Jee Mains
Let $\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}| \leq 5\}, \mathrm{S}_{2}=\left\{\mathrm{z} \in \mathrm{C}: \operatorname{Im}\left(\frac{\mathrm{z}+1-\sqrt{3} \mathrm{i}}{1-\sqrt{3} \mathrm{i}}\right) \geq 0\right\}$ and $\mathrm{S}_{3}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{z}) \geq 0\}$. Then the area of the region $\mathrm{S}_{1} \cap \mathrm{~S}_{2} \cap \mathrm{~S}_{3}$ is
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Solution
According to the question, $\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 25$
\[ \begin{aligned} & \mathrm{S}_{2}: \operatorname{Im}\left(\frac{\mathrm{z}+(1-\sqrt{3} \mathrm{i})}{(1-\sqrt{3} \mathrm{i})}\right) \geq 0 \Rightarrow \operatorname{Im}\left(\frac{\mathrm{x}+\mathrm{iy}}{1-\sqrt{3} \mathrm{i}}+1\right) \geq 0 \Rightarrow \operatorname{Im}\left(\frac{(\mathrm{x}+\mathrm{iy})(1+\sqrt{3} \mathrm{i})}{4}+1\right) \geq 0 \\ & \Rightarrow \frac{\mathrm{y}+\sqrt{3} \mathrm{x}}{4} \geq 0 \Rightarrow \sqrt{3} \mathrm{x}+\mathrm{y} \geq 0 \quad \ldots(2) \end{aligned} \]\[ \mathrm{S}_{3}: \mathrm{x} \geq 0 \quad \ldots(3) \text { Shaded area }=\left(\frac{150^{\circ}}{360^{\circ}}\right) \pi(5)^{2}=\frac{5}{12}\left(\pi(5)^{2}\right) \]
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