array}{l}
f^{(1)}=f(z), f^{(2)}(z)=f\left(f^{(1)}(z)\right), \cdots, \\
f^{(k+1)}(z)=f\left(f^{(k)}(z)\right), k=2,3, \cdots .
\end{array}
$$

If $c \in S$, such that
$$
f^{(1)}(c) \neq c, \cdots, f^{(n-1)}(c) \neq c, f^{(n)}(c)=c,
$$

then $c$ is called an $n_{-}$periodic point of $f$. Let $m$ be a natural number greater than 1, and $f$ is defined as follows:
$$
f(z)=z^{m}, z \in S \text {. }
$$

Try to find the total number of 1989-periodic points of $f$.
[Solution] Let the set of all 1989-periodic points of $f$ be denoted by $T$ and let $B_{j}=\{z \in S \mid f^{(j)}(z)=z\}, j=1,2, \cdots$.
(1) If $z_{0} \in S$ is a fixed point of $f^{(n)}$, i.e., $f^{(n)}\left(z_{0}\right)=z_{0}$, and $z_{0}$ is a
Suppose $n=p h+q, 0 \leqslant qk$, let $n=p k+q, 0 \leqslant q<k$, then $z_{0} \in B_{q}$, hence $z_{0} \in B_{k} \cap B_{q}, q<k$. By induction, we get $z_{0} \in B_{(n, k)}$. By the arbitrariness of $z_{0} \in B_{n} \cap B_{k}$, we have $B_{n} \cap B_{k} \subset B_{(n, k)}$. Therefore, $B_{n} \cap B_{k}=B_{(n, k)}$.
(4) From the result of (3), we get
$$
\begin{array}{l}
B_{663} \cap B_{153}=B_{51}, B_{663} \cap B_{117}=B_{39}, \\
B_{153} \cap B_{117}=B_{9}, B_{663} \cap B_{153} \cap B_{117}=B_{3} .
\end{array}
$$

Substituting (3) into (2), we get
$$
\begin{aligned}
\left|B_{663} \cup B_{153} \cup B_{117}\right|= & \left|B_{663}\right|+\left|B_{153}\right|+\left|B_{117}\right|-\left|B_{51}\right|- \\
& \left|B_{39}\right|-\left|B_{9}\right|+\left|B_{3}\right| .
\end{aligned}
$$
(5) From the given information,
$$
f^{(n)}(z)=z^{m^{n}} .
$$

It follows that, $z \in B_{n}$ if and only if
$$
z \in S, \quad z=f^{(n)}(z)=z^{m^{n}},
$$

which means $z$ is an $(m^{n}-1)$-th root of unity. Therefore, $\left|B_{n}\right|=m^{n}-1$. Using (4) and (1), we get
$$
|T|=m^{1989}-m^{663}-m^{153}-m^{117}+m^{51}+m^{39}+m^{9}-m^{3}
$$