The inequality $|z-4|<|z-2|$ represents the region given by
Step-by-Step Solution
Key Concept: $|z-a|<|z-b|$ is the open half-plane of points closer to $a$ than $b$, bounded by the perpendicular bisector of $ab$. Here the bisector is $x=3$, giving $\text{Re}(z)>3$.
**Step 1: Expand the inequality**
Let $z=x+iy$. $(x-4)^2+y^2<(x-2)^2+y^2 \Rightarrow x^2-8x+16<x^2-4x+4 \Rightarrow -4x<-12 \Rightarrow x>3$.
**Step 2: Compare with options**
The region is $\text{Re}(z)>3$. This is not listed among options (a)–(c). Answer: None of these.
Correct Answer: 4