Let `z_1, z_2` be two complex numbers represented by points on the circle `|z_1|= and and |z_2|=2` are then

Options

A. `max.|2z_(1)+z_(2)|=4`
B. `min.|z_(1)+z_(2)|=1`
C. `|z_(2)+(1)/(z_(1))|le3`
D. `|z_(1)+(2)/(z_(2))|le2`

Answer

To solve the problem step by step, we need to analyze the given complex numbers \( z_1 \) and \( z_2 \) which lie on circles defined by their moduli.

### Step 1: Understanding the Moduli
Given:
- \( |z_1| = 1 \) (This means \( z_1 \) lies on the unit circle)
- \( |z_2| = 2 \) (This means \( z_2 \) lies on a circle of radius 2)

### Step 2: Analyzing Maximum of \( |2z_1 + z_2| \)
To find the maximum value of \( |2z_1 + z_2| \):
- We can use the triangle inequality:
\[ |2z_1 + z_2| \leq |2z_1| + |z_2| = 2|z_1| + |z_2| = 2 \cdot 1 + 2 = 4 \]
- Thus, the maximum value of \( |2z_1 + z_2| \) is \( 4 \).

### Step 3: Analyzing Minimum of \( |z_2 - z_1| \)
To find the minimum value of \( |z_2 - z_1| \):
- The minimum distance occurs when \( z_2 \) and \( z_1 \) are aligned in the same direction:
\[ |z_2 - z_1| \geq ||z_2| - |z_1|| = |2 - 1| = 1 \]

### Step 4: Analyzing \( |z_2 + \frac{1}{z_1}| \)
To find \( |z_2 + \frac{1}{z_1}| \):
- Since \( |z_1| = 1 \), we have \( |\frac{1}{z_1}| = 1 \).
- Therefore:
\[ |z_2 + \frac{1}{z_1}| \leq |z_2| + |\frac{1}{z_1}| = 2 + 1 = 3 \]

### Step 5: Analyzing \( |z_1 + 2z_2| \)
To find \( |z_1 + 2z_2| \):
- Using the triangle inequality:
\[ |z_1 + 2z_2| \leq |z_1| + |2z_2| = 1 + 2 \cdot 2 = 5 \]
- However, we need to check if it is less than or equal to 2, which is not true.

### Conclusion
From the analysis:
1. Maximum \( |2z_1 + z_2| = 4 \) (True)
2. Minimum \( |z_2 - z_1| = 1 \) (True)
3. \( |z_2 + \frac{1}{z_1}| \leq 3 \) (True)
4. \( |z_1 + 2z_2| \leq 2 \) (False)

### Final Answers
- The correct options are 1, 2, and 3.