If $(a + i b) (c + i d) (e + i f) (g + i h) = A + i B$
then show that $(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2})$
$= A^{2} + B^{2}$

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Solution

If $(a + i b) (c + i d) (e + i f) (g + i h) = A + i B$

Taking modulus on both sides

$| (a + i b) (c + i d) (e + i f) (g + i h) | = | A + i B |$

$\Rightarrow | a + i b | | c + i d | | e + i f | | g + i h | = | A + i B |$

$\Rightarrow (\sqrt{a^{2} + b^{2}}) (\sqrt{c^{2} + d^{2}}) (\sqrt{e^{2} + f^{2}}) (\sqrt{g^{2} + h^{2}})$

$= \sqrt{A^{2} + B^{2}}$

Squaring both sides

$(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}$

Suggest Corrections

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