6-105 Let $S$ be the unit circle in the complex plane (i.e., the set of complex numbers with modulus 1), and $f$ be a mapping from $S$ to $S$. For any $z \in S$, define
$$
\begin{array}{l}
f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \cdots, \\
f^{(k)}(z)=\underbrace{f(f(\cdots(f}_{k \uparrow f}(z) \cdots))),
\end{array}
$$

If $c \in S$, such that
$$
\begin{array}{l}
f^{(1)}(C) \neq C, f^{(2)}(C) \neq C, \cdots, \\
f^{(n-1)}(C) \neq C, f^{(n)}(C)=C .
\end{array}
$$

we say that $C$ is an $n$-periodic point of $f$.
Let $m$ be a natural number greater than 1, and $f$ is defined as follows:
$$
f(z)=z^{m}(z \in S) .
$$

Try to calculate the total number of 1989-periodic points of $f$.
[Solution] We first prove:
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 an $m$-periodic point of $f$, then $m \mid n$.
Indeed, let $n=p m+q, 0 \leqslant q<m$, then we have
$$
\begin{aligned}
z_{0} & =f^{(n)}\left(z_{0}\right) \\
& =f^{(p m+q)}\left(z_{0}\right) \\
& =f^{(q)}\left(f^{(m)}\left(f^{(m)}\left(\cdots\left(f^{(m)}\left(z_{0}\right) \cdots\right)\right)\right)\right. \\
& =f^{(q)}\left(z_{0}\right) .
\end{aligned}
$$

If $q \neq 0$, then the above equation implies that $z_{0}$ is a $q$-periodic point of $f$, which contradicts the fact that $z_{0}$ is an $m$-periodic point of $f$. Therefore, we must have $q=0$, i.e., $m \mid n$.
Next, we prove that if $(n, k)$ denotes the greatest common divisor of $n$ and $k$, and we define
$$
B_{n}=\left\{z \in S \mid f^{(n)}(z)=z\right\},
$$

then we have
$$
B_{(n, k)}=B_{n} \cap B_{k} .
$$

Indeed, by definition, it is directly known that when $m \mid n$, $B_{m} \subseteq B_{n}$, thus
$$
B_{(n, k)} \subseteq B_{n} \cap B_{k} .
$$

On the other hand, if $z_{0} \in B_{n} \cap B_{k}$, let
$$
n=p k+q, 0 \leqslant q<k \text {. }
$$

Then we have $\quad z_{0}=f^{(q)}\left(z_{0}\right)$.
This means $z_{0} \in B_{q} \cap B_{k}$.
By induction, we know $z_{0} \in B_{(n, k)}$,

thus $\quad B_{n} \cap B_{k} \subseteq B_{(n, k)}$,
i.e., $B_{(n, k)}=B_{n} \cap B_{k}$.
We can also prove:
If $T$ denotes the set of all 1989-periodic points of $f$, and since $1989=3^{2} \times 13 \times 17$, we must have
$$
T=B_{1989} \backslash\left(B_{117} \cup B_{153} \cup B_{663}\right) .
$$

Indeed, it is easy to see that
$$
T \subseteq B_{1989} \backslash\left(B_{117} \cup B_{153} \cup B_{663}\right) .
$$

Therefore, we only need to prove the opposite inclusion.
For any $z_{0} \in B_{1989} \backslash\left(B_{117} \cup B_{153} \cup B_{663}\right)$, clearly, $z_{0}$ is a fixed point of $f^{(1989)}$, if $z_{0}$ is a $k$-periodic point of $f$, and
$$
k<1989 \text {. Then } \quad k \text { divides } 1989 \text {. }
$$

Thus $k$ must be a divisor of at least one of the numbers 117, 153, 663, so
$$
z_{0} \in B_{117} \cup B_{153} \cup B_{663} .
$$

This leads to a contradiction. Therefore, $k=1989$, i.e., $z_{0} \in T$. So
$$
T \supseteq B_{1989} \backslash\left(B_{117} \cup B_{153} \cup B_{663}\right),
$$

i.e.,
$$
T=B_{1989} \backslash\left(B_{117} \cup B_{153} \cup B_{663}\right) .
$$

Next, we calculate the number of elements in $B_{n}$.
Since
$$
f^{(n)}(z)=z,
$$

and
$$
f^{(n)}(z)=z^{m^{n}},
$$

we have
$$
z^{m^{n}}=z \text {, }
$$

i.e.,
$$
z^{m^{n}-1}=1 \text {. }
$$

Thus, $z$ is a root of unity of $z^{m^{n}-1}-1=0$, and $z \in B_{n}$, so the number of elements in $B_{n}$ is
$$
\left|B_{n}\right|=m^{n}-1 .
$$

Finally, we calculate the total number of 1989-periodic points of $f$.
By the principle of inclusion-exclusion, we get
$$
\begin{aligned}
& \left|B_{117} \cup B_{153} \cup B_{663}\right| \\
= & \left|B_{117}\right|+\left|B_{153}\right|+\left|B_{663}\right|-\left|B_{117} \cap B_{153}\right|-\left|B_{117} \cap B_{663}\right|- \\
& \left|B_{153} \cap B_{663}\right|+\left|B_{117} \cap B_{153} \cap B_{663}\right| \\
= & \left|B_{117}\right|+\left|B_{153}\right|+\left|B_{663}\right|-\left|B_{9}\right|-\left|B_{39}\right|-\left|B_{51}\right|+\left|B_{3}\right| \\
= & m^{117}-1+m^{153}-1+m^{663}-1-m^{9}+1-m^{39}+1-m^{51} \\
& +1+m^{3}-1 \\
= & m^{117}+m^{153}+m^{663}-m^{9}-m^{39}-m^{51}+m^{3}-1 .
\end{aligned}
$$

Also, $\quad\left|B_{1989}\right|=m^{1989}-1$.
Therefore, the total number of 1989-periodic points of $f$ is
$$
m^{1989}-m^{117}-m^{153}-m^{663}+m^{51}+m^{39}+m^{9}-m^{3} .
$$