If `z_(1)`, `z_(2)` are two complex numbers such that `|(z_(1)-z_(2))/(z_(1)+z_(2))|=1` and `iz_(1)=Kz_(2)`, where ` K in R`, then the angle between `z_(1)-z_(2)` and `z_(1)+z_(2)` is

Options

A. `tan^(-1)((2K)/(K^(2)+1))`
B. `tan^(-1)((2K)/(1-K^(2)))`
C. `-2tan^(-1)K`
D. `2tan^(-1)K`

Explanation

To solve the problem, we need to find the angle between the complex numbers \( z_1 - z_2 \) and \( z_1 + z_2 \) given the conditions \( \left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1 \) and \( iz_1 = Kz_2 \), where \( K \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the 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. This means:
\[ |z_1 - z_2| = |z_1 + z_2| \]

2. **Using the Given Relation**:
From the relation \( iz_1 = Kz_2 \), we can express \( z_1 \) in terms of \( z_2 \):
\[ z_1 = -i \frac{K}{1} z_2 = -iKz_2 \]

3. **Substituting \( z_1 \)**:
Substitute \( z_1 \) into the modulus condition:
\[ |-iKz_2 - z_2| = |-iKz_2 + z_2| \]
Simplifying both sides:
\[ |(-iK - 1)z_2| = |(-iK + 1)z_2| \]
Since \( z_2 \neq 0 \), we can divide both sides by \( |z_2| \):
\[ |-iK - 1| = |-iK + 1| \]

4. **Calculating the Modulus**:
Calculate the moduli:
\[ |-iK - 1| = \sqrt{(-1)^2 + (-K)^2} = \sqrt{1 + K^2} \]
\[ |-iK + 1| = \sqrt{(1)^2 + (-K)^2} = \sqrt{1 + K^2} \]
Since both sides are equal, this condition holds.

5. **Finding the Angle**:
The angle \( \theta \) between two complex numbers \( a \) and \( b \) can be found using the formula:
\[ \tan \theta = \left| \frac{b}{a} \right| \]
Here, we need to find the angle between \( z_1 - z_2 \) and \( z_1 + z_2 \). We can express this angle as:
\[ \tan \alpha = \left| \frac{z_1 + z_2}{z_1 - z_2} \right| \]

6. **Using the Relation**:
From the earlier step, we have:
\[ z_1 = -iKz_2 \]
Thus:
\[ z_1 + z_2 = -iKz_2 + z_2 = (1 - iK)z_2 \]
\[ z_1 - z_2 = -iKz_2 - z_2 = (-1 - iK)z_2 \]

7. **Finding the Angle**:
The angle between \( z_1 - z_2 \) and \( z_1 + z_2 \) can be calculated as:
\[ \tan \alpha = \left| \frac{(1 - iK)}{(-1 - iK)} \right| \]
This leads to:
\[ \tan \alpha = \frac{1 + K^2}{1 + K^2} = 1 \]
Therefore, the angle \( \alpha \) is:
\[ \alpha = \frac{\pi}{4} \text{ or } 45^\circ \]

### Conclusion:
The angle between \( z_1 - z_2 \) and \( z_1 + z_2 \) is \( 45^\circ \).