Example 10 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 $f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \cdots, f^{(k)}(z)=$ $\underbrace{f(f(\cdots f}_{k \uparrow f}(z)) \cdots))$.
(4th China 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, there must be a smallest positive integer $m$ such that $c$ is an $m-$ periodic point. In this case, $m \mid n$ must hold; 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 obtain: If the number of $n-$ fixed points and $n-$ periodic points of $f$ are denoted by $F_{n}$ and $P_{n}$, respectively, then
$$
P_{n}=F_{n}-\sum_{\substack{d \mid n \\ d \neq n}} P_{d} .
$$