Question

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

Answer 1

$(a+ib)(c+id)(e+if)(g+ih)=A+iB$

$∴|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|$

$⇒|(a+ib)|×|(c+id)|×|(e+if)|×|(g+ih)|=|A+iB|$ $[|z_1z_2|=|z_1||z_2|]$

$⇒ \sqrt a^2+b^2×\sqrt c^2+d^2×\sqrt e^2+f^2×\sqrt g^2+h^2=\sqrt A^2+B^2$

$\text{On squaring both sides, we obtain}$

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

$\text{Hence, proved.}$