Let $z=x+i y$ be a complex number $(x, y \in R)$. Let $A$ and $B$ be two sets such that $A=\{z:|z| \leq 2\}$ and $B=\{z:(z+2 y)+\bar{z} \geq 4\}$ the area of region $A \cap B$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } Z=x+i y, \forall x, y \in R \ & A=\{z:|z| \leq 2\} 	ext { and } B=\{z:(z+2 y)+\bar{z} \geq 4\} \ & 	herefore|z| \leq 2 	ext { and } z+2 y+\bar{z} \geq 4\end{aligned}$ $\begin{aligned} & \sqrt{x^2+y^2} \leq 2 	ext { and } x+i y+2 y+x-i y \geq 4 \ & x^2+y^2 \leq 4 	ext { and } 2 x+2 y \geq 4 \ & \Rightarrow x+y \geq 2\end{aligned}$ Point of intersection $ \begin{aligned} x^2+y^2=4 	ext { and } x+y & =2 \Rightarrow y=2-x \ x^2+(2-x)^2 & =4 \ x^2+4-4 x+x^2 & =4 \ 2 x^2-4 x & =0 \Rightarrow 2 x(x-2)=0 \ x=0, x & =2 \Rightarrow y=2, y=0 \end{aligned} $ $	herefore \quad A \cap B=$ Shaded Area $ \begin{aligned} = & \int_0^2 y_{	ext {circle }}-y_{	ext {line }}=\int_0^2\left\{\sqrt{4-x^2}-(2-x)ight\} d x \ = & {\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}-2 x+\frac{x^2}{2}ight]_0^2 } \ \Rightarrow \quad & \left(2 \sin ^{-1} 1-4+\frac{4}{2}ight)=\pi-2 \end{aligned} $

Let $z=x+i y$ be a complex number $(x, y \in R)$. Let $A$ and $B$ be two sets such that $A=\{z:|z| \leq 2\}$…

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