If (a + ib)(c + id)(e + if)(g + ih) = A + iB, | vedclass

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(A) Given that $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ ..... $(i)$Taking the conjugate on both sides,we get $(a - ib)(c - id)(e - if)(g - ih) = A

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${A^2} + {B^2}$

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(A) Given that $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ ..... $(i)$Taking the conjugate on both sides,we get $(a - ib)(c - id)(e - if)(g - ih) = A - iB$ ..... $(ii)$Multiplying equation $(i)$ and equation $(ii)$,we get:$[(a + ib)(a - ib)] 	imes [(c + id)(c - id)] 	imes [(e + if)(e - if)] 	imes [(g + ih)(g - ih)] = (A + iB)(A - iB)$Using the property $(x + iy)(x - iy) = x^2 + y^2$,we have:$({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) = {A^2} + {B^2}$

If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$,then $({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) =$

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