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

$(a)\;A^{2}+B^{2}\qquad(b)\;A^{2}-B^{2}\qquad(c)\;2A^{2}-B^{2}\qquad(d)\;A^{2}+2B^{2}$

Answer : $\;A^{2}+B^{2}$
Explanation :
$\;(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2}) = A^{2}+B^{2}$
$\;|(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2})| = |A^{2}+B^{2}|$
$\;|(a^{2}+b^{2})| \times |(c^{2}+d^{2})| \times |(e^{2}+f^{2})| \times |(g^{2}+h^{2})| = |A^{2}+B^{2}|$
$\;\sqrt{(a^{2}+b^{2})} \times \sqrt{(c^{2}+d^{2})} \times \sqrt{(e^{2}+f^{2})} \times \sqrt{(g^{2}+h^{2})} = \sqrt{A^{2}+B^{2}}$
On squaring both sides , we obtain
$\;(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2}) = A^{2}+B^{2}$