If `z_(1)` and `z_(2)` are two complex numbers such that `|(z_(1)-z_(2))/(z_(1)+z_(2))|=1`, then

Options

A. `z_(1)=kz_(2), k in R`
B. `z_(1)=ikz_(2), k in R`
C. `z_(1)=z_(2)`
D. none of these

Answer

To solve the problem, we start with the given condition:

\[ \left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1 \]

### Step 1: Analyze the modulus condition
The condition \(\left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1\) implies that the modulus of the numerator is equal to the modulus of the denominator. Therefore, we can write:

\[ |z_1 - z_2| = |z_1 + z_2| \]

### Step 2: Square both sides
To eliminate the modulus, we square both sides:

\[ |z_1 - z_2|^2 = |z_1 + z_2|^2 \]

### Step 3: Expand both sides
Using the property of modulus for complex numbers, we expand both sides:

\[ (z_1 - z_2)(\overline{z_1 - z_2}) = (z_1 + z_2)(\overline{z_1 + z_2}) \]

This gives us:

\[ (z_1 - z_2)(\overline{z_1} - \overline{z_2}) = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) \]

Expanding both sides, we have:

\[ |z_1|^2 - z_1 \overline{z_2} - z_2 \overline{z_1} + |z_2|^2 = |z_1|^2 + z_1 \overline{z_2} + z_2 \overline{z_1} + |z_2|^2 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ - z_1 \overline{z_2} - z_2 \overline{z_1} = z_1 \overline{z_2} + z_2 \overline{z_1} \]

Combining like terms results in:

\[ -2(z_1 \overline{z_2} + z_2 \overline{z_1}) = 0 \]

### Step 5: Conclusion
This implies:

\[ z_1 \overline{z_2} + z_2 \overline{z_1} = 0 \]

This can be interpreted as:

\[ z_1 = k i z_2 \]

where \(k\) is a real number. Thus, we conclude that \(z_1\) is directly proportional to \(i z_2\).

### Final Answer
\[ z_1 = k i z_2 \quad \text{(where \(k\) is a real number)} \]

---