Example 9 Let $S$ be the unit circle in the complex plane (i.e., the set of complex numbers with modulus 1), and let $f$ be a mapping from $S$ to $S$. For any $z \in S$, define
$$
\begin{array}{l}
f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \cdots, \\
f^{(k)}(z)=\underbrace{f(\cdots(f(z)) \cdots) .}_{k \uparrow}
\end{array}
$$

If $C \in S$ and a natural number $n$ exist such that
$$
\begin{array}{l}
f^{(1)}(C) \neq C, f^{(2)}(C) \neq C, \cdots, \\
f^{(n-1)}(C) \neq C, f^{(n)}(C)=C,
\end{array}
$$

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 calculate the number of 1989-periodic points of $f$.
(4th Russian Mathematical Olympiad)
When $f^{(n)}(C)=C$, it is known that $C$ is an $n$-fixed point of $f$. Clearly, an $n$-periodic point must be an $n$-fixed point. Conversely, if $C$ is an $n$-fixed point, then there must be a smallest positive integer $m$ such that $C$ is an $m$-periodic point, and at this time, $m \mid n$. Otherwise, let $n=k m+r(0<r<m)$, then $C=f^{(n)}(C)=f^{(r)}(C)$, which contradicts the minimality of $m$. From this, we get:

If the number of $n$-fixed points and $n$-periodic points of $f$ are denoted as $F_{n}$ and $P_{n}$, respectively, then
$$
P_{n}=F_{n}-\sum_{\substack{i \mid n \\ i \neq n}} P_{i} .
$$

Thus, it is easy to prove by induction that $f^{(k)}(z)=z^{m^{k}}$.
Hence, from $f^{(k)}(z)=z$, we know that $F_{n}=m^{k}-1$.
Since $1989=3^{2} \times 13 \times 17$, its (less than 1989) factors are 11:
$$
1,3,9,13,17,39,51,117,153,221,663 \text {. }
$$

Therefore, we can deduce (or use the principle of inclusion-exclusion to calculate):
$$
\begin{array}{l}
P_{1}=F_{1}=m-1 ; \\
P_{3}=F_{3}-P_{1}=m^{3}-m ; \\
P_{9}=F_{9}-P_{1}-P_{3}=m^{9}-m^{3} ; \\
P_{13}=m^{13}-m ; \\
P_{17}=m^{17}-m ; \\
P_{39}=m^{39}-m^{13}-m^{3}+m ; \\
P_{51}=m^{51}-m^{17}-m^{3}+m ; \\
P_{117}=m^{117}-m^{39}-m^{9}+m^{3} ; \\
P_{153}=m^{153}-m^{51}-m^{9}+m^{3} ; \\
P_{21}=m^{21}-m^{17}-m^{13}+m ; \\
P_{663}=m^{663}-m^{21}-m^{51}-m^{39}+m^{17}+ \\
\quad m^{13}+m^{3}-m . \\
P_{1989}=F_{1989}-\sum_{i \mid 1989} P_{i} \\
=m^{1989}-m^{663}-m^{153}-m^{117}+m^{51}+ \\
m^{39}+m^{9}-m^{3} .
\end{array}
$$

This is the desired result.
2.4 Using Fixed Points of Functions to Solve Sequence Problems

Theorem 1 Let $f(x)=a x+b(a \neq 1, a \neq 0)$. The sequence $\left\{a_{n}\right\}$ is determined by the initial term $a_{0} \neq f\left(a_{0}\right)$ and $a_{n}=$ $f\left(a_{n-1}\right)$, then, if and only if $x_{0}$ is a fixed point of $f(x)$, the sequence $\left\{a_{n}-x_{0}\right\}$ is a geometric sequence with common ratio $a$ (proof omitted).