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

Let $$z = x + iy$$.

• $$|z - 1| \le 2 \implies (x-1)^2 + y^2 \le 4$$. This is a circle centered at $$(1, 0)$$ with radius $$R=2$$.

• $$(z + \bar{z}) + i(z - \bar{z}) \le 2 \implies (2x) + i(2iy) \le 2 \implies 2x - 2y \le 2 \implies \mathbf{x - y \le 1}$$.

• $$\text{Im}(z) \ge 0 \implies \mathbf{y \ge 0}$$.

The line $$x - y = 1$$ (or $$y = x - 1$$) passes through the center of the circle $$(1, 0)$$. Since the line passes through the center, it divides the circle into two equal semicircles.

• The area of the full circle is $$\pi(2)^2 = 4\pi$$.

• The area of the semicircle above the $$x$$-axis ($$\text{Im}(z) \ge 0$$) is $$2\pi$$.

• The line $$y = x - 1$$ further bisects this upper semicircle because it passes through the center at an angle of $$45^\circ$$.

The region is the upper semi-circle restricted by the line $$y \ge x - 1$$. Looking at the geometry, this covers $$3/4$$ of the upper semi-circle's area, or specifically, the sector area.

the line $$x-y=1$$ and the x-axis ($$y=0$$) meet at the center $$(1,0)$$. The region defined is a circular sector of $$135^\circ$$ (or $$3\pi/4$$ radians).

$$\text{Area} = \frac{1}{2} R^2 \theta = \frac{1}{2} (2^2) \left(\frac{3\pi}{4}\right) = \frac{3\pi}{2}$$

Correct Option: D ($$3\pi/2$$)