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 $A=z:|z|\le 4$ and $B=z:(z+\overline{z})-i(z-\overline{z})\ge 8$. Find the area of region $A\cap B$.

• A. $4\pi -8$
• B. $2\pi -4$
• C. $4\pi -4$
• D. $\pi -2$

Answer 1

The correct option is A $4\pi -8$

As, $z=x+iy$
Let's plot given sets on x-y plane,
$A=z:|z|\le 4$
$\Rightarrow \sqrt{x^2+y^2}\le 4$

$\Rightarrow x^2+y^2\le 16$
$\Rightarrow z$ lies inside or on the circle $x^2+y^2=16$

$B=z:(z+\overline{z})-i(z-\overline{z})\ge 8$
$\Rightarrow (x+iy+x-iy)-i(x+iy-x+iy)\ge 8$
$\Rightarrow 2x-i\left(2iy\right)\ge 8$
$\Rightarrow 2x+2y\ge 8$
$\Rightarrow x+y\ge 4$

Area of region given by $A\cap B$ is shown by shaded region,
So, the area of shaded region is given by
Area =$\frac{\pi (4{)}^2}{4}-\frac{1}{2}\left(4\right)\left(4\right)$
Area =$4\pi -8$