If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that


If $(a+i b)(c+i d)(e+i l)(g+i h)=A+i B$, then show that



$(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$

$\therefore|(a+i b)(c+i d)(e+i f)(g+i h)|=|\mathrm{A}+i \mathrm{~B}|$

$\Rightarrow|(a+i b)| \times|(c+i d)| \times|(e+i f)| \times|(g+i h)|=|\mathrm{A}+i \mathrm{~B}| \quad\left[\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|\right]$

$\Rightarrow \sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}}=\sqrt{A^{2}+B^{2}}$

On squaring both sides, we obtain


Hence, proved.


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