If $|z^2-1|=|z{|}^2+1,$ then $z$ lies on

Given $|z^2-1|[cit{e}_{s}tart]=|z{|}^2+1$.
Squaring both sides:
$|z^2-1{|}^2=(|z{|}^2+1{)}^2$.
Using the property $|w{|}^2=w\overline{w}$:
$(z^2-1)({\overline{z}}^2-1)=|z{|}^4+2|z{|}^2+1$.
$z^2{\overline{z}}^2-z^2-{\overline{z}}^2+1=|z{|}^4+2|z{|}^2+1$.
Since $z^2{\overline{z}}^2=|z{|}^4$:
$-(z^2+{\overline{z}}^2)=2|z{|}^2$.
$-(z^2+{\overline{z}}^2)=2z\overline{z}$.
$z^2+{\overline{z}}^2+2z\overline{z}=0$.
$(z+\overline{z}{)}^2=0\Rightarrow z+\overline{z}=0$.
This implies $2\text{Re}(z)=0\Rightarrow \text{Re}(z)=0$.
Thus, $z$ lies on the imaginary axis.