If (a + ib) (c +

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