If `( 5 z_(2))/( 7 z_(1))` is purely imaginary number, then `|( 2 z_(1) + 3 z_(2))/( 2 z_(1) - 3 z_(2))|` is equal to

Options

A. `5//7`
B. `7//5`
C. `25//49`
D. none of these

Answer

To solve the problem, we start with the given condition that \(\frac{5 z_2}{7 z_1}\) is a purely imaginary number. This implies that the real part of \(\frac{5 z_2}{7 z_1}\) must be zero.

### Step 1: Set up the equation
Let \( z_1 = x + iy \) and \( z_2 = u + iv \), where \( x, y, u, v \) are real numbers. Then we have:
\[ \frac{5 z_2}{7 z_1} = \frac{5 (u + iv)}{7 (x + iy)} = \frac{5u + 5iv}{7x + 7iy} \]
To express this in a standard form, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{5 (u + iv)(7x - 7iy)}{(7x + 7iy)(7x - 7iy)} = \frac{5 (7ux + 7vy + i(5v7x - 5u7y))}{49(x^2 + y^2)} \]
This simplifies to:
\[ \frac{5(7ux + 7vy)}{49(x^2 + y^2)} + i \frac{5(7vy - 7ux)}{49(x^2 + y^2)} \]
For this to be purely imaginary, the real part must be zero:
\[ 7ux + 7vy = 0 \implies ux + vy = 0 \tag{1} \]

### Step 2: Find the modulus
Next, we need to find the modulus of \(\frac{2 z_1 + 3 z_2}{2 z_1 - 3 z_2}\):
\[ \frac{2 z_1 + 3 z_2}{2 z_1 - 3 z_2} \]
Substituting \(z_1\) and \(z_2\):
\[ = \frac{2(x + iy) + 3(u + iv)}{2(x + iy) - 3(u + iv)} = \frac{(2x + 3u) + i(2y + 3v)}{(2x - 3u) + i(2y - 3v)} \]

### Step 3: Calculate the modulus
The modulus of a complex number \(\frac{a + ib}{c + id}\) is given by:
\[ \left| \frac{a + ib}{c + id} \right| = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \]
Thus, we have:
\[ \left| \frac{2 z_1 + 3 z_2}{2 z_1 - 3 z_2} \right| = \frac{\sqrt{(2x + 3u)^2 + (2y + 3v)^2}}{\sqrt{(2x - 3u)^2 + (2y - 3v)^2}} \]

### Step 4: Use the relation from Step 1
From equation (1), we know \(ux + vy = 0\). This implies that \(u = -\frac{vy}{x}\) (assuming \(x \neq 0\)). We can substitute this into our modulus expression.

### Final Calculation
After substituting and simplifying, we find that the modulus simplifies to a constant value. After careful calculations, we find that:
\[ \left| \frac{2 z_1 + 3 z_2}{2 z_1 - 3 z_2} \right| = 1 \]

### Conclusion
Thus, the final answer is:
\[ \left| \frac{2 z_1 + 3 z_2}{2 z_1 - 3 z_2} \right| = 1 \]