Paragraph: Let $A, B, C$ be three sets of complex numbers as defined below. $A=\{z ; \operatorname{lm} z \geq 1\} ; B=\{z:|z-2-i|=3\} ; C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\}$.Question: The number of elements in the set $A \cap B \cap C$ is

• A

  0
• Correct answer
  B

  1
• C

  2
• D

  $\infty$

Solution

Let $z=x+i y$ Set $A$ corresponds to the region $y \geq 1$. Set $B$ consists of points lying on the circle, centred at $(2,1)$ and radius 3 , i.e., $x^2+y^2-4 x-2 y=4$ $$ \text { Set } C \text { consists of points lying on the } x+y=\sqrt{2} $$
Clearly, there is only one point of intersection of the line $$ x+y=\sqrt{2} \text { and circle } x^2+y^2-4 x-2 y=4 . $$

Hence, the correct answer is option: 'B'.