Question

Let $z=x+i y$ be a complex number, $A=\{z|| z \mid \leq 2\}$ and $B =\{z /(1-i) z+(1+i) \bar{z} \geq 4\}$ 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. $2+2 i$

Answer 1

Answer: (A)

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