(3) 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

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

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 $f$.
Solve: Let $f(z)=z^{m}$, by induction it is easy to know $f^{(n)}(z)=z^{m^{n}} (n=1,2,3, \cdots)$.

For natural number $n$, let $B_{n}=\left\{z \in S \mid f^{(n)}(z)=z\right\}$. Since $B_{n}$ is the set of all $(m^{n}-1)$-th roots of unity, we have $\left|B_{n}\right|=m^{n}-1 (|B|$ represents the number of elements in set $B)$.

For natural numbers $k$ and $z \in B_{k}$, it is easy to see that for natural number $r$, $f^{(k)}(z)=f^{(r-1) r)}\left(f^{(k)}(z)\right)=f^{(r-1) k)}(z)=\cdots=f^{(k)}(z)=z$. Therefore, when $k, n$ are natural numbers and $k \mid n$, $B_{k} \subset B_{n}$.

Furthermore, if $k, n$ are natural numbers, $n=\tau k+j (1 \leqslant j \leqslant k-1)$, and $z \in B_{k}$, we have
$$
f^{(n)}(z)=f^{(j)}\left(f^{(\tau)}(z)\right)=f^{(j)}(z)
$$

Thus, if $z \in B_{k} \cap B_{n}$, then by the above equation, $z \in B_{j}$. Using the Euclidean algorithm, we know $B_{k} \cap B_{n} \subset B_{(k, n)} ((k, n)$ represents the greatest common divisor of $k$ and $n)$.
Combining the above conclusions, we get
$$
B_{k} \cap B_{n}=B_{(k, n)}
$$

By definition, $\{f$'s 1989-periodic points $\}=B_{1989} \backslash\left(\bigcup_{k=1}^{1288} B_{k}\right)$. Therefore, it equals
$$
B_{1989} \backslash\left(\bigcup_{k=1}^{1988}\left(B_{1989} \cap B_{k}\right)\right)=B_{1989} \backslash\left(\bigcup_{B_{6}}\right)
$$
(here $t \in T=\{t \mid t$ divides $1989, t<1989\}$ )
Since $1989=3^{2} \cdot 221=3^{2} \cdot 13 \cdot 17$, the proper divisors of $1989$ must be at least one of the divisors 663 $(=3 \cdot 13 \cdot 17)$, 153 $(=3^{2} \cdot 17)$, 117 $(=3^{2} \cdot 13)$. Therefore,
$$
\{f \text{'s } 1989 \text{ -periodic points}\}=B_{1989} \backslash\left(B_{663} \cup B_{153} \cup B_{117}\right)
$$

By the principle of inclusion-exclusion,
$$
\begin{array}{l}
\left|B_{663} \cup B_{153} \cup B_{117}\right|= \\
\left|B_{663}\right|+\left|B_{153}\right|+\left|B_{117}\right|-\left|B_{663} \cap B_{153}\right|- \\
\left|B_{663} \cap B_{117}\right|-\left|B_{153} \cap B_{117}\right|+\left|B_{663} \cap B_{153} \cap B_{117}\right|= \\
\left|B_{663}\right|+\left|B_{153}\right|+\left|B_{117}\right|-\left|B_{51}\right|-\left|B_{39}\right|-\left|B_{9}\right|+\left|B_{3}\right|
\end{array}
$$

Thus, the total number of $f$'s 1989-periodic points is
$$
m^{1989}-m^{663}-m^{153}-m^{117}+m^{51}+m^{39}+m^{9}-m^{3}
$$

Note: In the solution of this problem, equation (1) is important. Furthermore, by the properties of the three divisors 663, 153, 117 of 1989 (1989 has three prime divisors 3, 13, 17, and these three numbers are the quotients of 1989 divided by these three prime divisors), we get equation (2). Some contestants derived a conclusion equivalent to (1) through discussions on roots of unity, and some even pointed out the properties of the three divisors of 1989, but very few wrote a conclusion equivalent to (2). Most directly wrote "By the principle of inclusion-exclusion, we get the total number of $f$'s 1989-periodic points is $\cdots \cdots$", but the results they calculated were different. The notation $B_{n}$ in this solution makes the use of the principle of inclusion-exclusion clear and explicit.