If the real part of the complex number ` z = ( 3 + 2 i cos theta)/( 1 - 3 i cos theta) , theta in (0, (pi)/(2))` is zero, then the value of ` sin^(2) 3 theta + cos ^(2) theta ` is equal to _____

Answer

To solve the problem, we need to find the value of \( \sin^2(3\theta) + \cos^2(\theta) \) given that the real part of the complex number \( z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta} \) is zero.

### Step-by-Step Solution:

1. **Identify the Complex Number**:
\[ z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta} \]

2. **Rationalize the Denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(3 + 2i \cos \theta)(1 + 3i \cos \theta)}{(1 - 3i \cos \theta)(1 + 3i \cos \theta)} \]

3. **Calculate the Denominator**:
\[ (1 - 3i \cos \theta)(1 + 3i \cos \theta) = 1^2 - (3i \cos \theta)^2 = 1 + 9 \cos^2 \theta \]

4. **Calculate the Numerator**:
\[ (3 + 2i \cos \theta)(1 + 3i \cos \theta) = 3 + 9i \cos \theta + 2i \cos \theta - 6 \cos^2 \theta = 3 - 6 \cos^2 \theta + 11i \cos \theta \]

5. **Combine Results**:
Thus,
\[ z = \frac{3 - 6 \cos^2 \theta + 11i \cos \theta}{1 + 9 \cos^2 \theta} \]

6. **Separate Real and Imaginary Parts**:
The real part of \( z \) is:
\[ \text{Re}(z) = \frac{3 - 6 \cos^2 \theta}{1 + 9 \cos^2 \theta} \]
The imaginary part is:
\[ \text{Im}(z) = \frac{11 \cos \theta}{1 + 9 \cos^2 \theta} \]

7. **Set the Real Part to Zero**:
For the real part to be zero:
\[ 3 - 6 \cos^2 \theta = 0 \]

8. **Solve for \( \cos^2 \theta \)**:
Rearranging gives:
\[ 6 \cos^2 \theta = 3 \implies \cos^2 \theta = \frac{1}{2} \]
Thus,
\[ \cos \theta = \frac{1}{\sqrt{2}} \quad (\text{since } \theta \in (0, \frac{\pi}{2})) \]

9. **Determine \( \theta \)**:
From \( \cos \theta = \frac{1}{\sqrt{2}} \), we find:
\[ \theta = \frac{\pi}{4} \]

10. **Calculate \( \sin^2(3\theta) + \cos^2(\theta) \)**:
First, compute \( 3\theta \):
\[ 3\theta = 3 \times \frac{\pi}{4} = \frac{3\pi}{4} \]
Now calculate:
\[ \sin^2(3\theta) = \sin^2\left(\frac{3\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
And:
\[ \cos^2(\theta) = \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

11. **Combine Results**:
\[ \sin^2(3\theta) + \cos^2(\theta) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer:
The value of \( \sin^2(3\theta) + \cos^2(\theta) \) is \( \boxed{1} \).