Question 19:

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Answer

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

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

⇒ |(a + ib) ×|(c + id)| × |(e + if | × |(g + ih)| = |A + iB|   [|z₁z₂| = |z₁||z₂|]

On squaring both sides, we obtain

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

Hence, proved.

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