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Q)

If $A+iB=(a_{1}+ib_{1})(a_{2}+ib_{2})(a_{3}+ib_{3}) $ then $A^{2}+B^{2}$is

\[\]$(1)a^{2}_{1}+b^{2}_{1}+a^{2}_{2}+b^{2}_{2}+a^{2}_{3}+b^{2}_{3}$\[\](2)$(a_{1}+a_{2}+a_{3})^{2}+(b_{1}+b_{2}+b_{3})^{2}$\[\]$(3)(a^{2}_{1}+b^{2}_{1})(a^{2}_{2}+b^{2}_{2})(a^{2}_{3}+b^{2}_{3})$\[\](4)$(a^{2}_{1}+a^{2}_{2}+a^{2}_{3})(b^{2}_{1}+b^{2}_{2}+b^{2}_{3})$

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A)
$A+iB=(a_1+ib_1)(a_2 + ib_2)(a_3+ib_3)$
$|A+iB=(a_1+ib_1)(a_2 + ib_2)(a_3+ib_3)|$
$|A+iB|=|(a_1+ib_1)| |(a_2 + ib_2)| |(a_3+ib_3)|$
$\sqrt {A^2+B^2}=\sqrt{a_1+b_1^2} \sqrt {a_2 +b_2^2} \sqrt{a_3+b_3^2}$
Squaring on both sides
$A^2+B^2=(a^{2}_{1}+b^{2}_{1})(a^{2}_{2}+b^{2}_{2})(a^{2}_{3}+b^{2}_{3})$
Hence 3 is the correct answer.
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