If a complex number coincides with its conjugate,

Let $z=x+i y$ and $\bar{z}=\overline{x+i y}$

$\bar{z}=x-i y$

Since $z=\bar{z}$ (given)

$\Rightarrow x+i y=x-i y$

$\Rightarrow i y=-i y$

$\Rightarrow 2 i y=0$

i.e $y=0$

Then $z$ lies an $x$-axis.