If $\left|z-4\right|<\left|z-2\right|$ then its solution is given by

Given $\left|z-4\right|<\left|z-2\right|$

$\text{let}z=x+iy$

$\Rightarrow \left|\left(x-4\right)+iy\right|<\left|\left(x-2\right)+iy\right|$

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

 $\Rightarrow x^2-8x+16<x^2-4x+4$

$\Rightarrow 12<4x$

 $\begin{array}{c}\Rightarrow x>3\\ \Rightarrow \mathrm{Re}\left(z\right)>3\end{array}$