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If $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B$, then show that $\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)$ $=A^2+B^2$
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We have $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B \quad \ldots$ (i)
Replacing $i$ by $-i$ both side, we get
$(a-i b)(c-i d)(e-i f)(g-i h)=A-i B \quad \ldots$ (ii)
Multiplying eqns. (i) \& (ii), we get
$[(a+i b)(a-i b)][(c+i d)(c-i d)]$
$[(e+i f)(e-i f)][(g+i h)(g-i h)]$
$\begin{aligned} &=(A+i B)(A-i B) \\ \text { or } \quad &\left(a^2-i^2 b^2\right)\left(c^2-i^2 d^2\right)\left(e^2-i^2 f^2\right)\left(g^2-i^2 h^2\right) \\ &=\left(A^2-i^2 B^2\right) \\\left(a^2+\right.&\left.b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right) \\ &=\left(A^2+B^2\right)\left(\because i^2=-1\right) \text {. Hence proved } \end{aligned}$
Replacing $i$ by $-i$ both side, we get
$(a-i b)(c-i d)(e-i f)(g-i h)=A-i B \quad \ldots$ (ii)
Multiplying eqns. (i) \& (ii), we get
$[(a+i b)(a-i b)][(c+i d)(c-i d)]$
$[(e+i f)(e-i f)][(g+i h)(g-i h)]$
$\begin{aligned} &=(A+i B)(A-i B) \\ \text { or } \quad &\left(a^2-i^2 b^2\right)\left(c^2-i^2 d^2\right)\left(e^2-i^2 f^2\right)\left(g^2-i^2 h^2\right) \\ &=\left(A^2-i^2 B^2\right) \\\left(a^2+\right.&\left.b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right) \\ &=\left(A^2+B^2\right)\left(\because i^2=-1\right) \text {. Hence proved } \end{aligned}$
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