Let `z_(1)` and `z_(2)` be two non -zero complex number such that `|z_(1)+z_(2)| = |z_(1) | = |z_(2)|` . Then `(z_(1))/(z_(2))` can be equal to (`omega` is imaginary cube root of unity).

Options

A. `1 + omega`
B. `1+ omega^(2)`
C. `omega`
D. `omega^(2)`

Explanation

To solve the problem, we start with the given conditions for the complex numbers \( z_1 \) and \( z_2 \):

1. **Given Condition**:
\[ |z_1 + z_2| = |z_1| = |z_2| \]

2. **Let** \( |z_1| = |z_2| = r \). Thus, we can express \( z_1 \) and \( z_2 \) in polar form:
\[ z_1 = r e^{i\theta_1}, \quad z_2 = r e^{i\theta_2} \]

3. **Using the condition** \( |z_1 + z_2| = |z_1| \):
\[ |z_1 + z_2| = |r e^{i\theta_1} + r e^{i\theta_2}| = r |e^{i\theta_1} + e^{i\theta_2}| \]
This implies:
\[ |e^{i\theta_1} + e^{i\theta_2}| = 1 \]

4. **Using the cosine rule**:
The magnitude \( |e^{i\theta_1} + e^{i\theta_2}| \) can be expressed as:
\[ |e^{i\theta_1} + e^{i\theta_2}| = \sqrt{2 + 2\cos(\theta_2 - \theta_1)} = 1 \]
Squaring both sides gives:
\[ 2 + 2\cos(\theta_2 - \theta_1) = 1 \]
Simplifying this results in:
\[ 2\cos(\theta_2 - \theta_1) = -1 \quad \Rightarrow \quad \cos(\theta_2 - \theta_1) = -\frac{1}{2} \]

5. **Finding the angle**:
The angle \( \theta_2 - \theta_1 \) that satisfies \( \cos(\theta_2 - \theta_1) = -\frac{1}{2} \) is:
\[ \theta_2 - \theta_1 = \pm \frac{2\pi}{3} \quad \text{(120 degrees or -120 degrees)} \]

6. **Expressing the ratio**:
The ratio \( \frac{z_1}{z_2} \) can be calculated:
\[ \frac{z_1}{z_2} = \frac{r e^{i\theta_1}}{r e^{i\theta_2}} = e^{i(\theta_1 - \theta_2)} = e^{-i(\theta_2 - \theta_1)} \]
Substituting \( \theta_2 - \theta_1 = \frac{2\pi}{3} \) or \( -\frac{2\pi}{3} \):
\[ \frac{z_1}{z_2} = e^{-i\frac{2\pi}{3}} \quad \text{or} \quad e^{i\frac{2\pi}{3}} \]

7. **Identifying the roots of unity**:
The values \( e^{-i\frac{2\pi}{3}} \) and \( e^{i\frac{2\pi}{3}} \) correspond to the cube roots of unity:
\[ \omega = e^{i\frac{2\pi}{3}}, \quad \omega^2 = e^{-i\frac{2\pi}{3}} \]

Thus, we conclude that:
\[ \frac{z_1}{z_2} = \omega \quad \text{or} \quad \frac{z_1}{z_2} = \omega^2 \]