The given expression is
x−iy= a−ib c−id (1)
On multiplying the numerator and denominator by c+id in R.H.S., we get
x−iy= a−ib c−id × c+id c+id = ( ac+bd )+i( ad−bc ) c 2 + d 2
Now, squaring equation (1), we get
( x−iy ) 2 = ( ac+bd )+i( ad−bc ) c 2 + d 2 x 2 − y 2 −2ixy= ( ac+bd )+i( ad−bc ) c 2 + d 2
By comparing the real and imaginary terms, we get
x 2 − y 2 = ac+bd c 2 + d 2 , −2xy= ad−bc c 2 + d 2 (2)
We know that, ( x 2 + y 2 ) 2 = ( x 2 − y 2 ) 2 +4 x 2 y 2 [ ∵ ( a+b ) 2 = ( a−b ) 2 +4ab ]
Solve for ( x 2 + y 2 ) 2 using equation (2)
( x 2 + y 2 ) 2 = ( ac+bd c 2 + d 2 ) 2 + ( ad−bc c 2 + d 2 ) 2 ( x 2 + y 2 ) 2 = a 2 c 2 + b 2 d 2 +2abcd+ a 2 d 2 + b 2 c 2 −2abcd ( c 2 + d 2 ) 2 ( x 2 + y 2 ) 2 = a 2 ( c 2 + d 2 )+ b 2 ( c 2 + d 2 ) ( c 2 + d 2 ) 2
Simplifying the above expression further,
( x 2 + y 2 ) 2 = ( c 2 + d 2 )( a 2 + b 2 ) ( c 2 + d 2 ) 2 ( x 2 + y 2 ) 2 = a 2 + b 2 c 2 + d 2
Hence, proved.