`z in C, S_1 ={ abs(z-1) lt sqrt2} , S_2={Re(z(1-i)ge1} , S_3={Im (z) lt 1 }`. Then `n(S_1capS_2capS_3)`

Options

A. is a singletonset
B. Infinite set
C. Has exactly 2 elements
D. Null Set

Answer

To solve the problem, we need to analyze the sets \( S_1 \), \( S_2 \), and \( S_3 \) defined in the complex plane and find the number of points in their intersection.

### Step 1: Analyze \( S_1 \)

The set \( S_1 \) is defined as:
\[ S_1 = \{ z \in \mathbb{C} : |z - 1| < \sqrt{2} \} \]
This describes a circle in the complex plane centered at \( z = 1 \) (which corresponds to the point \( (1, 0) \) in the Cartesian plane) with a radius of \( \sqrt{2} \). The inequality \( |z - 1| < \sqrt{2} \) indicates that we are interested in the interior of this circle.

### Step 2: Analyze \( S_2 \)

The set \( S_2 \) is defined as:
\[ S_2 = \{ z \in \mathbb{C} : \text{Re}(z(1 - i)) \geq 1 \} \]
Let \( z = x + iy \) where \( x \) is the real part and \( y \) is the imaginary part. Then:
\[ z(1 - i) = (x + iy)(1 - i) = x + iy - ix - y = (x + y) + i(y - x) \]
The real part of this expression is \( x + y \). Therefore, the condition becomes:
\[ x + y \geq 1 \]
This represents the region above (and including) the line \( x + y = 1 \) in the Cartesian plane.

### Step 3: Analyze \( S_3 \)

The set \( S_3 \) is defined as:
\[ S_3 = \{ z \in \mathbb{C} : \text{Im}(z) < 1 \} \]
This describes the region below the line \( y = 1 \) in the complex plane.

### Step 4: Find the Intersection \( S_1 \cap S_2 \cap S_3 \)

Now, we need to find the intersection of these three sets:
1. The interior of the circle centered at \( (1, 0) \) with radius \( \sqrt{2} \).
2. The region above the line \( x + y = 1 \).
3. The region below the line \( y = 1 \).

#### Step 4.1: Circle Equation

The equation of the circle can be expressed as:
\[ (x - 1)^2 + y^2 < 2 \]

#### Step 4.2: Line Equations

The line \( x + y = 1 \) can be rewritten as:
\[ y = 1 - x \]
The line \( y = 1 \) is a horizontal line.

#### Step 4.3: Finding the Intersection Points

To find the intersection points, we need to check where the line \( y = 1 - x \) intersects the circle:
1. Substitute \( y = 1 - x \) into the circle equation:
\[ (x - 1)^2 + (1 - x)^2 < 2 \]
Expanding this gives:
\[ (x - 1)^2 + (1 - x)^2 = (x - 1)^2 + (x - 1)^2 = 2(x - 1)^2 < 2 \]
Simplifying leads to:
\[ (x - 1)^2 < 1 \implies -1 < x - 1 < 1 \implies 0 < x < 2 \]
Therefore, \( x \) can take values between \( 0 \) and \( 2 \).

2. For \( y \), substituting \( x \) gives:
\[ y = 1 - x \]
Thus, \( y \) will range from \( 1 \) to \( -1 \) as \( x \) varies from \( 0 \) to \( 2 \).

### Conclusion

The intersection \( S_1 \cap S_2 \cap S_3 \) is a region that contains infinitely many points, as it is bounded by the lines and the circle but does not restrict to a finite number of points.

### Final Answer

Thus, the number of points in the intersection \( n(S_1 \cap S_2 \cap S_3) \) is infinite.