Let `z_(1)` and `z_(2)` be any two non-zero complex numbers such that `3|z_(1)|=2|z_(2)|. "If "z=(3z_(1))/(2z_(2)) + (2z_(2))/(3z_(1))`, then

Options

A. `|z| = (1)/(2) sqrt((17)/(2))`
B. Re(z) = 0
C. `|z| = sqrt((5)/(2))`
D. Im(z) = 0

Answer

To solve the problem, we need to analyze the expression given for \( z \) and the condition involving the complex numbers \( z_1 \) and \( z_2 \).

### Step-by-Step Solution:

1. **Given Condition**:
We have the condition \( 3|z_1| = 2|z_2| \). This can be rewritten as:
\[ \frac{|z_1|}{|z_2|} = \frac{2}{3} \]

2. **Expression for \( z \)**:
We need to evaluate:
\[ z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1} \]

3. **Let \( a = \frac{3z_1}{2z_2} \)**:
From the condition, we can find the modulus of \( a \):
\[ |a| = \left| \frac{3z_1}{2z_2} \right| = \frac{3|z_1|}{2|z_2|} = \frac{3 \cdot \frac{2}{3}|z_2|}{2|z_2|} = 1 \]
Thus, \( |a| = 1 \).

4. **Finding \( \frac{2z_2}{3z_1} \)**:
We can express \( \frac{2z_2}{3z_1} \) in terms of \( a \):
\[ \frac{2z_2}{3z_1} = \frac{1}{a} \]

5. **Substituting back into \( z \)**:
Now substituting \( a \) and \( \frac{1}{a} \) into the expression for \( z \):
\[ z = a + \frac{1}{a} \]

6. **Using Polar Form**:
Since \( |a| = 1 \), we can express \( a \) in polar form:
\[ a = e^{i\theta} = \cos\theta + i\sin\theta \]
Therefore,
\[ \frac{1}{a} = e^{-i\theta} = \cos\theta - i\sin\theta \]

7. **Adding the two expressions**:
Now, we compute:
\[ z = (\cos\theta + i\sin\theta) + (\cos\theta - i\sin\theta) = 2\cos\theta \]
The imaginary parts cancel out, leaving us with a real number.

8. **Conclusion**:
Since \( z \) is a real number, the imaginary part of \( z \) is zero. Therefore, we conclude that:
\[ \text{Imaginary part of } z = 0 \]

### Final Answer:
The imaginary part of \( z \) is zero.