If `S={z in c: 1 le abs(z-(1+i)) le 2}` and `A={z in s: abs(z-(1-i))=1}` then A is

Options

A. null set
B. singleton set
C. exactly two element
D. infinite elements

Answer

To solve the problem, we need to analyze the sets defined in the question.

### Step 1: Understand the set \( S \)

The set \( S \) is defined as:
\[ S = \{ z \in \mathbb{C} : 1 \leq |z - (1+i)| \leq 2 \} \]
This means that \( S \) consists of all complex numbers \( z \) whose distance from the point \( (1+i) \) (which is the point \( (1, 1) \) in the complex plane) is between 1 and 2.

### Step 2: Visualize the set \( S \)

The inequality describes an annular region (ring-shaped area) in the complex plane:
- The inner circle has a radius of 1 centered at \( (1, 1) \).
- The outer circle has a radius of 2 centered at \( (1, 1) \).

### Step 3: Understand the set \( A \)

The set \( A \) is defined as:
\[ A = \{ z \in S : |z - (1-i)| = 1 \} \]
This means \( A \) consists of all complex numbers \( z \) that are exactly 1 unit away from the point \( (1-i) \) (which is the point \( (1, -1) \) in the complex plane).

### Step 4: Visualize the set \( A \)

The equation \( |z - (1-i)| = 1 \) describes a circle:
- The circle has a radius of 1 centered at \( (1, -1) \).

### Step 5: Analyze the intersection of sets \( S \) and \( A \)

Now, we need to find the intersection of the annular region defined by \( S \) and the circle defined by \( A \):
- The center of the circle for \( A \) is at \( (1, -1) \) and has a radius of 1.
- The inner circle of \( S \) is at \( (1, 1) \) with a radius of 1, and the outer circle of \( S \) is at \( (1, 1) \) with a radius of 2.

### Step 6: Determine the relationship between the circles

- The center of the circle for \( A \) is below the center of the circle for \( S \).
- The distance between the centers of the two circles is 2 units (from \( (1, 1) \) to \( (1, -1) \)).
- The radius of circle \( A \) is 1, and the radius of the inner circle of \( S \) is also 1.

### Conclusion

The circle defined by \( A \) is tangent to the inner circle of \( S \) at the point \( (1, 0) \) and lies entirely within the annular region of \( S \). Therefore, the set \( A \) contains points that are infinitely many as they form a circle.

Thus, the answer is that \( A \) is an infinite set of points.