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

(N/A) Given: $(a+ib)(c+id)(e+if)(g+ih)=A+iB$
Taking the modulus on both sides:
$|(a+ib)(c+id)(e+if)(g+ih)| = |A+iB|$
Using the property $|z_{1}z_{2}z_{3}z_{4}| = |z_{1}||z_{2}||z_{3}||z_{4}|$,we get:
$|a+ib| \times |c+id| \times |e+if| \times |g+ih| = |A+iB|$
Since $|x+iy| = \sqrt{x^{2}+y^{2}}$,we have:
$\sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}} = \sqrt{A^{2}+B^{2}}$
Squaring both sides:
$(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2}) = A^{2}+B^{2}$
Hence proved.