Question
If $(a + ib) (c + id) (e+ if) (g + ih) = A + iB$ then show that
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$
✓
Answer
Here $(a + ib) (c + id) (e+ if) (g + ih) = A + iB$
Taking modulus on both sides
$| (a + ib) (c + id) (e+ if) (g + ih) | = |A + iB|$
$ \Rightarrow |a + ib | |c + id | | e + if | |g + ih| = | A + iB|$
$ \Rightarrow \left( {\sqrt {{a^2} + {b^2}} } \right)\left( {\sqrt {{c^2} + {d^2}} } \right)\left( {\sqrt {{e^2} + {f^2}} } \right)$$\left( {{g^2} + {h^2}} \right) = \sqrt {{A^2} + {B^2}} $
Squaring both sides
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$
Taking modulus on both sides
$| (a + ib) (c + id) (e+ if) (g + ih) | = |A + iB|$
$ \Rightarrow |a + ib | |c + id | | e + if | |g + ih| = | A + iB|$
$ \Rightarrow \left( {\sqrt {{a^2} + {b^2}} } \right)\left( {\sqrt {{c^2} + {d^2}} } \right)\left( {\sqrt {{e^2} + {f^2}} } \right)$$\left( {{g^2} + {h^2}} \right) = \sqrt {{A^2} + {B^2}} $
Squaring both sides
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$
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