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The problem asks you to determine the graph of the open set in the complex plane \(\mathbb{C}\) that satisfies the inequality:

\[
\left|\frac{z - 3}{z + 3}\right| < 2
\]

Here, \(z\) is a complex number.

### Step-by-Step Solution:

1. **Express \(z\) as a Complex Number**:  
   Let \(z = x + iy\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit.

2. **Substitute \(z\) in the Inequality**:  
   Substitute \(z = x + iy\) into the expression:

   \[
   \frac{z - 3}{z + 3} = \frac{(x - 3) + iy}{(x + 3) + iy}
   \]

3. **Simplify the Expression**:  
   The expression can be simplified by multiplying the numerator and the denominator by the conjugate of the denominator:

   \[
   \frac{(x - 3) + iy}{(x + 3) + iy} \cdot \frac{(x + 3) - iy}{(x + 3) - iy} = \frac{((x - 3) + iy)((x + 3) - iy)}{(x + 3)^2 + y^2}
   \]

   Simplify the numerator using the distributive property:

   \[
   (x - 3)(x + 3) - iy(x + 3) + iy(x - 3) - i^2y^2
   \]

   Since \(i^2 = -1\), the expression becomes:

   \[
   \frac{x^2 - 9 - y^2 + 2iy(x - 3)}{(x + 3)^2 + y^2}
   \]

   The modulus of this complex fraction is:

   \[
   \left|\frac{(x^2 - 9 - y^2) + 2iy(x - 3)}{(x + 3)^2 + y^2}\right|
   \]

   Simplifying further:

   \[
   \frac{\sqrt{(x^2 - 9 - y^2)^2 + (2y(x - 3))^2}}{(x + 3)^2 + y^2}
   \]

4. **Simplifying the Inequality**:  
   We need to determine when:

   \[
   \frac{\sqrt{(x^2 - 9 - y^2)^2 + 4y^2(x - 3)^2}}{(x + 3)^2 + y^2} < 2
   \]

   Instead of further expanding this directly, consider instead using a known geometric property of these types of inequalities in the complex plane.

5. **Geometric Interpretation**:  
   The inequality of the form \(\left|\frac{z - a}{z - b}\right| < k\) represents an open set in the complex plane, which is a disk (or circular region).

   - The open set described by this inequality represents a disk centered at a point on the complex plane.
   - Specifically, \(\frac{z - 3}{z + 3}\) implies a transformation where the center of the disk lies between \(z = 3\) and \(z = -3\).

   To find the exact center and radius:
   
   - **Center of the Disk**: This lies on the line joining the points \(3\) and \(-3\).
   - **Radius**: Derived from the inequality.

   The inequality tells us that the center of the disk lies at a point on the real line between \(z = 3\) and \(z = -3\), and the radius is determined based on the modulus and the constant 2.

### Conclusion:
The solution set to the inequality is a disk centered on the real axis (along the line segment connecting 3 and -3) in the complex plane, with a specific radius derived from the constant in the inequality. The specific center and radius can be calculated directly from the inequality.

Would you like to dive deeper into the geometric reasoning, calculate the exact center and radius of the disk, or have any other questions? Here are some related questions that might help:

1. How do you compute the exact center of a disk in a complex inequality?
2. What is the geometric meaning of \(\left|\frac{z - a}{z - b}\right| = k\) in the complex plane?
3. How does the radius of a disk relate to the constant in a complex inequality?
4. How do transformations in the complex plane affect the position of geometric figures?
5. What role do conjugates play in simplifying complex inequalities?

**Tip**: Always remember that complex inequalities often translate into geometric shapes, such as disks or half-planes, which makes visualization a powerful tool for solving these problems.

Explore the graph of the open set in the complex plane that satisfies the modulus inequality |(z - 3)/(z + 3)| < 2. Understand complex numbers, inequalities, and geometric transformations involved in solving this problem.

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