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. Define sets $A$ and $B$ based on the given conditions. - Set $A$ is defined by the equation $|z - 2 - i| = 3$, which translates to $(x - 2)^2 + (y - 1)^2 = 9$. This represents a circle centered at $(2, 1)$ with a radius of 3. - Set $B$ is defined by the equation $\text{Re}(z - iz) = 2$, which simplifies to $x + y = 2$. This represents a straight line. 2. Solve the system of equations formed by the conditions of sets $A$ and $B$. The system is: \[ \begin{cases} (x - 2)^2 + (y - 1)^2 = 9 \\ x + y = 2 \end{cases} \] From the second equation, substitute $y = 2 - x$ into the first equation: \[ (x - 2)^2 + ((2 - x) - 1)^2 = 9 \] \[ (x - 2)^2 + (1 - x)^2 = 9 \] Expanding and simplifying: \[ x^2 - 4x + 4 + 1 - 2x + x^2 = 9 \] \[ 2x^2 - 6x + 5 = 9 \] \[ 2x^2 - 6x - 4 = 0 \] Divide by 2: \[ x^2 - 3x - 2 = 0 \] Using the quadratic formula to solve for $x$: \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \] Find the corresponding $y$ values using $y = 2 - x$: If $x = \frac{3 + \sqrt{17}}{2}$, then $y = 2 - \frac{3 + \sqrt{17}}{2} = \frac{4 - (3 + \sqrt{17})}{2} = \frac{1 - \sqrt{17}}{2}$. If $x = \frac{3 - \sqrt{17}}{2}$, then $y = 2 - \frac{3 - \sqrt{17}}{2} = \frac{4 - (3 - \sqrt{17})}{2} = \frac{1 + \sqrt{17}}{2}$. The intersection points (solutions in $S$) are $z_1 = \frac{3 + \sqrt{17}}{2} + i \frac{1 - \sqrt{17}}{2}$ and $z_2 = \frac{3 - \sqrt{17}}{2} + i \frac{1 + \sqrt{17}}{2}$. 3. Calculate the sum of the squared magnitudes of the solutions in $S$. For $z = x + iy$, $|z|^2 = x^2 + y^2$. For $z_1 = \frac{3 + \sqrt{17}}{2} + i \frac{1 - \sqrt{17}}{2}$: $|z_1|^2 = \left(\frac{3 + \sqrt{17}}{2}\right)^2 + \left(\frac{1 - \sqrt{17}}{2}\right)^2$ $= \frac{9 + 6\sqrt{17} + 17}{4} + \frac{1 - 2\sqrt{17} + 17}{4} = \frac{26 + 6\sqrt{17} + 18 - 2\sqrt{17}}{4} = \frac{44 + 4\sqrt{17}}{4} = 11 + \sqrt{17}$. For $z_2 = \frac{3 - \sqrt{17}}{2} + i \frac{1 + \sqrt{17}}{2}$: $|z_2|^2 = \left(\frac{3 - \sqrt{17}}{2}\right)^2 + \left(\frac{1 + \sqrt{17}}{2}\right)^2$ $= \frac{9 - 6\sqrt{17} + 17}{4} + \frac{1 + 2\sqrt{17} + 17}{4} = \frac{26 - 6\sqrt{17} + 18 + 2\sqrt{17}}{4} = \frac{44 - 4\sqrt{17}}{4} = 11 - \sqrt{17}$. The sum is $\sum_{z \in S} |z|^2 = |z_1|^2 + |z_2|^2 = (11 + \sqrt{17}) + (11 - \sqrt{17}) = 22$. Alternatively, using the given calculation: \[ \sum_{z \in S} |z|^2 = \left( \frac{3 + \sqrt{17}}{2} \right)^2 + \left( \frac{1 - \sqrt{17}}{2} \right)^2 + \left( \frac{3 - \sqrt{17}}{2} \right)^2 + \left( \frac{1 + \sqrt{17}}{2} \right)^2 \] The terms are the squared magnitudes of the two intersection points. $(\frac{3 + \sqrt{17}}{2})^2 = \frac{9 + 6\sqrt{17} + 17}{4} = \frac{26 + 6\sqrt{17}}{4}$ $(\frac{1 - \sqrt{17}}{2})^2 = \frac{1 - 2\sqrt{17} + 17}{4} = \frac{18 - 2\sqrt{17}}{4}$ $(\frac{3 - \sqrt{17}}{2})^2 = \frac{9 - 6\sqrt{17} + 17}{4} = \frac{26 - 6\sqrt{17}}{4}$ $(\frac{1 + \sqrt{17}}{2})^2 = \frac{1 + 2\sqrt{17} + 17}{4} = \frac{18 + 2\sqrt{17}}{4}$ Sum = $\frac{26 + 6\sqrt{17} + 18 - 2\sqrt{17} + 26 - 6\sqrt{17} + 18 + 2\sqrt{17}}{4} = \frac{26+18+26+18}{4} = \frac{88}{4} = 22$. Therefore, the correct answer is (1) 22.