The number of points in the complex plane satisfying both $|z-2|=2$ and $z(1-i)+\bar{z}(1+i)=4$ (where $i=\sqrt{-1}$) is
Step-by-Step Solution
Key Concept: Convert $|z-2|=2$ to Cartesian: circle $(x-2)^2+y^2=4$. Convert the line equation: $z(1-i)+\bar{z}(1+i)=2\text{Re}(z\bar{(1+i)})=2(x+y)=4$, i.e., $x+y=2$.
Circle centre $(2,0)$, radius $2$. Line: $x+y=2$. Centre satisfies the line, so chord passes through centre → 2 intersection points.
Correct Answer: (C) 2