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 start with the given condition and expression for \( z \).

### Step 1: Understand the given condition
We are given that \( 3 |z_1| = 2 |z_2| \). This can be rearranged to express the ratio of the moduli of \( z_1 \) and \( z_2 \):
\[ \frac{|z_1|}{|z_2|} = \frac{2}{3} \]

### Step 2: Express \( z \) in terms of \( z_1 \) and \( z_2 \)
The expression for \( z \) is given as:
\[ z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1} \]

### Step 3: Substitute the ratio of moduli
Let \( k = \frac{z_1}{z_2} \). Then we can express \( z \) in terms of \( k \):
\[ z = \frac{3k}{2} + \frac{2}{3k} \]

### Step 4: Find a common denominator
To combine the terms, we find a common denominator:
\[ z = \frac{3k \cdot 3k + 2 \cdot 2}{6k} = \frac{9k^2 + 4}{6k} \]

### Step 5: Separate real and imaginary parts
Assuming \( k \) is a complex number, we can write \( k = r e^{i\theta} \) where \( r = |k| \) and \( \theta \) is the argument of \( k \). Thus:
\[ z = \frac{9r^2 e^{2i\theta} + 4}{6r e^{i\theta}} \]

### Step 6: Rationalize the expression
To simplify further, we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(9r^2 e^{2i\theta} + 4) \cdot e^{-i\theta}}{6r} \]
This gives:
\[ z = \frac{9r^2 e^{i\theta} + 4 e^{-i\theta}}{6r} \]

### Step 7: Identify the real and imaginary parts
The real part of \( z \) is:
\[ \text{Re}(z) = \frac{9r^2 \cos(\theta) + 4 \cos(-\theta)}{6r} = \frac{9r^2 \cos(\theta) + 4 \cos(\theta)}{6r} = \frac{(9r^2 + 4) \cos(\theta)}{6r} \]
The imaginary part of \( z \) is:
\[ \text{Im}(z) = \frac{9r^2 \sin(\theta) - 4 \sin(-\theta)}{6r} = \frac{(9r^2 + 4) \sin(\theta)}{6r} \]

### Step 8: Set the imaginary part to zero
For the imaginary part to be zero:
\[ (9r^2 + 4) \sin(\theta) = 0 \]
This implies either \( \sin(\theta) = 0 \) or \( 9r^2 + 4 = 0 \) (which is impossible since \( r \) is non-zero).

### Conclusion
Thus, \( \sin(\theta) = 0 \), which means \( \theta = n\pi \) for \( n \in \mathbb{Z} \). Therefore, the imaginary part of \( z \) is indeed zero.

The final answer is that the imaginary part of \( z \) is zero.