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

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Aicharya · Accepted Answer

$S_1: x^2+y^2 \leq 25...(1)$ $\begin{aligned} & S_2: \operatorname{lm} \text { of } \frac{z+(1-\sqrt{3} i)}{(1-\sqrt{3} i)} \geq 0 \\ & \operatorname{lm} \text { of }\left(\frac{x+i y}{1-\sqrt{3} i}+1\right) \geq 0 \\ & \operatorname{lm} \text { of }\left(\frac{(x+i y)(1+\sqrt{3} i)}{4}\right) \geq 0 \\ & \Rightarrow \sqrt{3} x+y \geq 0 ...(2)\\ & S_3: x \geq 0 ...(3)\\ & \text { Area }=\frac{5}{12}\left(\pi(5)^2\right)\end{aligned}$

Let $S_1=\{z \in C:|z| \leq 5\}, S_2=\left\{z \in C: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3}…

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