Question

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

• A. 9
• B. 2
• C. 3
• D. 4

Hint:

The modulus of a complex number \( z = a + bi \) is given by: \[ |z| = \sqrt{a^2 + b^2} \]

Answer 1

The Correct Option is C

Step 1: {Convert given information into mathematical form}
\[ |z + 3 - i| = 1 \] Step 2: {Rewrite as a circle equation}
\[ (x + 3)^2 + (y - 1)^2 = 1 \] Step 3: {Find modulus}
Since \( \arg(z) = \pi \), the point lies on the negative real axis: \[ z = -3 + 0i \] Step 4: {Calculate modulus}
\[ |z| = \sqrt{(-3)^2 + 0^2} = 3 \]

Answer 2

Step 1: Understand the given conditions
The complex number \( z \) satisfies two conditions:
1) \( |z + 3 - i| = 1 \)
2) \( \arg(z) = \pi \)

Step 2: Interpret the argument condition
\(\arg(z) = \pi\) means \( z \) lies on the negative real axis.
So, \( z = -r \), where \( r \) is a positive real number.

Step 3: Express \( z \) and apply the modulus condition
Let \( z = -r + 0i \). Then:
\[ |z + 3 - i| = |-r + 3 - i| = \sqrt{(-r + 3)^2 + (-1)^2} = 1 \]

Step 4: Set up the equation and solve for \( r \)
\[ \sqrt{(-r + 3)^2 + 1} = 1 \implies (-r + 3)^2 + 1 = 1^2 = 1 \implies (-r + 3)^2 = 0 \]
\[ \Rightarrow -r + 3 = 0 \implies r = 3 \]

Step 5: Find the modulus of \( z \)
Since \( z = -r = -3 \), its modulus is:
\[ |z| = 3 \]

Final Answer: 3