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Question
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
Solution
(a + ib) (c + id) (e + if)(g + ih) = A + iB ......(1)
By placing i in place of i,
(a – ib) (c – id) (e – if)(g – ih) = A – iB ......(2)
On multiplying equations (1) and (2),
[(a + ib) (a – ib)] [(c + id) (c – id)] [(e + if)(e – if)][(g + ih)(g – ih)] = (A + iB)(A – iB)
⇒ `(a^2 - i^2b^2)(c^2 - i^2d^2)(e^2 - i^2 f^2)(g^2 - i^2h^2) = A^2 - i^2B^2`
⇒ `(a^2 + b^2)(c^2 + d^2) (e^2 + f^2) (g^2 + h^2) = A^2 + B^2`
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