If `abs(z+4) le 3`, the maximum value of `abs(z+1)` is

Options

A. 4
B. 10
C. 6
D. 0

Answer

To solve the problem, we need to find the maximum value of \( |z + 1| \) given that \( |z + 4| \leq 3 \).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We start with the inequality \( |z + 4| \leq 3 \). This means that the complex number \( z + 4 \) lies within or on the boundary of a circle centered at -4 with a radius of 3 in the complex plane.

2. **Expressing the Inequality**:
The inequality \( |z + 4| \leq 3 \) can be interpreted as:
\[ -3 \leq z + 4 \leq 3 \]
This means that \( z + 4 \) can take values between -3 and 3.

3. **Rearranging the Inequality**:
To find the bounds for \( z \), we subtract 4 from all parts of the inequality:
\[ -3 - 4 \leq z \leq 3 - 4 \]
Simplifying this gives:
\[ -7 \leq z \leq -1 \]

4. **Finding \( z + 1 \)**:
Now, we need to find the bounds for \( z + 1 \):
\[ z + 1 = z + 1 \]
Adding 1 to the entire inequality:
\[ -7 + 1 \leq z + 1 \leq -1 + 1 \]
This simplifies to:
\[ -6 \leq z + 1 \leq 0 \]

5. **Finding the Maximum Value of \( |z + 1| \)**:
The expression \( |z + 1| \) will reach its maximum when \( z + 1 \) is at its furthest point from the origin (0). The maximum value occurs at the endpoint of the interval:
- The maximum value of \( |z + 1| \) occurs when \( z + 1 = -6 \):
\[ |z + 1| = |-6| = 6 \]

### Conclusion:
Thus, the maximum value of \( |z + 1| \) is \( 6 \).