If `z_1,z_2` are complex numbers such that, `(2z_1)/(3z_2)` is purely imaginary number then find `|(z_1-z_2)/((z_1+z_2))|`.

Answer

To solve the problem, we start with the given condition that \(\frac{2z_1}{3z_2}\) is a purely imaginary number.

### Step 1: Set up the equation
Since \(\frac{2z_1}{3z_2}\) is purely imaginary, we can express this condition mathematically:
\[ \frac{2z_1}{3z_2} = -\frac{2z_1}{3z_2}^* \]
where \(z^*\) denotes the complex conjugate of \(z\).

### Step 2: Cross-multiply
Cross-multiplying gives:
\[ 2z_1 = -2z_1^* \frac{z_2}{3} \]
This simplifies to:
\[ 2z_1 z_2^* + 2z_1^* z_2 = 0 \]
Dividing by 2, we get:
\[ z_1 z_2^* + z_1^* z_2 = 0 \]
This is our equation (1).

### Step 3: Define the expression to find
We need to find:
\[ \left| \frac{z_1 - z_2}{z_1 + z_2} \right| \]
Let this expression be \(x\):
\[ x = \frac{z_1 - z_2}{z_1 + z_2} \]

### Step 4: Square the expression
Squaring both sides gives:
\[ x^2 = \left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 = \frac{|z_1 - z_2|^2}{|z_1 + z_2|^2} \]

### Step 5: Calculate the numerator
Using the property of modulus:
\[ |z_1 - z_2|^2 = (z_1 - z_2)(z_1 - z_2)^* = (z_1 - z_2)(z_1^* - z_2^*) \]
Expanding this gives:
\[ |z_1|^2 - z_1 z_2^* - z_1^* z_2 + |z_2|^2 \]

### Step 6: Calculate the denominator
Similarly,
\[ |z_1 + z_2|^2 = (z_1 + z_2)(z_1 + z_2)^* = (z_1 + z_2)(z_1^* + z_2^*) \]
Expanding this gives:
\[ |z_1|^2 + z_1 z_2^* + z_1^* z_2 + |z_2|^2 \]

### Step 7: Substitute into the expression for \(x^2\)
Now substituting the numerator and denominator into the expression for \(x^2\):
\[ x^2 = \frac{|z_1|^2 - z_1 z_2^* - z_1^* z_2 + |z_2|^2}{|z_1|^2 + z_1 z_2^* + z_1^* z_2 + |z_2|^2} \]

### Step 8: Use equation (1)
From equation (1), we know that:
\[ z_1 z_2^* + z_1^* z_2 = 0 \]
Thus, we can substitute \(z_1 z_2^* + z_1^* z_2 = 0\) into our expression:
\[ x^2 = \frac{|z_1|^2 + |z_2|^2}{|z_1|^2 + |z_2|^2} = 1 \]

### Step 9: Conclusion
Taking the square root gives:
\[ |x| = 1 \]
Thus, the final result is:
\[ \left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1 \]