The area (in sq. units) of the region S = {z | vedclass

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(B) Let $z = x + iy$. Given $|z-1| \leq 2$, we have $(x-1)^2 + y^2 \leq 2^2$, which represents a disk with center $(1, 0)$ and radius $r = 2$. Given $

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$\frac{3 \pi}{2}$

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(B) Let $z = x + iy$. Given $|z-1| \leq 2$, we have $(x-1)^2 + y^2 \leq 2^2$, which represents a disk with center $(1, 0)$ and radius $r = 2$. Given $(z+\overline{z}) + i(z-\overline{z}) \leq 2$, we substitute $z = x+iy$ and $\overline{z} = x-iy$: $(x+iy + x-iy) + i(x+iy - (x-iy)) \leq 2$ $2x + i(2iy) \leq 2$ $2x - 2y \leq 2 \Rightarrow x - y \leq 1 \Rightarrow y \geq x - 1$. Given $\operatorname{Im}(z) \geq 0$, we have $y \geq 0$. The region is the intersection of the disk $(x-1)^2 + y^2 \leq 4$, the half-plane $y \geq x-1$, and the upper half-plane $y \geq 0$. The line $y = x-1$ passes through $(1, 0)$ and makes an angle of $45^\circ$ (or $\pi/4$ radians) with the positive $x$-axis. The area is the area of the semi-circle above the $x$-axis minus the area of the circular sector cut off by the line $y = x-1$ in the first quadrant. Area of semi-circle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = 2\pi$. The sector corresponds to the region between the line $y=x-1$ and the $x$-axis within the circle. The angle subtended by this sector at the center $(1, 0)$ is $\pi/4$. Area of sector = $\frac{1}{2} r^2 	heta = \frac{1}{2} (2)^2 (\pi/4) = \pi/2$. Required area = $2\pi - \pi/2 = \frac{3\pi}{2}$.

The area (in sq. units) of the region $S = \{z \in \mathbb{C} : |z-1| \leq 2, (z+\overline{z}) + i(z-\overline{z}) \leq 2, \operatorname{Im}(z) \geq 0\}$ is

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