If `z_1`,`z_2` are nonreal complex and `|(z_1+z_2)/(z_1-z_2)|`=1 then `(z_1)/(z_2)` is

Options

A. real positive
B. purely imaginary
C. negative real
D. none of these

Answer

To solve the problem, we need to find the value of \( \frac{z_1}{z_2} \) given that \( | \frac{z_1 + z_2}{z_1 - z_2} | = 1 \).

### Step-by-Step Solution:

1. **Let \( \frac{z_1}{z_2} = w \)**:
We can express \( z_1 \) in terms of \( z_2 \):
\[ z_1 = w z_2 \]

2. **Substituting into the given condition**:
Substitute \( z_1 \) into the expression \( | \frac{z_1 + z_2}{z_1 - z_2} | = 1 \):
\[ | \frac{w z_2 + z_2}{w z_2 - z_2} | = 1 \]
Simplifying this, we get:
\[ | \frac{(w + 1) z_2}{(w - 1) z_2} | = 1 \]

3. **Cancelling \( z_2 \)**:
Since \( z_2 \) is non-zero (as it is a non-real complex number), we can cancel \( z_2 \) from the numerator and denominator:
\[ | \frac{w + 1}{w - 1} | = 1 \]

4. **Using the property of modulus**:
The condition \( | \frac{w + 1}{w - 1} | = 1 \) implies that:
\[ |w + 1| = |w - 1| \]

5. **Squaring both sides**:
Squaring both sides gives:
\[ (w + 1)(\overline{w + 1}) = (w - 1)(\overline{w - 1}) \]
Where \( \overline{w} \) is the complex conjugate of \( w \).

6. **Let \( w = x + iy \)**:
Substitute \( w = x + iy \):
\[ |(x + 1) + iy| = |(x - 1) + iy| \]

7. **Expanding the moduli**:
This gives:
\[ \sqrt{(x + 1)^2 + y^2} = \sqrt{(x - 1)^2 + y^2} \]

8. **Squaring both sides again**:
Squaring both sides results in:
\[ (x + 1)^2 + y^2 = (x - 1)^2 + y^2 \]

9. **Cancelling \( y^2 \)**:
The \( y^2 \) terms cancel out:
\[ (x + 1)^2 = (x - 1)^2 \]

10. **Expanding both sides**:
Expanding gives:
\[ x^2 + 2x + 1 = x^2 - 2x + 1 \]

11. **Simplifying**:
Cancelling \( x^2 + 1 \) from both sides leads to:
\[ 2x = -2x \]
Thus:
\[ 4x = 0 \implies x = 0 \]

12. **Conclusion**:
Since \( x = 0 \), we have:
\[ w = iy \]
This means that \( \frac{z_1}{z_2} \) is purely imaginary.

### Final Answer:
\[ \frac{z_1}{z_2} \text{ is purely imaginary.} \]