Let `z_(1),z_(2)` and `z_(3)` be three points on `|z|=1`. If `theta_(1), theta_(2)` and `theta_(3)` be the arguments of `z_(1),z_(2),z_(3)` respectively, then `cos(theta_(1)-theta_(2))+cos(theta_(2)-theta_(3))+cos(theta_(3)-theta_(1))`

Options

A. `ge-3/2`
B. `le -3/2`
C. `ge 3/2`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( \cos(\theta_1 - \theta_2) + \cos(\theta_2 - \theta_3) + \cos(\theta_3 - \theta_1) \) given that \( z_1, z_2, z_3 \) are points on the unit circle, meaning \( |z_1| = |z_2| = |z_3| = 1 \).

### Step-by-Step Solution:

1. **Understanding the Points on the Unit Circle**:
Since \( z_1, z_2, z_3 \) lie on the unit circle, we can express them in terms of their arguments:
\[ z_1 = e^{i\theta_1}, \quad z_2 = e^{i\theta_2}, \quad z_3 = e^{i\theta_3} \]

2. **Using the Cosine Difference Identity**:
We can use the cosine difference identity:
\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]
Therefore, we can rewrite the expression:
\[ \cos(\theta_1 - \theta_2) = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \]
\[ \cos(\theta_2 - \theta_3) = \cos \theta_2 \cos \theta_3 + \sin \theta_2 \sin \theta_3 \]
\[ \cos(\theta_3 - \theta_1) = \cos \theta_3 \cos \theta_1 + \sin \theta_3 \sin \theta_1 \]

3. **Combining the Cosine Terms**:
Now, we can combine these:
\[ \cos(\theta_1 - \theta_2) + \cos(\theta_2 - \theta_3) + \cos(\theta_3 - \theta_1 = (\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) + (\cos \theta_2 \cos \theta_3 + \sin \theta_2 \sin \theta_3) + (\cos \theta_3 \cos \theta_1 + \sin \theta_3 \sin \theta_1) \]

4. **Using the Properties of Complex Numbers**:
Since \( |z_1| = |z_2| = |z_3| = 1 \), we know:
\[ z_1 + z_2 + z_3 = e^{i\theta_1} + e^{i\theta_2} + e^{i\theta_3} \]
The modulus of this sum is bounded. By the triangle inequality, we have:
\[ |z_1 + z_2 + z_3| \leq |z_1| + |z_2| + |z_3| = 3 \]
The minimum value occurs when the points are evenly spaced on the circle.

5. **Applying the Cauchy-Schwarz Inequality**:
We can apply the Cauchy-Schwarz inequality to the terms:
\[ (1^2 + 1^2 + 1^2)(\cos^2(\theta_1 - \theta_2) + \cos^2(\theta_2 - \theta_3) + \cos^2(\theta_3 - \theta_1) \geq (\cos(\theta_1 - \theta_2) + \cos(\theta_2 - \theta_3) + \cos(\theta_3 - \theta_1))^2 \]
This gives us a lower bound.

6. **Final Result**:
After evaluating, we find:
\[ \cos(\theta_1 - \theta_2) + \cos(\theta_2 - \theta_3) + \cos(\theta_3 - \theta_1) \geq -\frac{3}{2} \]
Thus, the final answer is:
\[ \cos(\theta_1 - \theta_2) + \cos(\theta_2 - \theta_3) + \cos(\theta_3 - \theta_1) \geq -\frac{3}{2} \]