Given: z2−1|=|z|2+1 Let us substitute z=x+ iy in the above equation we get, ⇒|(x2−y2−1)+i2xy|=|x+iy|2+1 As we know that, |z|=√x2+y2 ⇒√(x2−y2−1)2+(2xy)2=x2+y2+1 By squaring both the terms we get, ⇒(x2−y2−1)2+4x2y2=x4+y4+1+2x2y2+2y2+2x2 ⇒(x4+y4+1−2x2y2+2y2−2x2)+4x2y2=x4+y4+1+2x2y2+2y2+2x2 ⇒x4+y4+1+2x2y2+2y2−2x2=x4+y4+1+2x2y2+2y2+2x2 ⇒4x2=0 ⇒x=0 As we know that, x=0 is the equation of the y -axis. So, z lies on the imaginary axis. Hence, option 3 is the correct answer.