Question

Let \(α=8−14𝑖\), \(𝐴=\{z∈C:\frac {αz−\bar α\bar z}{z^2−(\bar z)^2−112i }=1\}\) and \(𝐵={𝑧∈C:|𝑧+3𝑖|=4}\). Then \(∑_{𝑧∈𝐴∩𝐵 }(\text {𝑅𝑒}\ ⁡𝑧−\text {𝐼𝑚⁡}\ 𝑧)\) is equal to _____ .

Answer 1

Correct Answer: 14

(A)Simplify \( |z^2 - \alpha^2| = |z^2 - \overline{\alpha}^2| \): \[ \alpha = 8 - 14i, \quad \overline{\alpha} = 8 + 14i. \] Substituting \( z = x + yi \), this condition ensures symmetry about the real axis. (B)The set \( A \) represents the locus of \( z \) in the complex plane where the distances of \( z^2 \) from \( \alpha^2 \) and \( \overline{\alpha}^2 \) are equal. This is the perpendicular bisector of the segment joining \( \alpha^2 \) and \( \overline{\alpha}^2 \). (C)The set \( B \) represents a circle with center \( (0, -3) \) and radius \( 4 \). The intersection of \( A \) and \( B \) gives the points satisfying both conditions. (D)Solve for intersection points: \[ z = x + yi, \quad \operatorname{Re}(z) - \operatorname{Im}(z) = c_1, \quad \text{where } c_1 \text{ is derived from the intersection.} \] (E) Summing \( \operatorname{Re} z - \operatorname{Im} z \) over the intersection points yields: \[ \sum (\operatorname{Re} z - \operatorname{Im} z) = 14. \]