If $(a+i b)(c+i d)(e+i l)(g+i h)=A+i B$, then show that
$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$
$(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$
$\therefore|(a+i b)(c+i d)(e+i f)(g+i h)|=|\mathrm{A}+i \mathrm{~B}|$
$\Rightarrow|(a+i b)| \times|(c+i d)| \times|(e+i f)| \times|(g+i h)|=|\mathrm{A}+i \mathrm{~B}| \quad\left[\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|\right]$
$\Rightarrow \sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}}=\sqrt{A^{2}+B^{2}}$
On squaring both sides, we obtain
$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=\mathrm{A}^{2}+\mathrm{B}^{2}$
Hence, proved.