JEE Mathematics 2023 Question with Solution

Given: $\alpha = 8 - 14i$, $A = \{z \in \mathbb{C} : |z^2 - \alpha^2| = |z^2 - \overline{\alpha}^2| \}$ and $B = \{z \in \mathbb{C} : |z + 3i| = 4\}$.

Find: $\sum_{z \in A \cap B}(\operatorname{Re} z - \operatorname{Im} z)$.

From the solution, the condition

$$|z^2 - \alpha^2| = |z^2 - \overline{\alpha}^2|$$

means that $z^2$ is equidistant from $\alpha^2$ and $\overline{\alpha}^2$. Hence, the set $A$ is the perpendicular bisector of the segment joining $\alpha^2$ and $\overline{\alpha}^2$ in the $z^2$-plane.

Also, the set $B$ is the circle with center $(0,-3)$ and radius $4$.

The solution states that solving the intersection condition and summing over the required points gives

$$\sum (\operatorname{Re} z - \operatorname{Im} z) = 14$$

Therefore, the required numerical value is $14$.

Given: $\alpha = 8 - 14i$ and $\overline{\alpha} = 8 + 14i$.

Find: the sum of $\operatorname{Re} z - \operatorname{Im} z$ over all points in $A \cap B$.

The provided explanation identifies:

1. $A$ as the locus where $z^2$ is at equal distance from $\alpha^2$ and $\overline{\alpha}^2$.
2. $B$ as the circle

$$|z+3i|=4$$

with center $(0,-3)$ and radius $4$.

Thus, the required points come from the intersection of these two conditions. The source solution does not show the intermediate algebra explicitly, but it concludes that the sum

$$\sum_{z \in A \cap B}(\operatorname{Re} z - \operatorname{Im} z)$$

is

$$14$$

Hence, the final answer is $14$.