Question

Let $z=x+iy$ be a complex number, where $x$ and $y$ are real numbers. Let $A$ and $B$ be the sets defined by $A=\{z:|z|\le 2\}$ and $B=\{z:(1-i)z+(1+i)\overline{z}\ge 4\}$. Then the area of region $A\cap B$ is

• A. $2$ sq.units
• B. $\pi$ sq.units
• C. $\pi -2$ sq.units
• D. $\pi +2$ sq.units

Answer 1

The correct option is C $\pi -2$ sq.units

$z=x+iyA=\{z:|z|\le 2\}$
$\Rightarrow x^2+y^2\le 4$
$\Rightarrow z$ lies on or inside the circle $x^2+y^2=4$
$B=\{z:(1-i)z+(1+i)\overline{z}\ge 4\}$
$\Rightarrow (1-i)(x+iy)+(1+i)(x-iy)\ge 4$
$\Rightarrow x+iy-ix+y+x-iy+ix+y\ge 4$
$\Rightarrow x+y\ge 2$
Area of region $A\cap B$ is shaded region of the diagram

$\therefore$ Required area $=\frac{\pi (2{)}^2}{4}-\frac{1}{2}\times 2\times 2=(\pi -2)\text{sq.units}$