If `z_1 and z_2` are two distinct non-zero complex number such that `|z_1|= |z_2|`, then `(z_1+ z_2)/(z_1 - z_2)` is always

Options

A. purely real
B. purely imaginary
C. equal to zero
D. None of these

Answer

To solve the problem, we need to analyze the expression \(\frac{z_1 + z_2}{z_1 - z_2}\) given that \(z_1\) and \(z_2\) are distinct non-zero complex numbers with equal magnitudes, i.e., \(|z_1| = |z_2|\).

### Step-by-Step Solution:

1. **Express \(z_1\) and \(z_2\) in terms of their real and imaginary parts:**
Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\), where \(x_1, y_1, x_2, y_2\) are real numbers.

2. **Use the condition \(|z_1| = |z_2|\):**
The magnitudes of \(z_1\) and \(z_2\) can be expressed as:
\[ |z_1| = \sqrt{x_1^2 + y_1^2}, \quad |z_2| = \sqrt{x_2^2 + y_2^2} \]
Since \(|z_1| = |z_2|\), we have:
\[ x_1^2 + y_1^2 = x_2^2 + y_2^2 \]

3. **Calculate \(z_1 + z_2\) and \(z_1 - z_2\):**
\[ z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) \]
\[ z_1 - z_2 = (x_1 - x_2) + i(y_1 - y_2) \]

4. **Form the expression \(\frac{z_1 + z_2}{z_1 - z_2}\):**
\[ \frac{z_1 + z_2}{z_1 - z_2} = \frac{(x_1 + x_2) + i(y_1 + y_2)}{(x_1 - x_2) + i(y_1 - y_2)} \]

5. **Rationalize the denominator:**
Multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x_1 + x_2) + i(y_1 + y_2))((x_1 - x_2) - i(y_1 - y_2))}{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]

6. **Expand the numerator:**
Using the distributive property:
\[ = \frac{(x_1 + x_2)(x_1 - x_2) - i(y_1 + y_2)(y_1 - y_2) + i(y_1 + y_2)(x_1 - x_2) + (x_1 + x_2)(-i(y_1 - y_2))}{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]

7. **Combine real and imaginary parts:**
The real part will be:
\[ \text{Real part} = \frac{(x_1 + x_2)(x_1 - x_2) + (y_1 + y_2)(y_1 - y_2)}{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
The imaginary part will be:
\[ \text{Imaginary part} = \frac{(y_1 + y_2)(x_1 - x_2) - (x_1 + x_2)(y_1 - y_2)}{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]

8. **Show that the real part cancels out:**
Since \(x_1^2 + y_1^2 = x_2^2 + y_2^2\), the real part simplifies to zero, leaving only the imaginary part.

9. **Conclusion:**
Therefore, \(\frac{z_1 + z_2}{z_1 - z_2}\) is purely imaginary.

### Final Answer:
The expression \(\frac{z_1 + z_2}{z_1 - z_2}\) is always purely imaginary.