Let z=x+iy be a complex number (x, y in R). L | vedclass

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(D) Given $z=x+iy$,where $x, y \in R$. Set $A=\{z:|z| \leq 2\}$ represents the interior and boundary of a circle with center $(0,0)$ and radius $2$,i.

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$\pi-2$

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(D) Given $z=x+iy$,where $x, y \in R$. Set $A=\{z:|z| \leq 2\}$ represents the interior and boundary of a circle with center $(0,0)$ and radius $2$,i.e.,$x^2+y^2 \leq 4$. Set $B=\{z:(z+2y)+\bar{z} \geq 4\}$. Substituting $z=x+iy$ and $\bar{z}=x-iy$: $(x+iy+2y)+(x-iy) \geq 4$ $2x+2y \geq 4 \Rightarrow x+y \geq 2$. This represents the region on or above the line $x+y=2$. The intersection $A \cap B$ is the region bounded by the circle $x^2+y^2=4$ and the line $x+y=2$ in the first quadrant. The points of intersection are found by solving $x^2+(2-x)^2=4$: $x^2+4-4x+x^2=4$ $\Rightarrow 2x^2-4x=0$ $\Rightarrow 2x(x-2)=0$. So,$x=0$ (giving $y=2$) and $x=2$ (giving $y=0$). The area is given by $\int_0^2 (y_{	ext{circle}} - y_{	ext{line}}) dx = \int_0^2 (\sqrt{4-x^2} - (2-x)) dx$. $= \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}ight) - 2x + \frac{x^2}{2} ight]_0^2$ $= (0 + 2\sin^{-1}(1) - 4 + 2) - (0 + 0 - 0 + 0) = 2(\frac{\pi}{2}) - 2 = \pi-2$.

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

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