Question

Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to

Hint:

Solve the system of equations to find the intersection points of the sets.

Answer 1

Correct Answer: 22

1. Identify the sets $A$ and $B$: - $A: |z - 2 - i| = 3$ \[ |(x - 2) + (y - 1)i| = 3 \] \[ (x - 2)^2 + (y - 1)^2 = 9 \] - $B: \text{Re}(z - iz) = 2$ \[ \text{Re}((x + y) + i(y - x)) = 2 \] \[ x + y = 2 \]
2. Solve the system of equations: \[ \begin{cases} (x - 2)^2 + (y - 1)^2 = 9 x + y = 2 \end{cases} \] - Substitute $y = 2 - x$ into the first equation: \[ (x - 2)^2 + (2 - x - 1)^2 = 9 \] \[ (x - 2)^2 + (1 - x)^2 = 9 \] \[ x^2 - 4x + 4 + 1 - 2x + x^2 = 9 \] \[ 2x^2 - 6x + 5 = 9 \] \[ 2x^2 - 6x - 4 = 0 \] \[ x^2 - 3x - 2 = 0 \] \[ x = \frac{3 \pm \sqrt{17}}{2} \] - Corresponding $y$ values: \[ y = 2 - x = 2 - \frac{3 \pm \sqrt{17}}{2} = \frac{1 \mp \sqrt{17}}{2} \]
3. Calculate $\sum_{z \in S} |z|^2$: \[ \sum_{z \in S} |z|^2 = \left( \frac{3 + \sqrt{17}}{2} \right)^2 + \left( \frac{1 - \sqrt{17}}{2} \right)^2 + \left( \frac{3 - \sqrt{17}}{2} \right)^2 + \left( \frac{1 + \sqrt{17}}{2} \right)^2 \] \[ = \frac{1}{4} \left[ 2 \times 26 + 2 \times 18 \right] = \frac{88}{4} = 22 \] Therefore, the correct answer is (1) 22.