Let A = {z : (frac{z - bar{z}}{2i})^2 leqslan | vedclass

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(D) Let $z = x + iy$. Then $\frac{z - \bar{z}}{2i} = y$.For set $A$, the condition is $y^2 \leqslant 2y$, which simplifies to $y^2 - 2y \leqslant 0$,

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$13$

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(D) Let $z = x + iy$. Then $\frac{z - \bar{z}}{2i} = y$.For set $A$, the condition is $y^2 \leqslant 2y$, which simplifies to $y^2 - 2y \leqslant 0$, giving $0 \leqslant y \leqslant 2$.For set $B$, the condition is $|z| \leqslant \sqrt{5}$, which implies $x^2 + y^2 \leqslant 5$.We need to find points $(x, y)$ where $x, y \in \mathbb{Z}$ such that $0 \leqslant y \leqslant 2$ and $x^2 + y^2 \leqslant 5$.Case $y = 0$: $x^2 \leqslant 5 \implies x \in \{-2, -1, 0, 1, 2\}$. Points: $(-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0)$ ($5$ points).Case $y = 1$: $x^2 + 1 \leqslant 5 \implies x^2 \leqslant 4 \implies x \in \{-2, -1, 0, 1, 2\}$. Points: $(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 1)$ ($5$ points).Case $y = 2$: $x^2 + 4 \leqslant 5 \implies x^2 \leqslant 1 \implies x \in \{-1, 0, 1\}$. Points: $(-1, 2), (0, 2), (1, 2)$ ($3$ points).Total number of points = $5 + 5 + 3 = 13$.

Let $A = \{z : (\frac{z - \bar{z}}{2i})^2 \leqslant 2(\frac{z - \bar{z}}{2i})\}$ where $i = \sqrt{-1}$ and $B = \{z : |z| \leqslant \sqrt{5}\}$. The number of points with integral real and imaginary parts of $z$ lying in $A \cap B$ is -

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