Consider the given that,
(a+ib)(c+id)=x+iy........(1)
Part (1)
prove that, (a−ib)(c−id)=x−iy
Proof:- By equation (1),
Take conjugate and we get,
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(a+ib)(c+id)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯x+iy
⇒¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(a+ib)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(c+id)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯x+iy
⇒a−ibc−id=x−iyHenceproved.......(2)
Part (2),
Multiplying (1) and (2) to and we get,
(a+ib)(c+id)×(a−ib)(c−id)=(x+iy)(x−iy)
a2−i2b2c2−i2d2=x2−i2y2
a2+b2c2+d2=x2+y2Henceproved.