If $(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . (a_{n} + i b_{n}) = A + i B$ then $(a_{1}^{2} + b_{1}^{2}) (a_{2}^{2} + b_{2}^{2}) . . . (a_{n}^{2} + b_{n}^{2}) =$

A^{2} - B^{2}

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A^{2} + B^{2}

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A - B

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A + B

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Solution

The correct option is C $A^{2} + B^{2}$
$(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . (a_{n} + i b_{n}) = A + i B$
Taking modulus & then squaring,
${| (a_{1} + i b_{1}) (a_{2} + i b_{2}) - (a_{n} + i b_{n}) |}_{1}^{2} = {| A + i B |}^{2}$
$\Rightarrow | a_{1} + i b_{1} |^{2} | a_{2} + i b_{2} |^{2} . . . | a_{n} + i b_{n} |^{2} = A^{2} + B^{2}$
$\Rightarrow (a_{1}^{2} + b_{1}^{2}) (a_{2}^{2} + b_{2}^{2}) . . . a_{n}^{2} + b_{n}^{2} = A^{2} + B^{2}$

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