If `|z_1-1|<1`, `|z_2-2|<2`,`|z_3-3|<3` then `|z_1+z_2+z_3|`

Options

A. is less than 6
B. is more than 3
C. is less than 12
D. lies between 6 and 12

Explanation

To solve the problem, we need to analyze the given inequalities involving the complex numbers \( z_1, z_2, \) and \( z_3 \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We have three inequalities:
\[ |z_1 - 1| < 1, \quad |z_2 - 2| < 2, \quad |z_3 - 3| < 3 \]
These inequalities indicate that \( z_1 \) is within a distance of 1 from 1, \( z_2 \) is within a distance of 2 from 2, and \( z_3 \) is within a distance of 3 from 3.

2. **Identifying the Regions**:
- The inequality \( |z_1 - 1| < 1 \) describes a circle centered at \( 1 \) with a radius of \( 1 \). Thus, \( z_1 \) lies in the region:
\[ 0 < z_1 < 2 \]
- The inequality \( |z_2 - 2| < 2 \) describes a circle centered at \( 2 \) with a radius of \( 2 \). Thus, \( z_2 \) lies in the region:
\[ 0 < z_2 < 4 \]
- The inequality \( |z_3 - 3| < 3 \) describes a circle centered at \( 3 \) with a radius of \( 3 \). Thus, \( z_3 \) lies in the region:
\[ 0 < z_3 < 6 \]

3. **Combining the Inequalities**:
We need to find \( |z_1 + z_2 + z_3| \). To do this, we can express \( z_1, z_2, \) and \( z_3 \) in terms of their distances from their respective centers:
\[ z_1 = 1 + r_1, \quad z_2 = 2 + r_2, \quad z_3 = 3 + r_3 \]
where \( |r_1| < 1 \), \( |r_2| < 2 \), and \( |r_3| < 3 \).

4. **Finding the Maximum Value of \( z_1 + z_2 + z_3 \)**:
The sum can be expressed as:
\[ z_1 + z_2 + z_3 = (1 + r_1) + (2 + r_2) + (3 + r_3) = 6 + (r_1 + r_2 + r_3) \]
The maximum possible value of \( r_1 + r_2 + r_3 \) occurs when \( r_1 \) is at its maximum (1), \( r_2 \) at its maximum (2), and \( r_3 \) at its maximum (3):
\[ |r_1 + r_2 + r_3| < 1 + 2 + 3 = 6 \]
Therefore:
\[ |z_1 + z_2 + z_3| < |6 + (r_1 + r_2 + r_3)| < 6 + 6 = 12 \]

5. **Conclusion**:
Hence, we conclude that:
\[ |z_1 + z_2 + z_3| < 12 \]

### Final Answer:
The maximum value of \( |z_1 + z_2 + z_3| \) is less than 12.

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