Key Concepts and Formulas

• Complex Numbers: A complex number $z$ can be expressed as $z = x + iy$, where $x = \text{Re}(z)$ is the real part and $y = \text{Im}(z)$ is the imaginary part. The modulus is given by $|z| = \sqrt{x^2 + y^2}$.
• Geometric Interpretation of Modulus: $|z - z_0|$ represents the distance between the points $z$ and $z_0$ in the complex plane. The inequality $|z - z_0| \le r$ represents a closed disk centered at $z_0$ with radius $r$.
• Geometric Interpretation of Inequalities: Inequalities involving $\text{Re}(z)$ or $\text{Im}(z)$ typically represent half-planes.

Step-by-Step Solution

Step 1: Analyze Set $S_1$

We are given $S_1 = \{z \in \mathbb{C} : |z - 1| \le \sqrt{2}\}$. Let $z = x + iy$. We want to express this inequality in terms of $x$ and $y$ to understand its geometric representation.

$|(x + iy) - 1| \le \sqrt{2}$ $|(x - 1) + iy| \le \sqrt{2}$ Using the definition of modulus: $\sqrt{(x - 1)^2 + y^2} \le \sqrt{2}$ Squaring both sides (since both sides are non-negative): $(x - 1)^2 + y^2 \le 2 \quad (\text{Equation 1})$ This represents a closed disk centered at $(1, 0)$ with radius $\sqrt{2}$.

Step 2: Analyze Set $S_2$

We are given $S_2 = \{z \in \mathbb{C} : \text{Re}((1 - i)z) \ge 1\}$. Let $z = x + iy$. We need to find the real part of $(1 - i)z$.

$(1 - i)(x + iy) = x + iy - ix - i^2y$ Since $i^2 = -1$: $(1 - i)(x + iy) = x + iy - ix + y = (x + y) + i(y - x)$ Now, take the real part: $\text{Re}((1 - i)z) = x + y$ Substituting this back into the inequality: $x + y \ge 1 \quad (\text{Equation 2})$ This represents a closed half-plane bounded by the line $x + y = 1$.

Step 3: Analyze Set $S_3$

We are given $S_3 = \{z \in \mathbb{C} : \text{Im}(z) \le 1\}$. Let $z = x + iy$. We need to find the imaginary part of $z$.

$\text{Im}(z) = y$ Substituting this into the inequality: $y \le 1 \quad (\text{Equation 3})$ This represents a closed half-plane bounded by the line $y = 1$.

Step 4: Find the Intersection Points

We need to find the intersection $S_1 \cap S_2 \cap S_3$. This involves finding the points that satisfy all three inequalities. The boundary equations are:

• Circle: $(x - 1)^2 + y^2 = 2$
• Line 1: $x + y = 1$
• Line 2: $y = 1$

We will find the intersection points of these boundaries.

Step 5: Intersection of Line 1 and Line 2

Substitute $y = 1$ into $x + y = 1$: $x + 1 = 1 \Rightarrow x = 0$ So, the point is $P_1 = (0, 1)$.

Step 6: Intersection of the Circle and Line 2

Substitute $y = 1$ into $(x - 1)^2 + y^2 = 2$: $(x - 1)^2 + 1^2 = 2$ $(x - 1)^2 = 1$ $x - 1 = \pm 1$ If $x - 1 = 1$, then $x = 2$, giving the point $P_2 = (2, 1)$. If $x - 1 = -1$, then $x = 0$, giving the point $P_1 = (0, 1)$ (already found).

Step 7: Intersection of the Circle and Line 1

From $x + y = 1$, we have $y = 1 - x$. Substitute into $(x - 1)^2 + y^2 = 2$: $(x - 1)^2 + (1 - x)^2 = 2$ $(x - 1)^2 + (x - 1)^2 = 2$ $2(x - 1)^2 = 2$ $(x - 1)^2 = 1$ $x - 1 = \pm 1$ If $x - 1 = 1$, then $x = 2$, so $y = 1 - 2 = -1$, giving the point $P_3 = (2, -1)$. If $x - 1 = -1$, then $x = 0$, so $y = 1 - 0 = 1$, giving the point $P_1 = (0, 1)$ (already found).

Step 8: Verification of the Intersection Points

We have three potential intersection points: $P_1(0, 1)$, $P_2(2, 1)$, and $P_3(2, -1)$. We need to check if these points satisfy all three inequalities.

• $P_1(0, 1)$:
  • $(0 - 1)^2 + 1^2 = 2 \le 2$ (Satisfies $S_1$)
  • $0 + 1 = 1 \ge 1$ (Satisfies $S_2$)
  • $1 \le 1$ (Satisfies $S_3$)
• $P_2(2, 1)$:
  • $(2 - 1)^2 + 1^2 = 2 \le 2$ (Satisfies $S_1$)
  • $2 + 1 = 3 \ge 1$ (Satisfies $S_2$)
  • $1 \le 1$ (Satisfies $S_3$)
• $P_3(2, -1)$:
  • $(2 - 1)^2 + (-1)^2 = 2 \le 2$ (Satisfies $S_1$)
  • $2 + (-1) = 1 \ge 1$ (Satisfies $S_2$)
  • $-1 \le 1$ (Satisfies $S_3$)

All three points satisfy the inequalities defining $S_1$, $S_2$, and $S_3$. Therefore, the intersection $S_1 \cap S_2 \cap S_3$ contains exactly three elements.

Common Mistakes & Tips

• Careless Algebra: Double-check calculations, especially when dealing with complex numbers and squaring terms.
• Misinterpreting Inequalities: Remember that $\le$ and $\ge$ include the boundary lines/curves.
• Assuming Infinitely Many Solutions: While intersections of continuous regions often have infinitely many points, look for specific intersection points when the answer choices are discrete numbers.

Summary

We analyzed the three sets $S_1$, $S_2$, and $S_3$ geometrically, converting the complex inequalities into Cartesian form. We then found the intersection points of the boundary curves and lines, and verified that these points satisfy all three original inequalities. This led us to the conclusion that the set $S_1 \cap S_2 \cap S_3$ contains exactly three elements.

Final Answer

The final answer is \boxed{has exactly three elements}, which corresponds to option (A).