Question

The region of the $z-$plane for which

$\left|\frac{z-a}{z+\overline{a}}\right|=1(Rea\ne 0)$ is $y-$axis.

Answer 1

$\left|\frac{z-a}{z+\overline{a}}\right|=1\Rightarrow {|z-a|}^2={|z+\overline{a}|}^2$
If $z=(x,y),a=(p,q)$ then $\overline{a}=(p,-q)$
$\therefore$ $(x-p{)}^2+(y-q{)}^2=(x+p{)}^2+(y-q{)}^2$
$\therefore$ $(x-p{)}^2-(x-p{)}^2=0$ or $4px=0$
$\therefore$ $x=0$ i.e. $y-$axis as $p\ne 0$This is the equation of $y-$axis.