Answer :

It is given in the question that,

(a + ib) (c + id) (e + if) (g + ih) = A + iB


Using, [|z1z2|=|z1||z2|]


|(a + ib)| × | (c + id) | × |(e + if) | × |(g + ih)| = |A + iB|



Now, we will square on both sides,


(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2


Hence, proved


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