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

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

Taking modulus both sides,

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

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

$\Rightarrow \sqrt{a^{2} + b^{2}} \times \sqrt{c^{2} + d^{2}} \times \sqrt{e^{2} + f^{2}} \times \sqrt{g^{2} + h^{2}} = \sqrt{A^{2} + B^{2}}$

On squaring both sides, we obtain

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

Hence, proved

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

(a + i b) (c + i d) = A + i B

(a^{2} + b^{2}) (c^{2} + d^{2}) = A^{2} + B^{2}

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a, b, c, d, p

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A^{- 1}

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