If x - iy = √(a - ib/c - id), prove that (x2 + y2)2 = a2 + b2/c2 + d2.
x - iy = a - ib/c - id
= a - ib/c - id × c + id/c + id
= (ac + bd) + i(ad - bc)/c2 + d2
∴ (x - iy) = (ac + bd) + i(ad - bc)/c2 + d2
⇒ x2 - y2 - 2ixy = (ac + bd) + i(ad - bc)/c2 + d2
On comparing real and imaginary parts, we obtain
x2 - y2 = ac + bd/c2 + d2, -2xy = ad - bc/c2 + d2 ... (1)
(x2 - y2)2 = (x2 + y2)2 + 4x2y2
= (ac + bd/c2 + d2)2 + (ad - bc/c2 + d2)2 [Using (1)]
= a2c2 + b2d2 + 2acbd + a2d2 + b2c2 - 2adbc/(c2 + d2)2
= a2c2 + b2d2 + a2d2 + b2c2/(c2 + d2)2
= a2(c2 + d2) + b2(c2 + d2)/(c2 + d2)2
= (c2 + d2)(a2 + b2)/(c2 + d2)2
= a2 + b2/c2 + d2
Hence, proved.