How many complex numbers $z$ such that $\left| z \right| < 30$ satisfy the equation
\[
  e^z = \frac{z - 1}{z + 1} \, ?
\]
1. **Substitute \( z = m + ni \) into the equation:**

   Given the equation:
   \[
   e^z = \frac{z - 1}{z + 1}
   \]
   Substitute \( z = m + ni \):
   \[
   e^{m+ni} = \frac{(m-1) + ni}{(m+1) + ni}
   \]

2. **Separate the exponential term:**

   Using Euler's formula \( e^{m+ni} = e^m \cdot e^{ni} \):
   \[
   e^m \cdot (\cos n + i \sin n) = \frac{(m-1) + ni}{(m+1) + ni}
   \]

3. **Compare magnitudes:**

   The magnitude of the left-hand side is:
   \[
   |e^m \cdot (\cos n + i \sin n)| = e^m
   \]
   The magnitude of the right-hand side is:
   \[
   \left| \frac{(m-1) + ni}{(m+1) + ni} \right| = \frac{\sqrt{(m-1)^2 + n^2}}{\sqrt{(m+1)^2 + n^2}}
   \]
   Equating the magnitudes:
   \[
   e^m = \frac{\sqrt{(m-1)^2 + n^2}}{\sqrt{(m+1)^2 + n^2}}
   \]

4. **Simplify the magnitude equation:**

   Squaring both sides:
   \[
   e^{2m} = \frac{(m-1)^2 + n^2}{(m+1)^2 + n^2}
   \]

5. **Analyze the equation for \( m \):**

   - If \( m > 0 \), then \( e^{2m} > 1 \), but \(\frac{(m-1)^2 + n^2}{(m+1)^2 + n^2} < 1\).
   - If \( m < 0 \), then \( e^{2m} < 1 \), but \(\frac{(m-1)^2 + n^2}{(m+1)^2 + n^2} > 1\).

   Therefore, \( m = 0 \).

6. **Substitute \( m = 0 \):**

   The equation simplifies to:
   \[
   e^{ni} = \frac{-1 + ni}{1 + ni}
   \]
   Using Euler's formula \( e^{ni} = \cos n + i \sin n \):
   \[
   \cos n + i \sin n = \frac{1 - n^2}{1 + n^2} + i \frac{2n}{1 + n^2}
   \]

7. **Equate real and imaginary parts:**

   - Real part:
     \[
     \cos n = \frac{1 - n^2}{1 + n^2}
     \]
   - Imaginary part:
     \[
     \sin n = \frac{2n}{1 + n^2}
     \]

8. **Find intersections of the functions:**

   The graphs of \( y = \sin x \) and \( y = \frac{2x}{1 + x^2} \) intersect at 10 points when \( |x| < 30 \).

The final answer is \(\boxed{10}\)