For `x_(1), x_(2), y_(1), y_(2) in R ` if `0 lt x_(1)lt x_(2)lt y_(1) = y_(2)` and `z_(1) = x_(1) + i y_(1), z_(2) = x_(2)+ iy_(2)` and `z_(3) = (z_(1) + z_(2))//2,`then ` z_(1) , z_(2) , z_(3)` satisfy :

Options

A. `abs(z_(1)) = abs( z_(2) ) = abs(z_(3))`
B. `abs(z_(1)) lt abs( z_(2) ) lt abs(z_(3))`
C. `abs(z_(1)) gt abs( z_(2) ) gt abs(z_(3))`
D. `abs(z_(1)) lt abs( z_(3) ) lt abs(z_(2))`

Answer

To solve the problem, we will analyze the given complex numbers \( z_1, z_2, z_3 \) and their moduli step by step.

### Step 1: Define the Complex Numbers
We have:
- \( z_1 = x_1 + i y_1 \)
- \( z_2 = x_2 + i y_2 \)
- \( z_3 = \frac{z_1 + z_2}{2} = \frac{x_1 + x_2}{2} + i \frac{y_1 + y_2}{2} \)

### Step 2: Calculate the Moduli
The modulus of a complex number \( z = a + ib \) is given by \( |z| = \sqrt{a^2 + b^2} \).

1. **For \( z_1 \)**:
\[ |z_1| = \sqrt{x_1^2 + y_1^2} \]

2. **For \( z_2 \)**:
\[ |z_2| = \sqrt{x_2^2 + y_2^2} \]

3. **For \( z_3 \)**:
Since \( y_1 = y_2 \), we have:
\[ |z_3| = \sqrt{\left(\frac{x_1 + x_2}{2}\right)^2 + \left(\frac{y_1 + y_2}{2}\right)^2} = \sqrt{\left(\frac{x_1 + x_2}{2}\right)^2 + \left(\frac{y_1 + y_1}{2}\right)^2} \]
Simplifying this gives:
\[ |z_3| = \sqrt{\frac{(x_1 + x_2)^2}{4} + \frac{(2y_1)^2}{4}} = \frac{1}{2} \sqrt{(x_1 + x_2)^2 + 4y_1^2} \]

### Step 3: Compare the Moduli
Given that \( 0 < x_1 < x_2 < y_1 = y_2 \), we can analyze the inequalities:

1. **Compare \( |z_1| \) and \( |z_2| \)**:
Since \( x_1 < x_2 \) and \( y_1 = y_2 \):
\[ |z_1|^2 = x_1^2 + y_1^2 < x_2^2 + y_1^2 = |z_2|^2 \]
Thus, \( |z_1| < |z_2| \).

2. **Compare \( |z_3| \) with \( |z_1| \) and \( |z_2| \)**:
We need to show that \( |z_1| < |z_3| < |z_2| \).

- To show \( |z_3| > |z_1| \):
\[ |z_3| = \frac{1}{2} \sqrt{(x_1 + x_2)^2 + 4y_1^2} \]
Since \( x_1 + x_2 > 2x_1 \) (because \( x_2 > x_1 \)), we have:
\[ |z_3|^2 = \frac{(x_1 + x_2)^2 + 4y_1^2}{4} > |z_1|^2 \]

- To show \( |z_3| < |z_2| \):
\[ |z_3|^2 = \frac{(x_1 + x_2)^2 + 4y_1^2}{4} < |z_2|^2 \]
Since \( x_1 + x_2 < 2x_2 \) (because \( x_1 < x_2 \)), we can conclude that:
\[ |z_3|^2 < |z_2|^2 \]

### Conclusion
From the inequalities derived, we conclude that:
\[ |z_1| < |z_3| < |z_2| \]