JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 15)
The area (in sq. units) of the region $$S=\{z \in \mathbb{C}:|z-1| \leq 2 ;(z+\bar{z})+i(z-\bar{z}) \leq 2, \operatorname{lm}(z) \geq 0\}$$ is
$$\frac{7 \pi}{4}$$
$$\frac{3 \pi}{2}$$
$$\frac{7 \pi}{3}$$
$$\frac{17 \pi}{8}$$
Explanation
$$|z-1| \leq 2 \quad \Rightarrow \quad(x-1)^2+y^2=4$$
$$\begin{aligned} & z+\bar{z}+i(z-\bar{z}) \leq 2 \\ \Rightarrow \quad & x-y \leq 1 \\ & \operatorname{Im}(z) \geq 0 \\ \Rightarrow \quad & y \geq 0 \end{aligned}$$
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$$\begin{aligned} & \text { Required area }=\left(\frac{\frac{3 \pi}{4}}{2 \pi}\right)(\pi)(2)^2 \\ & =\frac{3}{2} \pi \end{aligned}$$
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