Question

Let $C$ be the set of all complex numbers. Let
$S_{1}=\left\{z \in C|| z-3-\left.2 i\right|^{2}=8\right\}$
$S_{2}=\{z \in C \mid Re(z) \geq 5\}$ and
$S_{3}=\{z \in C|| z-\bar{z} \geq 8\}$
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)

$S_{1}:|z-3-2 i|^{2}=8$
$|z-3-2 i|=2 \sqrt{2}$
$(x-3)^{2}+(y-2)^{2}=(2 \sqrt{2})^{2}$
$S_{2}: x \geq 5$
$S_{3}:|z-\bar{z}| \geq 8$
$|2 i y| \geq 8$
$2|y| \geq 8 \therefore y \geq 4, y \leq-4$

$n S_{1} \cap S_{2} \cap S_{3}=1$