Let $S$ be the unit circle in the complex plane (i.e. the set of all complex numbers with their moduli equal to $1$). 
We define function $f:S\rightarrow S$ as follow: $\forall z\in S$,
$ f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \dots,$ 
$f^{(k)}(z)=f(f^{(k-1)}(z)) (k>1,k\in \mathbb{N}), \dots$
We call $c$ an $n$-[i]period-point[/i] of $f$ if $c$ ($c\in S$) and $n$ ($n\in\mathbb{N}$) satisfy:
$f^{(1)}(c) \not=c, f^{(2)}(c) \not=c, f^{(3)}(c) \not=c, \dots, f^{(n-1)}(c) \not=c, f^{(n)}(c)=c$.
Suppose that $f(z)=z^m$ ($z\in S; m>1, m\in \mathbb{N}$), find the number of $1989$-[i]period-point[/i] of $f$.
1. **Define the function and the unit circle:**
   Let \( S \) be the unit circle in the complex plane, i.e., the set of all complex numbers with modulus equal to 1. We define the function \( f: S \rightarrow S \) as \( f(z) = z^m \) for \( z \in S \) and \( m > 1 \), \( m \in \mathbb{N} \).

2. **Define the \( n \)-period-point:**
   A complex number \( c \in S \) is an \( n \)-period-point of \( f \) if:
   \[
   f^{(1)}(c) \neq c, \quad f^{(2)}(c) \neq c, \quad \dots, \quad f^{(n-1)}(c) \neq c, \quad f^{(n)}(c) = c
   \]
   where \( f^{(k)}(z) = f(f^{(k-1)}(z)) \) and \( f^{(1)}(z) = f(z) \).

3. **Express the function iteratively:**
   Given \( f(z) = z^m \), we have:
   \[
   f^{(k)}(z) = z^{m^k}
   \]
   Therefore, for \( z \) to be a 1989-period-point, we need:
   \[
   f^{(1989)}(z) = z^{m^{1989}} = z
   \]

4. **Solve the equation \( z^{m^{1989}} = z \):**
   Let \( z = e^{i\theta} \). Then:
   \[
   e^{i\theta m^{1989}} = e^{i\theta}
   \]
   This implies:
   \[
   i\theta m^{1989} = i\theta + 2\pi k \quad \text{for some integer } k
   \]
   Simplifying, we get:
   \[
   \theta (m^{1989} - 1) = 2\pi k
   \]
   \[
   \theta = \frac{2\pi k}{m^{1989} - 1}
   \]

5. **Determine the distinct values of \( \theta \):**
   Since \( z \) must be on the unit circle, \( \theta \) must be a multiple of \( \frac{2\pi}{m^{1989} - 1} \). Therefore, \( z \) can take the values:
   \[
   z = e^{i\frac{2\pi k}{m^{1989} - 1}} \quad \text{for } k = 0, 1, 2, \dots, m^{1989} - 2
   \]

6. **Count the valid \( 1989 \)-period-points:**
   For \( z \) to be a 1989-period-point, \( z \) must not satisfy \( f^{(k)}(z) = z \) for any \( k < 1989 \). This means \( z \) must not be a root of unity of any order dividing \( m^{1989} - 1 \) other than 1989. The number of such valid \( z \) is given by Euler's totient function \( \phi(m^{1989} - 1) \).

The final answer is \( \boxed{ \phi(m^{1989} - 1) } \).