Key Concepts and Formulas

• Complex Numbers in Cartesian Form: $z = x + iy$, where $x = \operatorname{Re}(z)$ and $y = \operatorname{Im}(z)$.
• Modulus of a Complex Number: $|z| = \sqrt{x^2+y^2}$, so $|z|^2 = x^2 + y^2$.
• Equation of a Circle: $|z - z_0| = R$ represents a circle centered at $z_0$ with radius $R$.
• Vieta's Formulas: For $ax^2 + bx + c = 0$ with roots $x_1, x_2$: $x_1 + x_2 = -b/a$ and $x_1 x_2 = c/a$. Also, $x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2$.

Step-by-Step Solution

Step 1: Expressing Conditions in Cartesian Coordinates

Let $z = x+iy$.

• Condition for Set A: $|z-2-i|=3$

  • What: Substitute $z=x+iy$ into the equation.
  • Why: To obtain an equation in terms of $x$ and $y$. $|(x+iy)-2-i|=3$ $|(x-2) + i(y-1)|=3$ $\sqrt{(x-2)^2 + (y-1)^2} = 3$ $(x-2)^2 + (y-1)^2 = 9 \quad (1)$
  • Explanation: This represents a circle centered at $(2,1)$ with radius $3$.

• Condition for Set B: $\operatorname{Re}(z-i z)=2$

  • What: Substitute $z=x+iy$ into the expression and find the real part.
  • Why: To obtain an equation in terms of $x$ and $y$. $z-iz = (x+iy) - i(x+iy) = x+iy -ix -i^2y = x+iy-ix+y = (x+y) + i(y-x)$ $\operatorname{Re}(z-iz) = x+y$ $x+y=2 \quad (2)$
  • Explanation: This represents a straight line.

Step 2: Finding the Intersection Points (Set S)

• What: Solve the system of equations (1) and (2) to find the intersection points.
  • Why: The intersection points are the elements of set $S = A \cap B$. From (2), $x = 2-y$. Substitute this into (1): $(2-y-2)^2 + (y-1)^2 = 9$ $(-y)^2 + (y-1)^2 = 9$ $y^2 + y^2 - 2y + 1 = 9$ $2y^2 - 2y - 8 = 0$ $y^2 - y - 4 = 0 \quad (3)$
  • Explanation: This quadratic equation gives the $y$-coordinates of the intersection points. Since the discriminant is positive, the line intersects the circle at two distinct points. Let these roots be $y_1$ and $y_2$.

Step 3: Calculating Sums of Squares using Vieta's Formulas

• What: Use Vieta's formulas to find the sum and product of the roots of equation (3).

  • Why: This allows us to calculate $y_1^2 + y_2^2$ and $x_1^2 + x_2^2$ without explicitly solving for $y_1$ and $y_2$.

  For the $y$-coordinates: $y_1+y_2 = 1$ $y_1y_2 = -4$ $y_1^2+y_2^2 = (y_1+y_2)^2 - 2y_1y_2 = (1)^2 - 2(-4) = 1 + 8 = 9$

  For the $x$-coordinates: Since $x = 2-y$, we have $x_1 = 2-y_1$ and $x_2 = 2-y_2$. $x_1+x_2 = (2-y_1) + (2-y_2) = 4 - (y_1+y_2) = 4 - 1 = 3$ $x_1x_2 = (2-y_1)(2-y_2) = 4 - 2(y_1+y_2) + y_1y_2 = 4 - 2(1) + (-4) = -2$ $x_1^2+x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = (3)^2 - 2(-2) = 9 + 4 = 13$

Step 4: Calculating the Final Sum

• What: Calculate the sum of the squared moduli of the intersection points.
  • Why: This is the desired quantity. $\sum_{z \in S}|z|^2 = |z_1|^2 + |z_2|^2 = (x_1^2+y_1^2) + (x_2^2+y_2^2) = (x_1^2+x_2^2) + (y_1^2+y_2^2)$ $\sum_{z \in S}|z|^2 = 13 + 9 = 22$

Common Mistakes & Tips

• Be careful when substituting and simplifying complex expressions. A small error can propagate through the entire solution.
• Vieta's formulas are extremely useful for finding sums and products of roots without solving the equation.
• Remember the relationship between $x$ and $y$ from equation (2), $x=2-y$, when calculating the sums of squares.

Summary

By converting the complex number conditions into Cartesian equations, we found the intersection of a circle and a line. Vieta's formulas allowed us to efficiently calculate the sums of squares of the coordinates of the intersection points. The final result is $\sum_{z \in S}|z|^2 = 22$.

Final Answer

The final answer is \boxed{22}.