Question

Let $C$ be the set of all complex numbers. Let
$S_1=\left\{z\in C||z-3-{2i|}^2=8\right\}$
$S_2=\{z\in C\mid Re(z)\ge 5\}$ and
$S_3=\{z\in C||z-\bar{z}\ge 8\}$
Then the number of elements in $S_1\cap S_2\cap S_3$ is equal to

• A. 1
• B. 0
• C. 2
• D. Infinite

Answer 1

Answer: (A)

$S_1:|z-3-2i{|}^2=8$
$|z-3-2i|=2\sqrt{2}$
$(x-3{)}^2+(y-2{)}^2=(2\sqrt{2}{)}^2$
$S_2:x\ge 5$
$S_3:|z-\bar{z}|\ge 8$
$|2iy|\ge 8$
$2|y|\ge 8\therefore y\ge 4,y\le -4$

$n{S}_1\cap S_2\cap S_3=1$