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 $A=\{z:\left|z\right|\le 2\}$ and $B=\{z:(1-i)z+(1+i)\overline{z}\ge 41\}$ . Find the area of region $A\cap B$.

• A. $3\pi -2$
• B. $3\pi +2$
• C. $\pi +2$
• D. $\pi -2$

Answer 1

Answer: (D)

$\pi -2$

$z=x+iy$
$A=\{z:\left|z\right|\le 2\}$
$⟹\sqrt{x^2+y^2}\le 2$
or $x^2+y^2\le 4$
$⟹z$ lies on or inside the circle $x^2+y^2=4$
$B=\{z:(1-i)z+(1+i)\overline{z}\ge 4\}$
$⟹(1-i)(x+iy)+(1+i)(x-iy)\ge 4$
$⟹x+iy-ix+y+x-iy+ix+y\ge 4$
$⟹x+y\ge 2$
Area of region $A\cap B$ is the shaded region shown in Fig. 3.19
Area = $\frac{\pi (2{)}^2}{4}-\frac{1}{2}\times 2\times 2=\pi -2$
Ans: D