The inequality $|Z-4|<|Z-2|$ represents the region given by:

1. None of these
2. $\text{Re}\left(z\right)<0$
3. $\text{Re}\left(z\right)\ge 0$
4. $\text{Re}\left(z\right)>0$

The correct option is D $\text{Re}\left(z\right)>0$
Given:
$|Z-4|<|Z-2|$

Let $Z=x+iy$
$\Rightarrow |(x-4)+iy|<|(x-2)+iy|$

$\Rightarrow (x-4{)}^2+y^2<(x-2{)}^2+y^2$

$\Rightarrow x^2+16-8x+y^2<x^2+4-4x+y^2$

$\Rightarrow -8x+16<4-4x$

$\Rightarrow 4x>12$

$\Rightarrow x>3$

$\text{Re}\left(z\right)>3$

Hence, Option (c) is correct.