UPSC NDA Math Model Paper 1

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.