The modulus of the complex number `z` such that `| z + 3 - i| = 1` and arg ` z=pi` is equal to

Options

A. 3
B. 1
C. 2
D. 9

Answer

To solve the problem, we need to find the modulus of the complex number \( z \) such that \( | z + 3 - i| = 1 \) and \( \arg z = \pi \).

### Step-by-Step Solution:

1. **Express \( z \) in terms of \( x \) and \( y \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Use the argument condition**:
The condition \( \arg z = \pi \) implies that the complex number \( z \) lies on the negative real axis. Therefore, \( y = 0 \) (since \( \tan(\pi) = 0 \)).

3. **Substitute \( y = 0 \) into the modulus condition**:
The modulus condition given is:
\[ | z + 3 - i | = 1 \]
Substituting \( z = x + 0i \) gives:
\[ |(x + 3) + (-1)i| = 1 \]

4. **Calculate the modulus**:
The modulus can be calculated as:
\[ |(x + 3) - i| = \sqrt{(x + 3)^2 + (-1)^2} = \sqrt{(x + 3)^2 + 1} \]
Setting this equal to 1 gives:
\[ \sqrt{(x + 3)^2 + 1} = 1 \]

5. **Square both sides**:
Squaring both sides results in:
\[ (x + 3)^2 + 1 = 1 \]

6. **Simplify the equation**:
Subtracting 1 from both sides leads to:
\[ (x + 3)^2 = 0 \]

7. **Solve for \( x \)**:
Taking the square root gives:
\[ x + 3 = 0 \implies x = -3 \]

8. **Determine \( z \)**:
Thus, we have:
\[ z = -3 + 0i = -3 \]

9. **Find the modulus of \( z \)**:
The modulus of \( z \) is:
\[ |z| = |-3| = 3 \]

### Final Answer:
The modulus of the complex number \( z \) is \( 3 \).