If |z2−1|=|z|2+1  then z lies on

|z2−1|=|z|2+1 ⇒|z2−1|2=(zz¯+1)2⇒(z2−1)(z¯2−1)=(zz¯+1)2 ⇒z2+2zz¯+z¯2=0⇒(z+z¯)2=0⇒z=−z¯ ⇒  z is purely imaginaryAlternatively ,let z=r(cosθ+isinθ) Then |z2−1|=|r2(cos2θ+isin2θ)−1|                         =r4−2r2cos2θ+1  and  |z2−1|2=(|z|2+1)2 ⇒r4−2r2cos2θ+1=r4+2r2+1 ⇒2cos2θ=0⇒cosθ=±π2 ∴ z lies on imaginary axis.