Let S_1 = {z in mathbb{C} : |z| leq 5}, S_2 = | vedclass

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(D) $S_1$ represents the interior and boundary of a circle with radius $r=5$ centered at the origin: $x^2 + y^2 \leq 25$.$S_2$ is defined by $\operato

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$\frac{125\pi}{12}$

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(D) $S_1$ represents the interior and boundary of a circle with radius $r=5$ centered at the origin: $x^2 + y^2 \leq 25$.$S_2$ is defined by $\operatorname{Im}\left(\frac{z}{1-\sqrt{3}i} + 1ight) \geq 0$. Since $\operatorname{Im}(1) = 0$, this is $\operatorname{Im}\left(\frac{x+iy}{1-\sqrt{3}i}ight) \geq 0$.Multiplying by the conjugate: $\operatorname{Im}\left(\frac{(x+iy)(1+\sqrt{3}i)}{4}ight) \geq 0 \implies \sqrt{3}x + y \geq 0$, which is the region above the line $y = -\sqrt{3}x$.$S_3$ is the region where $x \geq 0$ (the right half-plane).The intersection $S_1 \cap S_2 \cap S_3$ is a sector of the circle. The line $y = -\sqrt{3}x$ makes an angle of $-60^\circ$ (or $300^\circ$) with the positive $x$-axis. The region $S_2 \cap S_3$ covers the angular range from $-60^\circ$ to $90^\circ$, which is $150^\circ$ or $\frac{5\pi}{6}$ radians.Area $= \frac{	heta}{2\pi} 	imes \pi r^2 = \frac{5\pi/6}{2\pi} 	imes \pi(5)^2 = \frac{5}{12} 	imes 25\pi = \frac{125\pi}{12}$.

Let $S_1 = \{z \in \mathbb{C} : |z| \leq 5\}$, $S_2 = \{z \in \mathbb{C} : \operatorname{Im}\left(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i}\right) \geq 0\}$ and $S_3 = \{z \in \mathbb{C} : \operatorname{Re}(z) \geq 0\}$. Then the area of the region $S_1 \cap S_2 \cap S_3$ is:

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