If |z−6|<|z−2|, what is its solution given given by?

Step 1
The inequality is the same as
$|z-6{|}^2<|z-2{|}^2$
hence as
$(z-6)(\overline{z}-6)<(z-2)(\overline{z}-2)$
Expanding and simplifying we get
$4(z+\overline{z})>32$
that is
$\frac{z+\overline{z}}{2}>4$
So the solutions are all complex numbers whose real part is greater than 4

Step 1
Given
$$\begin{gathered} |z-6|<|z-2|\Rightarrow \\ \sqrt{(x-6{)}^2+y^2}<\sqrt{(x-2{)}^2+y^2} \end{gathered}$$
or
$(x-6{)}^2+y^2<(x-2{)}^2+y\Rightarrow 8x>32$or
$x>4$

Step 1 $z>4$ detalis There are 3 cases which are: First when $z-6\ge 0$ then $z-2>0$ and $|z-6|<|z-2|$ becomes $z-6<z-2$ subtract z gives $-6<-2$ So $|z-6|<|z-2|$ is true when $z\ge 6$ Second For $6>z\ge 2$ $|z-6|<|z-2|$ becomes $6-z<z-2$ $8<2z$ $4<z$ Combining the first two cases $z>4$ and $z\ge 6$ $z>4$ And finally for case three $z<2$ $|z-6|<|z-2|$ becomes $6-z<2-z$ add z $6<2$ which is false