If `(3 z_(1))/(5 z_(2))` is purely imaginary, then `|(2z_(1)-z_(2))/(2z_(1) + z_(2))|` is equal to _________

Answer

To solve the problem, we need to analyze the given condition that \(\frac{3z_1}{5z_2}\) is purely imaginary. This means that the real part of this expression must be zero. We will then find the value of \(\left|\frac{2z_1 - z_2}{2z_1 + z_2}\right|\).

### Step-by-Step Solution:

1. **Understanding Purely Imaginary Condition**:
Since \(\frac{3z_1}{5z_2}\) is purely imaginary, we can express this as:
\[ \frac{3z_1}{5z_2} = i \cdot k \quad \text{for some real number } k \]
This implies that the real part of \(\frac{3z_1}{5z_2}\) is zero.

2. **Setting Up the Ratio**:
Let \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) where \(r_1\) and \(r_2\) are the magnitudes and \(\theta_1\) and \(\theta_2\) are the arguments of \(z_1\) and \(z_2\) respectively. Then,
\[ \frac{3z_1}{5z_2} = \frac{3r_1 e^{i\theta_1}}{5r_2 e^{i\theta_2}} = \frac{3r_1}{5r_2} e^{i(\theta_1 - \theta_2)} \]
For this to be purely imaginary, the real part must be zero, which occurs when:
\[ \theta_1 - \theta_2 = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

3. **Finding the Modulus**:
We need to find:
\[ \left|\frac{2z_1 - z_2}{2z_1 + z_2}\right| \]
Substituting \(z_1\) and \(z_2\):
\[ \frac{2z_1 - z_2}{2z_1 + z_2} = \frac{2r_1 e^{i\theta_1} - r_2 e^{i\theta_2}}{2r_1 e^{i\theta_1} + r_2 e^{i\theta_2}} \]

4. **Using the Argument Condition**:
Since \(\theta_1 - \theta_2 = \frac{\pi}{2}\), we can write:
\[ e^{i\theta_1} = e^{i\theta_2} \cdot i \]
Thus, substituting this into our expression:
\[ \frac{2r_1 (i e^{i\theta_2}) - r_2 e^{i\theta_2}}{2r_1 (i e^{i\theta_2}) + r_2 e^{i\theta_2}} = \frac{(2r_1 i - r_2)e^{i\theta_2}}{(2r_1 i + r_2)e^{i\theta_2}} = \frac{2r_1 i - r_2}{2r_1 i + r_2} \]

5. **Calculating the Modulus**:
The modulus of a complex number \(\frac{a + bi}{c + di}\) is given by:
\[ \left|\frac{a + bi}{c + di}\right| = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \]
Here, let \(a = -r_2\), \(b = 2r_1\), \(c = r_2\), and \(d = 2r_1\):
\[ \left| \frac{2r_1 i - r_2}{2r_1 i + r_2} \right| = \frac{\sqrt{(-r_2)^2 + (2r_1)^2}}{\sqrt{(r_2)^2 + (2r_1)^2}} = \frac{\sqrt{r_2^2 + 4r_1^2}}{\sqrt{r_2^2 + 4r_1^2}} = 1 \]

### Final Answer:
Thus, the value of \(\left|\frac{2z_1 - z_2}{2z_1 + z_2}\right|\) is equal to **1**.