Question

Let \( S = \{ z \in \mathbb{C} : |z - 1 + i| = 1 \} \) represents:

• A. a circle with centre \( (-1, 1) \) and radius 1 unit
• B. a circle with centre \( (1, 2) \) and radius 5 units
• C. a circle with centre \( (-1, -1) \) and radius 1 unit
• D. an ellipse with centre \( (-1, -1) \)

Hint:

The equation \( |z - (a + bi)| = r \) represents a circle with center at \( (a, b) \) and radius \( r \).

Answer 1

The Correct Option is C

We are given the equation \( |z - (1 - i)| = 1 \), which describes a geometric figure in the complex plane. The general form for a circle in the complex plane is \( |z - c| = r \), where \( c \) is the center of the circle and \( r \) is the radius. Here, the equation is: \[ |z - (1 - i)| = 1 \] This represents a circle with center \( (1, -1) \) (as \( 1 - i \) corresponds to the point \( (1, -1) \) in the complex plane) and a radius of 1. Thus, the correct answer is option (3), which states that the circle has a center at \( (-1, -1) \) and a radius of 1 unit.

Answer 2

Step 1: Express \( z \) in terms of its components.
Let \( z = x + iy \), where \( x, y \in \mathbb{R} \).

Step 2: Rewrite the given equation carefully.
The condition is:
\[ |z - 1 + i| = 1. \]
This can be interpreted as:
\[ |z - (-1 - i)| = 1, \]
since subtracting \(-1 - i\) is equivalent to \(z - 1 + i\).

Step 3: Substitute \( z = x + iy \) and compute the modulus.
\[ | (x + iy) - (-1 - i) | = | (x + 1) + i(y + 1) | = \sqrt{(x + 1)^2 + (y + 1)^2}. \]

Step 4: Set the modulus equal to 1:
\[ \sqrt{(x + 1)^2 + (y + 1)^2} = 1. \]

Step 5: Square both sides:
\[ (x + 1)^2 + (y + 1)^2 = 1^2 = 1. \]

Step 6: Recognize the geometric figure.
This is the equation of a circle with center \( (-1, -1) \) and radius 1.

Hence, the set \( S \) represents a circle with center \((-1, -1)\) and radius 1 unit.