Question

Let $z=x+iy$ be a complex number, $A=\{z||z\mid \le 2\}$ and $B=\{z/(1-i)z+(1+i)\bar{z}\ge 4\}$ Then which one of the following options belongs to $A\cap B$ ?

• A. $\sqrt{3}+\frac{1}{2}i$
• B. $\frac{1}{2}+\frac{i}{2}$
• C. $\sqrt{2}+\frac{i}{2}$
• D. $2+2i$

Answer 1

Answer: (A)

We have
$\Rightarrow \sqrt{x^2+y^2}\le 2$
$\Rightarrow x^2+y^2\le 4\dots .(i)$
Again, $(1-i)z+(1+i)\bar{z}\ge 4$
$\Rightarrow (1-i)(x+iy)+(1+i)(x-iy)\ge 4$
$\Rightarrow x+iy-ix+y+x-iy+ix+y\ge 4$
$\Rightarrow 2x+2y\ge 4$
$\Rightarrow x+y\ge 2\dots (ii)$
So, $A\cap B=\left\{x^2+y^2\le 4\right\}\cap \{x+y\ge 2\}$
$\therefore z=\sqrt{3}+\frac{1}{2}i$